Class 11 Engineering Maths - Complex Numbers And Quadratic Equations

Number of roots of the quadratic equation $\displaystyle 8\sec ^{2}x-6\sec x+1= 0$ is/are:

On solving,

$\displaystyle \sec x= \frac{1}{2}\:or\:\frac{1}{4}$

$\displaystyle \Rightarrow \cos x= 2$ or 4
Both the values are not possible.
Hence there exist no solution of given equation.

$\left | z \right |\leq 5$ & $Arg\left ( z-1-i \right )> \pi /6$

1073129_332840_ans_73c243b9eafb4060b7476326326449b4.png

$Arg \left ( z-1-i \right )\geq \pi /3$

1073125_332839_ans_bf1e5a2764e24ac593d4ef26cbb301b1.png

Multiply $2\sqrt { -3 } +3\sqrt { -2 }$ by $4\sqrt { -3 } -5\sqrt { -2 }$

$(2\sqrt{-3}+3\sqrt{-2})\times(4\sqrt{-3}-5\sqrt{-2})$

$=(2\sqrt{3}\text i+3\sqrt{2}\text i)\times(4\sqrt{3}\text i-5\sqrt{2}\text i)$

$=-(2\sqrt{3}+3\sqrt{2})\times(4\sqrt{3}-5\sqrt{2})$

$=-(24-10\sqrt6+12\sqrt6-30)$

$=6-2\sqrt6$

If the value of the discriminant of the quadratic equation $a{ x }^{ 2 }+bx+c=0$ is less than $0$ , then the nature of the roots is ____

When

$a,b$ and

$c$ are real numbers ,

$a\neq0$ and discriminant is negative

$($ i.e.

$b^2-4ac<0$

$)$ , then the roots

$\alpha$ and

$\beta$ of the quadratic equation

$ax^2+bx+c=0$ are unequal and imaginary. Here the roots

$\alpha$ and

$\beta$ are pair of the complex conjugates.

If ${Z}_{1}=-1$ and ${Z}_{2}=i$ , then find $Arg\left( \cfrac { { Z }_{ 1 } }{ { Z }_{ 2 } } \right)$

$Arg\left( \cfrac { { Z }_{ 1 } }{ { Z }_{ 2 } } \right) =Arg\left(\dfrac{-1}{i} \right)$

$=Arg\left( \dfrac{i^2}{i}\right)=Arg(i)=\dfrac{\pi}{2}$

Since complex number

$i$ lie on the positive imaginary axis

Write an example for a quadratic polynomial that has no real zeroes.

The quadratic polynomial

$p(x)=x^2+1$ has not real zero. Even it has imaginary zeros

$-i,i$

Without solving the following quadratic equation, find the value of $m$ for which the given equation has real and equal roots.
$x^2+2(m-1)x+(m+5)=0$

The given equation is

$x^2 + 2(m-1)x + (m+5) = 0$

For the equation to have real and equal roots, the discriminant should be

$0$ .

$\therefore [2(m-1)]^2 -4(m+5) = 0$

$\Rightarrow 4(m^2 -2m+1) = 4(m+5)$

$\Rightarrow m^2 - 3m -4 = 0$

$\Rightarrow (m-4)(m+1) = 0$

$\therefore m = 4$ or

$m = -1$

Calculate $\displaystyle \sqrt{i}$

$z=i=re^{i\theta}$ (let)

$r=\sqrt{1+0}=1$

$\theta=\tan^{-1}(\cfrac{1}{0})=\pi/2$

$z=e^{i\pi/2}$

$z^2=e^{i\pi/4}$

$\implies (\cos \pi/4+i\sin \pi/4)$

$\implies \cfrac{1}{\sqrt 2}+\cfrac{i}{\sqrt 2}$

What is $cis\ 0$ ?

C is

$\theta$ =

$cos\theta +i sin\theta$

C is O

$= CosO+i sin0$

$= 1+i(0)$

$1$

Find the value of $k$ so that the following equation has equal roots.
$2x^{2} - (k - 2) x+1 = 0$

The given equation has equal roots

$\Rightarrow$ Discriminant

$={b}^{2}-4ac=0$

where

$a=2, b-\left(k-2\right),c=-1$

$\Rightarrow {\left(k-2\right)}^{2}-4\times 2\times 1=0$

$\Rightarrow {\left(k-2\right)}^{2}=8$

$\Rightarrow k-2=\pm\sqrt{8}=\pm2\sqrt{2}$

$\therefore k=2\pm2\sqrt{2}$

Calculate $\displaystyle \sqrt[3]{- \, i}$

Let

$z=-i=e^{-i\pi/2}$

$z^{1/3}=(-i)^{1/3}=e^{-i\pi/6}$

$\implies (\cos -\pi/6+i\sin -\pi/6)$

$\implies \cfrac{\sqrt 3}{2 }-\cfrac{i}{ 2}$

Represent $(-1-\sqrt 3i)$ in the polar form.

Let

$z=-1-\sqrt{3}i$

Since

$-1-\sqrt{3}i$ lies in third quadrant.

$\therefore$ Principal value of

$\arg{z}=-\pi+{\tan}^{-1}{\left|\dfrac{-\sqrt{3}}{-1}\right|}$

$-\pi+{\tan}^{-1}{\sqrt{3}}=-\pi+\dfrac{\pi}{3}=\dfrac{-3\pi+\pi}{3}=\dfrac{-2\pi}{3}$

and

$\left|z\right|=\sqrt{{\left(-1\right)}^{2}+{\left(-\sqrt{3}\right)}^{2}}=\sqrt{1+3}=2$

$\therefore$ Polar form of

$z=\left|z\right|\left[\cos{\left(\arg{z}\right)}+i\sin{\left(\arg{z}\right)}\right]$

Polar form of

$z=2\left[\cos{\left(\dfrac{-2\pi}{3}\right)}+i\sin{\left(\dfrac{-2\pi}{3}\right)}\right]$

For the real numbers x and y, if $(x-iy)(3+5i)$ is the conjugate of $-6-24i$ then find the value of $x$ and $y$ .

$(x-iy)(3+5i)=(3x+5y)+i(5x-3y)$

It is given that

$(x-iy)(3+5i)$ is the conjugate of

$-6-24i$

$(x-iy)(3+5i)=\overline {-6-24i}$

$(3x+5y)+i(5x-3y)=-6+24i$

$3x+5y=-6$ and

$5x-3y=24$

Solving these equations, we get,

$x=3$ and

$y=-3$ .

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle z^2 \, + \, | \, \overline{z} \, | \, = \, 0$

Let

$z=x+iy$

$|z|=\sqrt{x^2+y^2}$

$|z|+z^2=0$

$(\sqrt{x^2+y^2})$

$+x^2-y^2+2ixy=0$

$\implies (\sqrt{x^2+y^2})$

$+x^2-y^2=0$ (Real part) and

$2xy=0$ (imaginary part)

$\implies (x^2-y^2)^{2}=x^2+y^2$

$\implies x^4+y^4-2x^2y^2=x^2+y^2$

$\implies x^4-x^2=y^2-y^4$ (since

$xy=0$ )

$\implies x^2(-1+x^2)=y^2(1-y^2)$

Find the values of p for which the quadratic equation $4x^2+px+3=0$ has equal roots.

given quadratic equation

$4{ x }^{ 2 }+px+3=0$

condition for equal roots is

$\Delta =0$

$\Rightarrow{ p }^{ 2 }-\left( 4 \right) \left( 4 \right) \left( 3 \right) =0$

$\Rightarrow p=\pm 4\sqrt { 3 }$

Find the modulus of the complex number $2-5!$ .

Given complex number is

$2-5i$ ,

The modulus of this complex number is

$\sqrt { { 2 }^{ 2 }+{ 5 }^{ 2 } } \\ =\sqrt { 29 }$

Determine the nature of the roots of the following quadratic equation
$2x^2+ 6x + 3 = 0$

Consider the given equation.

$2{{x}^{2}}+6x+3=0$ …….. (1)

Therefore,

$x=\dfrac{-6\pm \sqrt{36-4\times 2\times 3}}{2\times 2}$

$x=\dfrac{-6\pm \sqrt{36-24}}{4}$

$x=\dfrac{-6\pm \sqrt{12}}{4}$

$x=\dfrac{-3\pm \sqrt{3}}{2}$

So, the roots are real and distinct.

Find the length of the segments connecting the points represented by the following pairs of numbers :
3, -4i

First we note that the length of segment joining the points

$z_1$ and

$z_2$ is |

$z_1 \, - \, z_2$ |. Therefore

$|3 - (- 4i)| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

locate the point representing the complex numbers $z$ on the Argand diagram for which
$\left| z \right| \ge 3,$

Here the point

$z$ lie outside and on the circle whose center
is origin and radius

$3$ .

Simplify the following :
$\dfrac{3 \, - \, i}{2 \, + \, i} \, + \, \dfrac{3 \, + \, i}{2 \, - \, i}$

$\cfrac{3-i}{2+i}$

$+\cfrac{3+i}{2-i}$

$\implies\cfrac{(3-i)(2-i)+(2+i)(3+i)}{2^2+1}$

$\implies\cfrac{6-5i-1+6+5i-1}{5}$

$\implies\cfrac{10}{5}=2$

Find the length of the segments connecting the points represented by the following pairs of numbers :
-1 - i, 2 + 3i

First we note that the length of segment joining the points

$z_1$ and

$z_2$ is |

$z_1 \, - \, z_2$ |. Therefore

$|(- 1 - i) - (2 + 3i)| = |- 3 - 4i|$

$= \, \sqrt{(-3)^2 + (-4)^2} = \sqrt{25} = 5$

$x \, +\, i \sqrt x^\, 4 \, +, x^2 \, +\, 1$

$=x \, +\, i \sqrt (x^\, 4 \, +, x^2 \, +\, 1)(x^\, 4 \, +, x^2 \, +\, 1)$

$= \frac{1}{2} \, [2x \, +\, 2i \, \sqrt (x^\, 4 \, +, x^2 \, +\, 1)(x^\, 4 \, -, x^2 \, +\, 1)]$

$= \frac{1}{2} \, \sqrt (x^\, 4 \, +, x^2 \, +\, 1) \sqrt (x^\, 4 \, -, x^2 \, +\, 1) \, \, 2i \,$

$\sqrt (x^\, 4 \, +, x^2 \, +\, 1)(x^\, 4 \, -, x^2 \, +\, 1)]$

$= \frac{1}{2} \, [\sqrt (x^\, 4 \, +, x^2 \, +\, 1) \, \,+ \, i \sqrt (x^\, 4 \, -, x^2 \, +\, 1)]$

$Hence[ x \, +\, i \sqrt (x^\, 4 \, +, x^2 \, +\, 1)(x^\, 4 \, +, x^2 \, +\, 1)] ^1/2$

$= \pm \frac{1}{\sqrt2}[\sqrt (x^\, 4 \, +, x^2 \, +\, 1) \, \,+ \, i \sqrt (x^\, 4 \, -, x^2 \, +\, 1)]$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
${x^2} = 9$

Given equation:

$x^2=9$

$\implies x^2-9=0$

$a=1, b=0,c=-9$

$D=b^2-4ac$

$\implies D=0-4(1)(-9)$

$\implies D=36$

Thus

$D>0$

$\therefore$ There are two distinct real zeros.

What is value of $(-i)^{12}$

${\textbf{Step -1: Analyzing powers of iota}.}$

${\text{We know that, i = }}\sqrt {{\text{ - 1}}}$

$\therefore {\text{ }}{{\text{i}}^{\text{2}}}{\text{ = - 1,}}$

$\Rightarrow {{\text{i}}^{\text{3}}}{\text{ = (}}{{\text{i}}^{\text{2}}}{\text{)i = - i, }}$

$\Rightarrow {{\text{i}}^{\text{4}}}{\text{ = }}{\left( {{{\text{i}}^{\text{2}}}} \right)^{\text{2}}}{\text{ = 1}}$

${\text{After this, the powers repeat itself, for example, }}{{\text{i}}^{\text{5}}}{\text{ = (}}{{\text{i}}^{\text{4}}}{\text{)i = }}{{\text{i}}^{\text{1}}}$

$\Rightarrow {\text{ }}{{\text{i}}^{\text{n}}}{\text{ = }}{{\text{i}}^{{\text{n% 4}}}}{\text{, where % is modulo }}$

${\textbf{Step -2: Calculating ( - i}}{{\textbf{)}}^{{\textbf{12}}}.}$

${{\text{( - i}}{\text{)}}^{{\text{12}}}}{\text{ = ( - 1}}{{\text{)}}^{{\text{12}}}}{{\text{(i)}}^{{\text{12}}}}$

$\Rightarrow {\text{ ( - i}}{{\text{)}}^{{\text{12}}}}{\text{ = }}{{\text{i}}^{{\text{12}}}}$

$\Rightarrow {\text{ ( - i}}{{\text{)}}^{{\text{12}}}}{\text{ = }}{\left( {{{\text{i}}^{\text{4}}}} \right)^{\text{3}}}$

$\Rightarrow {\text{ ( - i}}{{\text{)}}^{{\text{12}}}}{\text{ = }}{{\text{1}}^{\text{3}}}{\text{ = 1}}$

$\mathbf{{\text{Final Answer: The value of ( - i}}{{\text{)}}^{{\text{12}}}}{\text{ is 1}}.}$

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
${ x }^{ 2 }+5x+5=0$

The given equation is

${ x }^{ 2 }+5x+5=0$

Here,

$a=1,b=5,c=5$

$\therefore D={ b }^{ 2 }-4ac={ (25) }^{ }-4\times 1\times 5=5>0$

So, the given equation has real roots given by

$\alpha =\cfrac { -b+\sqrt { D } }{ 2a } =\cfrac { -1+\sqrt { 49 } }{ 2\times 6 } =\cfrac { 1 }{ 2 }$

$\beta =\cfrac { -b-\sqrt { D } }{ 2a } =\cfrac { -1-\sqrt { 49 } }{ 2\times 6 } =\cfrac { -2 }{ 3 }$

Determine the nature of the roots of the given quadratic equation
$2{ x }^{ 2 }+x-1=0$

The given quadratic equation is

$2{ x }^{ 2 }+x-1=0$

Here,

$a=2,b=1,c=-1$

$\therefore D={ b }^{ 2 }-4ac={ 1 }^{ 2 }-4\times 2\times -1=9$

Since

$D \neq0$ , therefore roots of the given equation are real and distinct.

Determine the nature of the roots of the given quadratic equation
$4{ x }^{ 2 }-4x+1=0\quad$

The given equation is

$4{ x }^{ 2 }-4x+1=0\quad$

Here,

$a=4,b=-4,c=1$

$\therefore D={ b }^{ 2 }-4ac={ (-4) }^{ 2 }-4\times 4\times 1=0$

since

$D=0$ , the roots of the given equation are real and equal.

Solve
$z = \left( {2 + i} \right)+\left( {1 - 3i} \right)$

$z = (2+i) + (1-3i)$

In complex number, addition of real number and imaginary numbers

$=(2+1) +i(1-3)$

$= 3-2i$

Find the modulus of: (i) ${{21} \over 5} - {{12} \over 5}i$ (ii) 3+4i

$\cfrac{21}{5} - \cfrac{12}{5}i$

Modulus =

$\sqrt{\cfrac{21^2}{5^2} + \cfrac{12^2}{5^2}}$

$=\cfrac{\sqrt {585}}{5}$

$3+4i$

Modulus =

$\sqrt{3^2 + 4^2} = 5$

Find the value of $(1+i)(1+{i}^{2})(1+{i}^{3})(1+{i}^{4})$ .

$(1+i)(1+i^{2})=1+i+i^{2}+i^{2}.i=1+i-1-i=0$

$(1+i^{3})(1+i^{4})=1+i^{2}.i+i^{2}.i^{2}+i^{2}.i^{2}.i^{2}.i$

$=1-i+1-i=2-2i$

$=0\times (2-2i)$

$=0$

Find the amplitude and modulus of following
(i) $\displaystyle {{21} \over 5} - {{12} \over 5}i$ (ii) $3 + 4{\rm{ i}}$

$(i)z=\cfrac { 21 }{ 5 } -\cfrac { 12i }{ 5 } \\ \left| z \right| =\sqrt { \cfrac { 441 }{ 25 } +\cfrac { 144 }{ 25 } } =\sqrt { \cfrac { 585 }{ 25 } } =\sqrt { \cfrac { 117 }{ 5 } }$

Amp(z)=Arg(z)

$=-\tan ^{ -1 }{ \left| \cfrac { Im(z) }{ Re(z) } \right| }$ (

$\therefore z$ is in

$4th$ quadrant)

$=-\tan ^{ -1 }{ (\cfrac { 12 }{ 21 } ) } \\ =-\tan ^{ -1 }{ (\cfrac { 4 }{ 7 } ) }$

$(ii)z=3+4i\\ \left| z \right| =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =\sqrt { 9+16 } =5;$

Amp(z)=Arg(z)

$=\tan ^{ -1 }{ \left| \cfrac { Im(z) }{ Re(z) } \right| }$

$=\tan ^{ -1 }{ (\cfrac { 4 }{ 3 } ) }$

Express the given complex number $3(7+i7)+i(7+i7)$ in the form $a+ib$ ,

$3(7+i7)+i(7+i7)=21+i21+i7+i\times i7$

$i\times i=-1$

$3(7+i7)+i(7+i7)=21+i21+i7-7$

$3(7+i7)+i(7+i7)=14+i28=14+28i$

Find the amplitude of $-4$ .

Since the point represented by the complex number

$-4$ lies on the negative half of

$x-$ axis.

So the amplitude of

$-4$ is

$\pi$ .

Find modulus and argument of $i$

Let

$z = i = 0 + i$

$\left| z \right| = \sqrt {0 + 1} = 1$

$\tan \theta = \left| {\dfrac{y}{x}} \right| = \dfrac{1}{0} = \infty$

$\Rightarrow \theta = \dfrac{\pi }{2}$

Therefore Argument of

$z$ is

$\dfrac{\pi }{2}$

Write the nature of roots of the quadratic equation $5x^2-2x+3=0$ .

$5x^2-2x+3=0$

$a=5,b=-2,c=3$

$D=b^2-4ac$

$D=-56 < 0$

So, the root are imaginary , i.e. it has two non real roots.

Write the argument of $(1+\sqrt {3})(1+i)(\cos \theta+i\sin \theta)$ .

we know that properties of,

$\begin{array}{l} ({ z_{ 1 } }.{ z_{ 2 } })=({ z_{ 1 } })+({ z_{ 2 } }) \\ \Rightarrow \, ({ z_{ 1 } }.{ z_{ 2 } }.{ z_{ 3 } })=({ z_{ 1 } })+({ z_{ 2 } })+arg({ z_{ 3 } }) \\ \Rightarrow \, (1+i\sqrt { 3 } )(1+i)(cos\theta +isin\theta )=\, (1+i\sqrt { 3 } )+arg(1+i)+arg(cos\theta +isin\theta ) \\ \\ Now, \\ argument\, find\, individually, \\ { z_{ 1 } }=1+i\sqrt { 3 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \tan \alpha =\left| { \frac { { Im({ z_{ 1 } }) } }{ { { { Re } }({ z_{ 1 } }) } } } \right| =\left| { \frac { { \sqrt { 3 } } }{ 1 } } \right| =\sqrt { 3 } ,\, \, \, \, \, \therefore \, \, \alpha =\frac { \pi }{ 6 } \, \, \, \, \, \, \, } \right. \\ \therefore \, \, \, \, \, ({ z_{ 1 } })=\theta =\alpha =\frac { \pi }{ 6 } \, \, \\ { z_{ 2 } }=1+i\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \tan \alpha =\frac { 1 }{ 1 } =1\, \, \, \Rightarrow \alpha =\frac { \pi }{ 4 } } \right. \\ \, \, \therefore \, \, \, ({ z_{ 2 } })=\theta =\alpha =\frac { \pi }{ 4 } \\ \, { z_{ 3 } }=1(\cos \theta +i\sin \theta ) \\ \, \therefore \quad ({ z_{ 3 } })=\theta \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \quad \quad \quad \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \therefore \, \, \, \, r(\cos } \right. \theta +i\sin \theta )=(z)=\theta ,\left| z \right| =\theta \\ \\ Substitute\, in\, equation\, of, \\ \, (1+i\sqrt { 3 } )(1+i)(cos\theta +isin\theta )=\dfrac { \pi }{ 6 } +\dfrac { \pi }{ 4 } +\theta \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (z)\, \, \, =\, \, \dfrac { { 5\pi } }{ { 12 } } +\theta \, \, \, \, \, \, \, \\ therefor\, \, \, \, (z)\, \, =\, \, \dfrac { { 5\pi } }{ { 12 } } +\theta \, \\ So,that\, the\, \, last\, complex\, is\, \, \, \dfrac { { 5\pi } }{ { 12 } } +\theta \end{array}$

The values of $x$ and $y$ satisfying the equation $\dfrac{(1+i)x - 2i}{3+ i} + \dfrac{(2 -3i)y + i}{3 - i} = i$ , are

Consider the given expressions.

$\dfrac{\left( 1+i \right)x-2i}{3+i}+\dfrac{\left( 2-3i \right)y+i}{3-i}=i$

$\dfrac{x+\left( x-2 \right)i}{3+i}+\dfrac{2y+\left( 1-3y \right)i}{3-i}=i$

$\dfrac{\left( x+\left( x-2 \right)i \right)\left( 3-i \right)+\left( 2y+\left( 1-3y \right)i \right)\left( 3+i \right)}{9-{{i}^{2}}}=i$

$x\left( 3-i \right)+i\left( x-2 \right)\left( 3-i \right)+2y\left( 3+i \right)+i\left( 1-3y \right)\left( 3+i \right)=\left( 9+1 \right)i$

$3x-ix+i\left( 3x-ix-6+2i \right)+6y+2iy+i\left( 3+i-9y-3yi \right)=10i$

$3x-ix+3xi-{{i}^{2}}x-6i+2{{i}^{2}}+6y+2iy+3i+{{i}^{2}}-9yi-3y{{i}^{2}}=10i$

$3x-ix+3xi+x-6i-2+6y+2iy+3i-1-9yi+3y=10i$

$4x+9y-3+2xi-7yi-13i=0$

$4x+9y-3+\left( 2x-7y-13 \right)i=0$

On comparing real part and imaginary part, we get

$4x+9y-3=0$ …… (1)

$2x-7y-13=0$ …… (2)

On solving both equations, we get

$x=3$

$y=-1$

Hence, the value of $x,y$ is $3, -1$ .

Solve for $x$ : $4\sqrt 3 x^{2} + 5x-2\sqrt 3 = 0$ Write about the nature of roots.

$4\sqrt 3 {x^2} + 5x - 2\sqrt 3 = 0$

$D = \,{b^2} - 4ac$

$= {\left( 5 \right)^2} - 4 \times 4\sqrt 3 \times - 2\sqrt 3$

$= 25 + 96$

$= 121 > 0$

So , roots of given equation are real and distinct

Express $(1-i)-(1+i6)$ as $a+ib$

$(1-i)-(1+i6)=1-i-1-6i=-7i$

Hence

$a=0$ and

$b=-7$

Thefore the number is purely imaginary.

If $\dfrac {a-ib}{a+ib}=\dfrac {1+i}{1-i}$ , then prove that $a+b=0$

We have,

$\dfrac{a-ib}{a+ib}=\dfrac{1+i}{1-i}$

$\left( a-ib \right)\left( 1-i \right)=\left( a+ib \right)\left( 1+i \right)$

$a-ai-ib+{{i}^{2}}b=a+ai+ib+{{i}^{2}}b$

$a-ai-ib-b=a+ai+ib-b$

$ai+ib=ai+ib$

$2ai+2bi=0$

$a+b=0$

Hence, proved.

Solve the equation $\left| z \right| + z = 2 + i$

Given expression:

$|z|+z=2+i$

Suppose that

$z=x+yi$ then

$|z|=\sqrt{(x^2+y^2)}$

On substituting the values in the given expression, we get

$\sqrt{(x^2+y^2)}+x+iy=2+iy$

On comparing the real and imaginary parts, we get

$y=1,$

$\sqrt{(x^2+y^2)}+x=2,$

$\sqrt{(x^2+1)}+x=2,$

$\sqrt{(x^2+1)}=2-x$

Taking square on both sides, we get

$x^2+1=4+x^2-2x,$

$2x=3,$

$x=\dfrac{3}{2}.$

The value of real part is

$\dfrac{3}{2}$ and imaginary part is

$1$ .

Convert the given complex number into the polar form: $z = -3$

$z=-3+0i$

putting

$-3=r\cos \theta\;,\;\;0=r\sin \theta$

$9+0=r^2(\sin^2\theta+\cos^2\theta)$

$r^2=9$

$r=3$

$\frac{r\sin\theta}{r\cos\theta}=\frac{0}{-3}$

$\tan \theta=0$

$\theta=0$

So,

$|z|=3$ , argument of

$z=0$

Therefore, polar form of

$z$ is

$z=r(\cos\theta+i\sin\theta)$

$=3(\cos 0^{\circ}+i\sin 0^{\circ})$

If ${z_1} = 1 + i = \sqrt 3 + i$ , then the principle arg $\left( {\frac{{{z_1}}}{{{z_2}}}} \right)$

$z_1 = 1+i$ ,

$z_2 = \sqrt3+i$

$\cfrac{z_1}{z_2} = \cfrac{1+i}{\sqrt3+i}$

$\cfrac{z_1}{z_2} = \cfrac{(1+i)(\sqrt3-i)}{(\sqrt3+i)(\sqrt3-i)}$

$\cfrac{z_1}{z_2} = \cfrac{\sqrt3 - i + \sqrt3 i +1}{3 - i^2}$

$\cfrac{z_1}{z_2} = \cfrac{\sqrt3+1}{4}+ i\cfrac{\sqrt3-1}{4}$

Hence,

$arg(\cfrac{z_1}{z_2}) =\cfrac{\pi}{12}$

If $a+ib=\cfrac { { (x+i) }^{ 2 } }{ 2{ x }^{ 2 }+1 }$ , prove that ${ a }^{ 2 }+{ b }^{ 2 }=\cfrac { { (x+i) }^{ 2 } }{ { (2{ x }^{ 2 }+1) }^{ 2 } }$ .

1374439_1081585_ans_73d1a99e787844eda1e9086ba829d714.jpg

Solve the problem:-
$\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right)$

Given,

$\left(\dfrac{1}{5}+i\dfrac{2}{5}\right)-\left(4+i\dfrac{5}{2}\right)$

$=\left ( \dfrac{1}{5}-4 \right )+i\left ( \dfrac{2}{5}-\dfrac{5}{2} \right )$

$=-\dfrac{19}{5}-i\dfrac{21}{10}$

For what value of $k$ , the equation $kx^2 - 6x - 2 = 0$ has equal roots.

The condition for a quadratic equation

$ax^{2}+bx+c=0$ to have equal roots, is

$b^{2}-4ac=0$

For the given equation

$kx^{2}-6x-2=0$

$b^{2}-4ac=(-6)^{2}-4k(-2)=0$

$36+8k=0$

$8k=-36$

$k=-\dfrac{36}{8}=-\dfrac{9}{2}$

Evaluate: $\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3$ .

Given,

$\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3$

$=\left(i^{18}+\dfrac{1}{i^{25}}\right)^3$

$=\left(-1+\dfrac{1}{i}\right)^3$

$=\left(-1\right)^3+3\left(-1\right)^2\dfrac{1}{i}+3\left(-1\right)\left(\dfrac{1}{i}\right)^2+\left(\dfrac{1}{i}\right)^3$

$=\dfrac{2}{i}+2$

$=-2i+2$

If $z=x+iy$ and $z_1=1+2i$ , determine the region in the complex plane represented by $1 < |z-z_1 |< 3$ . Represent it on the Argand plane.

1326003_1064951_ans_2c565c6bda67463796669a1360cc3881.jpg

Determine the nature of roots of the following quadratic equation : $2x^2+5x+5=0$

$2x^2+5x+5=0$

This is of the form

$ax^2+bx+c=0$ , where

$a=2,b=5,c=5$

Therefore,

$D=b^2-4ac=25-40=-15<0$

Therefore, the equation has no real roots.

If $z_1=2-i, z_2=1+i$ , find $\left|\dfrac{z_1+z_2+1}{z_1-z_2+i}\right|$ .

Given,

$z_1=2-i,z_2=1+i$

$\left | \dfrac{z_1+z_2+1}{z_1-z_2+i} \right |$

$=\left | \dfrac{2-i+1+i+1}{2-i-1-i+i} \right |$

$=\left | \dfrac{4}{1-i} \right |$

$=\dfrac{4\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}$

$=\dfrac{4\left(1+i\right)}{2}$

$=\left|2\left(1+i\right)\right|$

$=\sqrt{2^2+2^2}$

$=\sqrt{8}$

$=2\sqrt{2}$

Find the values of the following:
${ \left( 1+i\sqrt { 3 } \right) }^{ 3 }$

$(1+i\sqrt{3})^{3}$

$1^{3}+(i\sqrt{3})^{3}+3\sqrt{3}i(1+\sqrt{3}i)$

$=1+(-3\sqrt{3}i)+3\sqrt{3}i+(-9)$

$=-8$

Solve $(3-7i)+(-2+4i)$ .

$1-3i$

If $\alpha$ and $\beta$ are the zeroes of the equation $6{x^2} + x - 2 = 0$ , find $\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }$

Consider the given equation.

6x2+x−2=0 …….. (1)

Since, α,β are roots of given equation.

So,

α+β=−ba=−16

αβ=ca=−26=−13

Since,

=αβ+βα

=α2+β2αβ

=(α+β)2−2αβαβ

=(−16)2−2×−13−13 =\dfrac{{{\left( -\dfrac{1}{6} \right)}^{2}}-2\times -\dfrac{1}{3}}{-\dfrac{1}{3}}

=136+23−13

=1+2436−13

=−2512

Hence, the value is −2512.

$\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}$ .

$\cfrac{(2+5i)(3+2i) + (2-5i)(3-2i)}{(3-2i)(3+2i)}$

$\cfrac{-4+19i-4-19i}{13}$

$-\cfrac{8}{13}$

Write principal argument of $\dfrac {-\sqrt {11}i}{17}$ .

Since Real part = 0

Argument

$90^{\circ}(\pi/2)$

1240481_1135958_ans_a479b65a757b41999b15d0d34aa4c7f5.jpg

Find the modulus:
$-4-4i$

$z = - 4 - 4i$

$\left| z \right| = \sqrt {{{( - 4)}^2} + {{( - 4)}^2}}$

$= \sqrt {16 + 16}$

$= \sqrt {32}$

$= 4\sqrt 2$

If $a=\frac { -1+\sqrt { 3i } }{ 2 } ,b=\frac { -1-\sqrt { 3i } }{ 2 }$ then show that ${a}^{2}=b$ and ${b}^{2}=a$ .

$a=\dfrac{-1+\sqrt{3}i}{2}$

$a^2=\left(\dfrac{-1+\sqrt{3}i}{2}\right)^2$

$=\dfrac{\left(-1+\sqrt{3}i\right)^2}{2^2}$

$=\dfrac{-2-2\sqrt{3}i}{2^2}$

$=\dfrac{-1-1\sqrt{3}i}{2}$

$=b$

$b=\dfrac{-1-1\sqrt{3}i}{2}$

$b^2=\left ( \dfrac{-1-1\sqrt{3}i}{2} \right )^2$

$=\dfrac{(-1-1\sqrt{3}i)^2}{2^2}$

$=\dfrac{-2+2\sqrt{3}i}{2^2}$

$=\dfrac{-1+1\sqrt{3}i}{2}$

$=a$

Represent following complex numbers $z_1=1+2i$ and $z_2=5-7i$ by points in Argand's diagram and determine their amplitudes approximately.

R.E.F image

Argand plane

$amp(z_{1}) = tan^{-1}2 = 1.107$

$amp(z_{2}) = tan^{-1}(\dfrac{-7}{5})$

$= -0.95$

1135093_1136235_ans_bb011d5fdfda4ea78aa269609c686e2c.png

If $g\left( x \right) ={ x }^{ 4 }{ -x }^{ 3 }+{ x }^{ 2 }+3x-5$ , find $g(2+3i)$

$g(x)=x^4-x^3+x^2+3x-5$

$=x^2(x^2+1)-x(x^2-3)-5$

$x=2+3i$

$x^2=4-3+12i=1+12i$

$=(1+12i)(2+12i)-(1+12i)(-2+12i)-5$

$=2+12i+24i-144+2+24i-12i+144-5$

$=-1+48i$ .

1194723_1158223_ans_69d6b45325d44c019abd781b063cb1c6.jpg

What will be the nature of roots of the quadratic equation $2{x^2} + 4x + 7 = 0$ .

1302507_1141562_ans_99dcc1cd16f6426ba0940e960f3056d8.JPG

Find the modulus of $\dfrac {3 -4i}{5 + 7i}$ .

Given,

$\dfrac{3-4i}{5+7i}$

$=\dfrac{3-4i}{5+7i}\times \dfrac{5-7i}{5-7i}$

$=\dfrac{-23-41i}{74}$

$=-\dfrac{23}{74}-i\dfrac{41}{74}$

$modulus=\sqrt{\left ( -\dfrac{23}{74} \right )^2+\left ( -\dfrac{41}{74} \right )^2}$

$=\dfrac{\sqrt{2210}}{74}$

Write the nature of roots of quadratic equation $4{ x }^{ 2 }+6x+3=0$

According to question,

$4x^2+6x+3=0$

Discriminant,

$D=\sqrt{6^2-4(4)(3)}$

$\implies D=\sqrt{36-48}$

$\implies D=\sqrt{-16}$

So, Roots are imaginary

Find the amplitude of $1 + i \sqrt{3}$ .

$z=1+\sqrt{3}i$

Amplitude,

$\left | z \right |=\left | 1+\sqrt{3}i \right |$

$=\sqrt{1^{2}+(\sqrt{3})^{2}}$

$=\sqrt{1+3}$

$=\sqrt{4}=2$

Find the modulus of the complex number $z=2+3i$ .

Given the complex number is

$z=2+3i$ .

Then,

$|z|=\sqrt{2^2+3^2}$

$=\sqrt{4+9}$

$=\sqrt{13}$ .

Find discriminant of the equation $5x - 6 = \dfrac{1}{x}$ .

Given,

$5x-6=\dfrac{1}{x}$

$\implies5x^2-6x-1=0$

Comparing with the standard quadratic equation

$ax^2+bx+c=0$

$a=5, b=-6,c=-1$

( Discriminant)

$D=b^2-4ac=(-6)^2-4(5)(-1)=36+20=56$ .

If $a+ib=\dfrac{c+i}{c-i}$ , where $c$ is a real number, then prove that : $a^{2}+b^{2}=1$ and $\dfrac{b}{a}=\dfrac{2c}{c^{2}-1}$ .

Given,

$a+ib = \dfrac{c+i}{c-i}$

$= \dfrac{c+i}{c-i}\times\dfrac{c+i}{c+i}$ (On rationalisation)

$= \dfrac{c^{2}-1+2ci}{c^{2}+1}$

$= (\dfrac{c^{2}-1}{c^{2}+1}) + (\dfrac{2c}{c^{2}+1})i$

On comparison,

$a = \dfrac{c^{2}-1}{c^{2}+1}$ and

$b = \dfrac{2c}{c^{2}+1}$

Therefore,

$a^{2}+b^{2} = \dfrac{(c^{2}-1)^{2} + (2c)^{2}}{(c^{2}+1)^{2}}$

$= \dfrac{c^{4}-2c^{2}+1+4c^{2}}{(c^{2}+1)^{2}}$

$= \dfrac{(c^{2}+1)^{2}}{(c^{2}+1)^{2}}$ =

$1$

Now,

$\dfrac{b}{a} = \dfrac{\dfrac{2c}{c^{2}+1}}{\dfrac{c^{2}-1}{c^{2}+1}}$

$\dfrac{b}{a} = \dfrac{2c}{c^{2}-1}$

Write the nature of roots from the given values of the discriminants and complete the activity.

Value of discriminant	Nature of roots
$25$	Real and distinct roots
$-23$	No real roots
$0$	Real and equal roots
$10$	Real and distinct roots

Find the modulus and amplitude of $4+3i$ .

$z=4+3i$

Comapring with

$z=x+iy$

$x=4$ and

$y=3$

$\left|z\right|=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{4}^{2}+{3}^{2}}=\sqrt{25}=5$

Amplitude

$=\tan^{-1}{\left(\dfrac{y}{x}\right)}=\tan^{-1}{\left(\dfrac{3}{4}\right)}$

Write (i) $-1 -i$ (ii) $1 - i$ in polar form.

A complex number

$a+ib$ in polar form is written as

$r\cos \theta+i r\sin \theta$

So,

$\cos \theta=\dfrac{a}{r}$

$\sin\theta=\dfrac{b}{r}$

Hence,

$r=\sqrt{a^2+b^2}$

$Part\ (i)$

$-1-i$

$a=-1, b=-1$

So,

$r=\sqrt{(-1)^2+(-1)^2}$

$r=\sqrt{2}$

And hence

$\cos \theta=\dfrac{-1}{\sqrt 2}\Rightarrow \theta=\dfrac{3\pi}{4}$

Therefore, the polar form of

$-1-i$ is

$=\left(\sqrt 2, \dfrac{3\pi}{4}\right)$

$Part\ (ii)$

$1-i$

$a=1, b=-1$

So,

$r=\sqrt{(1)^2+(-1)^2}$

$r=\sqrt{2}$

And hence

$\cos \theta=\dfrac{1}{\sqrt 2}\Rightarrow \theta=\dfrac{\pi}{4}$

Therefore, the polar form of

$1-i$ is

$=\left(\sqrt 2, \dfrac{\pi}{4}\right)$

Hence, this is the answer.

Simplify and hence find the modulus and argument $\dfrac {(3 + i)(3 - i)}{2 + i}$ .

$\dfrac{(3+i)(3-i)}{2+i}$

$=\dfrac{(3)^2-(i)^2}{2+i}=\dfrac{9+1}{2+i}$

$=\dfrac{10}{2+i}\times \dfrac{2-i}{2-i}$

$=\dfrac{10(2-i)}{5}$

$=2(2-i)$

$z=4-2i$

$|z|=\sqrt{4^2+22}=2\sqrt{5}$

$arg=+n^{-1}\left(-\dfrac{1}{2}\right)$ .

1194592_1264241_ans_4e2ae08285c34990adb9ac89a93574bd.jpg

In $-3+3i$ , find amplitude and modulus

Given,

$z=-3+3i$

Comparing with

$z=x+iy$

$x=-3$ and

$y=3$

$\left | z \right |=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+3^{2}}= 3\sqrt{2}$

$Arg (z)= tan^{-1}\dfrac{3}{-3}= 135^{\circ}$

If the discriminant is 13, how many solutions and of what type?

Given discriminant is

$13$

Since the discriminant is positive and not equal to zero

The solutions will be real and distinct

The number of solutions are

$2$

Express $\dfrac{-1+i}{\sqrt{2}}$ in the polar form

$\frac{{ - 1 + i}}{{\sqrt 2 }} = r\left( {\cos \theta + i\sin \theta } \right)$

$r\cos \theta = d\frac{{ - 1}}{{\sqrt 2 }}$

$r\sin \theta = \dfrac{1}{{\sqrt 2 }}$

$\tan \theta = - 1$

$\theta = \dfrac{{3\pi }}{4}$

${r^2} = \dfrac{1}{2} + \dfrac{1}{2}$

${r^2} = 1$

$r = 1$

$So,\,polar\,form\,\,of\,\dfrac{{ - 1 + i}}{{\sqrt 2 }}\,is\,\left( { 1,\,\dfrac{{3\pi }}{4}} \right).$

If $z=(\cos \theta, \sin \theta))$ , find $\left(z-\dfrac{1}{z}\right)$

1171111_1313002_ans_2ac53b99070e440a98564c3b4bd49c7d.jpg

Find the nature of roots of the quadratic equation ${ y }^{ 2 }-7y+2=0$ .

Given: a quadratic equation.

$y^{2}-7y+2=0$

which is of the form

$ax^{2}+bx+c=0$

$\therefore a=1, b=-7, c=2.$

now, discriminant,

$D=b^{2}-4ac$

$=(-7)^{2}-4(1)(2)$

= 49-8

= 41 > 0

Since D > 0, the two roots are real and distinct.

Simplify: $\left(\dfrac{1}{3}+3i\right)^{3}$

${\left( \cfrac{1}{3} + 3i \right)}^{3}$

$= {\left( \cfrac{1}{3} \right)}^{3} + {\left( 3i \right)}^{3} + 3 \left( \cfrac{1}{3} \right) \left( 3i \right) \left( \cfrac{1}{3} + 3i \right)$

$= \cfrac{1}{27} + 27 {i}^{3} + 3i \left( \cfrac{1 + 9i}{3} \right)$

$= \cfrac{1}{27} - 27i + i + 9 {i}^{2}$

$= \cfrac{1}{27} - 26i - 9$

$= - \left( \cfrac{242}{27} + 26i \right)$

If $\omega =\frac { Z }{ \bar { Z } }$ , then $|\omega |=$

Here,

$\omega =\frac{z}{z}$

Let z = x+iy

$\bar{z}=x-iy$

$\left | \omega \right |=\left | \frac{z}{z} \right |\frac{\left | z \right |}{\left | z \right |}=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+(-y)^{2}}}=\frac{\sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}}=1$

1170896_1296533_ans_27befc15f7ab4709bd8be9af83bbbec6.JPG

If $z_1=5+3i\\z_2=2-3i$ Find $z_1+z_2$

$z_1=5+3i\\z_2=2-3i\\z_1+z_2=5+3i+2-3i=7$

$3 \sqrt { 2 } x ^ { 2 } - 2 \sqrt { 6 } x + 2$ find the nature of roots.

$a=3\sqrt 2\\b=-2\sqrt6\\c=2\\b^2-4ac\\(2\sqrt 6)^2-4(3\sqrt 2)(2)\\24-24\sqrt 2<0$

The roots are imaginary and complex

Check whether the following equation will have two distinct real or imaginary or two real equal roots.
$x^2-3x+5=0$ .

Given the quadratic equation is

$x^2-3x+5=0$ .

Now discriminant of this equation will be

$b^2-4ac=(-3)^2-4\times 5\times 1$

$=9-20$

$=-11$ .

As the discriminant

$<0$ then the quadratic equation will have imaginary roots.

Prove that the roots of the following quadratic equation has two real distinct roots.
$2x^2-6x+3=0$ .

Given the quadratic equation is

$2x^2-6x+3=0$

Now discriminant of this equation is

$(-6)^2-4\times 2\times 3$

$=36-24$

$=12$ .

Since the discriminant of this equation is

$>0$ .

Then the roots of the quadratic equation are real and distinct.

Find the condition for which the roots of the equation $ax^{2}+bx+c=0$ be real and distinct.

If the roots of the equation

$ax^{2}+bx+c=0$ be real and distinct then we must have the discriminant of the equation should be

$>0$ .

Then we must have,

$b^2-4ac>0$ .

This is the required condition.

Find the value of $\displaystyle \sum^{100}_{n=0}i^{n!}$ $(where, i=\sqrt {-1})$

$\displaystyle \sum_{n = 0}^{100} i^{n!} = i^{0!} +i ^{1!}+ i^{2!} + i^{3!} + i ^{4!} + ...$

subsequent terms would have 4 as factor and

$i^{4k}=1$

$=i^1 + i^1 + i^2 + i^6 + \underbrace{1 + 1 + ....}_{96 \, times}$

$= 2i -1-1+96$

$= 2i + 94$

1226971_1398531_ans_104078b97b6948b59b5571c10faaa85d.jpg

Check whether the roots of the following quadratic equations are real or not?
$3x^2-4\sqrt{3}x+4=0$ .

Given the quadratic equation is

$3x^2-4\sqrt{3}x+4=0$ .

Now discriminant of this equation

$(D)=b^2-4ac$

$D=(-4\sqrt{3})^2-4\times 3\times 4$

$=48-48$

$D=0$ .

Since the discriminant is

$0$ so the roots of the given quadratic equation are all real and equal.

Evaluate : $i^{373}$

$i^{373}$

$=i^{372} i$

$=(i^4)^{93} i$

$=1 \times i$ as

$i^4=1$

$=i$

Simplify: $\left(14+ 2i\right)\left(7 + 12i\right)$ where $i=\sqrt{-1}$

$(14+2i)(7+12i)=14\times 7+14\times 12i+2i\times 7+2i\times 12i$

$=98+168 i+14i-24$

$=74+182i$

1244937_1530157_ans_6bf4cc41de964cf9b8bd98d171f1cf32.PNG

Find the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

Let,

$z = \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

$= \dfrac{\left ( 1+i \right )^2-\left ( 1-i \right )^{2}}{\left ( 1-i \right )\left ( 1+i \right )}$

$= \dfrac{4i}{1-i^{2}}$

$= \dfrac{4i}{1+1} = 2i$

$\therefore \left | z \right | = 2$

Express the complex number $1 + i \sqrt{3}$ in modulus amplitude form.

$1+i{\sqrt{3}}=2\bigg(\dfrac{1}{2}+i\dfrac{\sqrt{3}}{2}\bigg)\\=2\bigg(\cos \dfrac{\pi}{3}+i\sin \dfrac{\pi}{3}\bigg)\\=2{e^{i\dfrac{\pi}{3}}}$

Simplify the following
$(2-5i)+(-3+4i)+(8-3i)$

$\left( 2 - 5i \right) + \left( -3 + 4i \right) + \left( 8 - 3i \right)$

$= 2 - 5i - 3 + 4i + 8 - 3i$

$= \left( 2 - 3 + 8 \right) + \left( -5 + 4 - 3 \right) i$

$= 7 - 4i$

Find z if , $\left| z \right| = 4$ and arg (z ) = ${{5\pi } \over 6}$

1322128_1553732_ans_43124d18795c417eba805731b3461966.jpeg

$x^{2}-2x-1=0$
Find the nature of roots

$x^2-2x-1=0\\a=1\\b=-2\\c=-1\\$

$\Delta=b^2-4ac\\(-2)^2-4(1)(-1)\\4+4=8>0$

So roots are real and distinct

Evaluate $\dfrac{1}{i^{78}}$

$\dfrac{1}{i^{78}}=\dfrac{1}{i^{78}}\times \dfrac{i^{2}}{i^{2}}=-1$

$\left[\because i^{80}=(i^{4})^{20}=1\right]$

Determine the nature of root of the quadratic equation.
$m^{2} + 2m + 1 = 0$

1815266_1851068_ans_324a3a0867a64bb8a86a0ae6eceb85e7.PNG

Find the modulus of the following ;
$-2-3i$

Let

$z = -2 - 3 i$

$\Rightarrow z = (-2) + (-3) i$
Modulus of

$z = \sqrt {(-2)^2 + (3)^2 }$

$= \sqrt {4 + 9 }= \sqrt {13}$
hence

$|z| = \sqrt {13}$

If $-5$ is a root of the quadratic equation $2{x}^{2}+px-15=0$ and the quadratic equation $p({x}^{2}+x)+k=0$ has equal roots, then find the value of $k$ .

Since

$-5$ is a root of the equation

$2{x}^{2}+px-15=0$

$\therefore$

$2{(-5)}^{2}+p(-5)-15=0$

$\Rightarrow$

$50-5p-15=0$ or

$5p=35$ or

$p=7$
Again

$p({x}^{2}+x)+k=0$ or

$7{x}^{2}+7x+k=0$ has equal roots

$D=0$
ie.,

${ b }^{ 2 }-4ac=0$ or

$49-4\times 7k=0$

$\Rightarrow$

$k=\cfrac{49}{28}=\cfrac{7}{4}$

For what value of $k$ , is $3$ a root of the equation $2{x}^{2}+x+k=0$ ?

$3$ is a root of

$2{x}^{2}+x+k=0$ , when

$2{(3)}^{2}+3+k=0$

$\Rightarrow 18+3+k=0\Rightarrow k=-21$

Evaluate : $i^{-50}$

$i^{-50}$

$=\dfrac{1}{i^{50}}$

$=\dfrac{1}{(i^4)^{12}i^2}$

$=\dfrac{1}{1\times {-1}}$ as

$i^2=-1,i^4=1$

$=-1$

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
$3{x}^{2}-4\sqrt { 3x } +4=0$

We have

$3{x}^{2}-4\sqrt { 3x } +4=0$
Here,

$a=3,b=-4\sqrt { 3x }$ and

$c=4$
Therefore,

$D={ b }^{ 2 }-4ac={(-4\sqrt { 3x } )}^{2}-4\times 3\times 4=48-48=0$
Hence, the given quadratic equation has real and equal roots
Thus,

$x=\cfrac { -b\pm \sqrt { D } }{ 2a } =\cfrac { -(-4\sqrt { 3 } \pm \sqrt { 0 } ) }{ 2\times 3 } =\cfrac { 2\sqrt { 3 } }{ 3 }$
Hence equal roots of given equation are

$\cfrac { 2\sqrt { 3 } }{ 3 } ,\cfrac { 2\sqrt { 3 } }{ 3 }$

Find the values of $k$ for which the quadratic equation $9{x}^{2}-3kx+k=0$ has equal roots.

For equal roots:

$D=0\Rightarrow {b}^{2}-4ac=0$

${(-3k)}^{2}-4x\times 9\times k=0$

$\Rightarrow 9{k}^{2}=36k\Rightarrow k=4$

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them:
$2{x}^{2}-6x+3=0$

We have

$2{x}^{2}-6x+3=0$
Here,

$a=2,b=-6,c=3$
Therefore,

$D={ b }^{ 2 }-4ac={(-6)}^{2}-4\times 2\times 3=36-24=12> 0$
Hence, given quadratic equation has real and distinct roots
Thus,

$x=\cfrac { -b\pm \sqrt { D } }{ 2a } =\cfrac { -(-6)\pm \sqrt { 12 } ) }{ 2\times 2 } =\cfrac { 3\pm \sqrt { 3 } }{ 2 }$
Hence roots of the given equation are

$\cfrac { 3+ \sqrt { 3 } }{ 2 }$ and

$\cfrac { 3- \sqrt { 3 } }{ 2 }$

Find the modulus of the following ;
$(4+i)$

let

$x = 4 + i$

$\Rightarrow z = 4 + 1. i$
Modulus of

$z = \sqrt { 4^2 + 1^2 }$

$= \sqrt { 16 + 1} = \sqrt {17}$
hence

$|z| = \sqrt {17}$

If $a$ and $b$ are real, then show that the principal value of arg $a$ is $0$ or $\pi$ according to $a$ is positive or negative and that of arg $b$ is $\pi$ /2 or $-\pi$ /2 according to $b$ is positive or negative.

Let

$a = r cos \alpha$ and

$0 = r sin \alpha$ ..... (i)
so that

$a^2 + 0^2 = r^2$

$\therefore r = |a|$
then

$a = |a| cos \alpha$ {from (i)}

$\therefore cos \theta = \pm 1$
Then

$cos \alpha = 1$ or

$= -1$ according as

$a$ is +ve or -ve and

$sin \alpha = 0$ . Hence

$\alpha = 0$ or

$\pi$ according as a is +ve and -ve.
Again let

$0 = r_1 cos \beta$ and

$b = r_1 sin \beta$ ....... (ii)
so that

$0^2 + b^2 = r_1^2$

$\therefore r_1 = |b|$
from (ii)

$b = |b| sin \beta$

$\therefore sin \beta = \pm 1$
Then

$sin \beta = 1$ or

$-1$ according as

$b$ is +ve or -ve and

$cos \beta = 0$ . Hence,

$\displaystyle \frac{\pi}{2}$ or

$- \displaystyle \frac{\pi}{2}$ according as

$b$ is +ve or -ve.

Ans: 1

If $z = 4 + i\sqrt{7}$ , then find the value of $|z^3 - 4z^2 - 9z + 91|$ .

$z = 4 + i\sqrt{7}$
or $z - 4 = i\sqrt{7}$

On squaring, we get $z^2 - 8z + 16 = -7$
$\Rightarrow z^2 - 8z + 23 = 0$
Now $|z^3 - 4z^2 - 9z + 91| = |(z^2 - 8z + 23)(z + 4) - 1| = 1$

Ans: 1

The number of complex numbers $z$ satisfies $Re(z^2) = 0, \left|z\right| = \sqrt{3}.$

$z = x + iy$
or

$z^2 = x^2 - y^2 + 2ixy$

$\Longrightarrow Re(z^2) = x^2 - y^2$
Also,

$\left|z\right| = \sqrt{x^2 + y^2}$

$\Longrightarrow x^2 -y^2 = 0, x^2 + y^2 = 3$

$\Longrightarrow x^2 = y^2 =\dfrac{3}{2}$

$\Longrightarrow x = \pm \sqrt{\dfrac{{3}}{2}}, y=\pm \sqrt{\dfrac {{3}}{2}}$

$\Longrightarrow z = \pm \sqrt{\dfrac{{3}}{2}} \pm \sqrt{\dfrac{{3}}{2}}i$
Thus, there are four complex numbers.

Ans: 4

If Arg $(z + i)\, -$ Arg $(z - i)$ $= \dfrac{\pi}{2}$ , then $z$ lies on a circle.
If statement is True, enter $1$ , else enter $0$

We have
Arg

$(z + i)\, -$ Arg

$(z - 1) =\displaystyle\frac{\pi}{2}$

$\Rightarrow$ Arg

$\displaystyle \left ( \frac{z + i}{z - i} \right ) = \frac{\pi}{2}$

$\therefore Re \displaystyle \left ( \frac{z + i}{z - i} \right ) = 0$

$\Rightarrow \displaystyle \frac{\left ( \frac{z + i}{z - i} \right) + \overline{\left ( \frac{z + i}{z - i} \right ) }}{2} = 0$

$\Rightarrow \displaystyle \left ( \frac{z + i}{z - i} \right ) + \left ( \frac{\overline{z} - i}{\overline{z} + i} \right ) = 0$

$\Rightarrow (z + i) (\overline{z} + i) + (z - i) (\overline{z} - i) = 0$

$\Rightarrow z \overline{z} + i ( \overline{z}+z)-1 + z \overline{z} - i (z + \overline{z}) - 1 = 0$

$\Rightarrow 2 (z \overline{z}) = 2$

$\Rightarrow z \overline{z} = 1$

$|z|^2 = 1$

$\Rightarrow |z| = 1$
The equation represents a circle centered at origin and radius

$1$ units.

Prove that he expression $\displaystyle \frac{\sqrt{(1 + m)} +i \sqrt{(1 - m)}}{\sqrt{(1 + m)} - i\sqrt{(1 - m)}} - \frac{\sqrt{(1 - m)} + i \sqrt{(1 + m)}}{\sqrt{(1 - m)} -i \sqrt{(1 + m)}} (m \in R)$ simplifies to 2m.

rationalise numerator and denominator in both the terms of expression.

on rationalisation we get.

$\frac { { (\sqrt { (1+m) } +i\sqrt { (1-m) } ) }^{ 2 } }{ 2 } -\frac { { (\sqrt { (1-m) } +i\sqrt { (1+m) } ) }^{ 2 } }{ 2 }$

$=\frac { (1+m-1+m+2*i\sqrt { (1+m)(1-m) } )-(1-m-1-m+2*i\sqrt { (1+m)(1-m) } ) }{ 2 }$

=2m

Write the correct letter from column II against the entry number in column I in your answer book, z $\neq$ 0 is a complex number

Let

$z=a+ib$

$Re\left( z \right) =0\Rightarrow a=0$
And

$z=ib\Rightarrow { z }^{ 2 }=-{ b }^{ 2 }\Rightarrow Im\left( { z }^{ 2 } \right) =0$

$arg\left( z \right) =\cfrac { \pi }{ 4 }$

$\Rightarrow z=r\left( cos\cfrac { \pi }{ 4 } +isin\cfrac { \pi }{ 4 } \right) \\ \Rightarrow { z }^{ 2 }={ r }^{ 2 }\left( { cos }^{ 2 }\cfrac { \pi }{ 4 } -{ sin }^{ 2 }\cfrac { \pi }{ 4 } \right) +2i{ r }^{ 2 }cos\cfrac { \pi }{ 4 } sin\cfrac { \pi }{ 4 } \\ =i{ r }^{ 2 }sin\cfrac { \pi }{ 2 } =i{ r }^{ 2 }\\ Re\left( { z }^{ 2 } \right) =0$

If $x = a + bi$ is a complex number such that $x^2 = 3 + 4i$ and $x^3 = 2 + 11i$ , where $i = \sqrt{-1}$ , then $(a+ b)$ equal to

given that $x^{ 2 }=3+4i$ ...(1)

and $x^{ 3 }=2+11i$ ...(2)

Dividing (2) by (1), we get

$\displaystyle x=\frac { 2+11i }{ 3+4i } =\frac { \left( 3-4i \right) \left( 2+11i \right) }{ { 3 }^{ 2 }+{ 4 }^{ 2 } } =\frac { 50+25i }{ 25 } =a+bi$

Therefore, $a+b=3$

Ans: 3

Let z and w be two nonzero complex numbers such that $\left|z\right| = \left|w\right|$ and $arg(z) + arg(w) = \pi$ .
Then prove that $z = -\overline{w}$ .

Let

$arg(w) = \theta$ . Then

$arg(z) = \pi - \theta.$

Therefore,

$w = \left|w\right| (cos + i sin \theta )$
and

$z = \left|z\right| (cos (\pi - \theta) + i sin (\pi - \theta)) = \left|w\right|(-cos \theta + i sin \theta)$ [

$\therefore \left|z\right| = \left|w\right|$ ]

$= -\left|w\right| (cos \theta - i sin \theta) = -\overline{w}$ .

Ans: 1

Solve $\displaystyle \sin 2x+ \cos 4x=2$

Given,

$\sin 2x+\cos 4x=2\\ \Rightarrow \sin { 2x } +1-2\sin ^{ 2 }{ 2x } =2\\ \Rightarrow 2\sin ^{ 2 }{ 2x } -\sin { 2x } +1=0$

$\displaystyle \Rightarrow \sin { 2x } =\frac { 1\pm \sqrt { 1-4 } }{ 4 }$
Imaginary roots hence no solution

Solve the equation ${ 5 }^{ 2x }-24.{ 5 }^{ x }-25=0$
find the number of roots .

Let

${ 5 }^{ x }=t,$ then the given equation can be reduce in the form.

${ t }^{ 2 }-24t-254=0\\ \Rightarrow \left( t-25 \right) \left( t+1 \right) =0\\ \Rightarrow t\neq -1\quad \quad \therefore t=25=5^x\therefore x=2$
Hence

$x=2$ is only root of the given equation.

State true or false:
$\displaystyle \sqrt{-2}.\sqrt{-3}= \sqrt{\left ( -2 \right )\left ( -3 \right )}= \sqrt{6}$ .
For false type 0 and for true type 1

The computation

$\displaystyle "\sqrt{-2}\sqrt{-3}= \sqrt{\left ( -2 \right )\left ( -3 \right )}= \sqrt{6}"$ is wrong since here both -2 and -3 are negative.
The correct computation is

$\displaystyle \sqrt{-2}\sqrt{-3}= \sqrt{2}\sqrt{-1}\sqrt{3}\sqrt{-1}= \sqrt{2}\sqrt{3}\left ( \sqrt{-1} \right )^{2}$

$\displaystyle = \sqrt{6}\left ( -1 \right )= -\sqrt{6}.$

Let $k$ be the value of $\lambda$ for which the given equation will have equal roots : $\displaystyle x^{2}+2\left ( 3\lambda +5 \right )x+2\left ( 9\lambda ^{2}+25 \right )=0$ .Find $6k$ ?

Discriminant of given quadratic is,

$D=4(3\lambda+5)^2-8(9\lambda^2+25)=-4\left ( 3\lambda -5 \right )^{2}.$
If

$\displaystyle \lambda \neq \frac{5}{3}$ their D is always-ive and
hence roots are complex.
But if

$\displaystyle \lambda=\frac{5}{3}$
then

$D=0$ and in that case the root will be equal.

What is the square of the modulus of the complex number 2+3 $i$

Modulus of z=x+

$iy$ is

$\sqrt{x^2+y^2}$

so modulus of

$2+3i$ is

$\sqrt13$

The equation $\displaystyle ax^{2}+bx+c=0$ where a, b, c are real numbers connected by the relation 4a+2b+c= 0 and ab > 0 has real roots.If you think this is true write 1 otherwise write 0 ?

$\displaystyle \Delta b^{2}16a^{2}+4a\left ( 4a+2b \right )$ by given relation

$\displaystyle =b^{2}+16a^{2}+8ab=+ive$
Hence both roots are real. Alt.x=2 is a real root as

$\displaystyle 4a+2b+c=0$
and since imaginary roots occur in pairs therefore the other root must also be real.

Find the value of the sum $\displaystyle \sum_{n=1}^5(i^n+i^{n+2})$ , where $i=\sqrt {-1}$

$\displaystyle \sum_{n=1}^5(i^n+i^{n+2})=\sum_{n=1}^5(i^n+i^{n}.i^2)=\sum_{n=1}^5(i^n+i^{n}.(-1))=\sum_{n=1}^5(0)=0$

Find the modulus and the argument of the complex number $\displaystyle z=-1-i\sqrt { 3 }$

$\displaystyle z=-1-i\sqrt { 3 }$
Let

$\displaystyle r\cos { \theta } =-1\quad and\quad r\sin { \theta } =-\sqrt { 3 }$
On squaring and adding, we obtain

$\displaystyle { \left( r\cos { \theta } \right) }^{ 2 }+{ \left( r\sin { \theta } \right) }^{ 2 }={ \left( -1 \right) }^{ 2 }+{ \left( -\sqrt { 3 } \right) }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }\left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) =1+3$

$\displaystyle \Rightarrow { r }^{ 2 }=4\quad \left[ { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta =1 \right]$

$\displaystyle \Rightarrow r=\sqrt { 4 } =2$ (Conventionally,r>0 )

$\displaystyle \therefore$ Modulus of

$z$ i.e

$|z| =2$

$\displaystyle \therefore \quad 2\cos { \theta } =-1\quad and\quad 2\sin { \theta } = -\sqrt { 3 }$
Since both the values of

$\displaystyle \sin { \theta }$ and

$\displaystyle \cos { \theta }$ are negative and

$\displaystyle \sin { \theta }$ and

$\displaystyle \cos { \theta }$ are negative in III quadrant,
Argument

$\displaystyle =-\left( \pi -\frac { \pi }{ 3 } \right) =\frac { -2\pi }{ 3 }$
Thus, the modules and argument of the complex number

$\displaystyle -1-\sqrt { 3i }$ are 2 and

$\displaystyle \frac { -2\pi }{ 3 }$ respectively.

Convert the given complex number in polar form : $\displaystyle -1+i$

Given,

$\displaystyle z=-(1-i)$
Let

$\displaystyle r\cos { \theta =1 }$ and

$r\sin { \theta } =-1$
On squaring and adding, we obtain

$\displaystyle { r }^{ 2 }{ \cos }^{ 2 }\theta +{ r }^{ 2 }{ \sin }^{ 2 }\theta ={ 1 }^{ 2 }+{ \left( -1 \right) }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }\left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) =1+1$

$\displaystyle \Rightarrow \quad { r }^{ 2 }=2$

$\displaystyle \Rightarrow \quad r=\sqrt { 2 }$ (Conventionally r>0 )

$\displaystyle \therefore \sqrt { 2 } \cos { \theta } =1$ and

$\sqrt { 2 } \sin { \theta } =-1$

$\displaystyle \Rightarrow \cos { \theta } =\frac { 1 }{ \sqrt { 2 } }$ and

$\sin { \theta } =-\dfrac { 1 }{ \sqrt { 2 } }$

$\displaystyle \therefore \theta =-\frac { \pi }{ 4 }$ [As

$\displaystyle \theta$ lies in the Ii quadrant]

So, the polar form of

$z=-(1-i)$ is

$\displaystyle \therefore -(1-i)=-(r\cos { \theta } +ir\sin { \theta }) =-\left(\sqrt { 2 } \cos { \left( -\frac { \pi }{ 4 } \right) +i\sqrt { 2 } \sin { \left( -\frac { \pi }{ 4 } \right) } }\right)$

$=-\sqrt { 2 } \left[- \cos { \left( \dfrac {3 \pi }{ 4 } \right) -i\sin { \left( \dfrac { 3\pi }{ 4 } \right) } } \right]$

$=\sqrt { 2 } \left[ \cos { \left( \dfrac {3 \pi }{ 4 } \right) +i\sin { \left( \dfrac { 3\pi }{ 4 } \right) } } \right]$

Convert the given complex number in polar form: $i$

$\displaystyle z=i$
Let

$\displaystyle r\cos { \theta =0 }$ and

$r\sin { \theta } =1$
On squaring and adding, we obtain

$\displaystyle { r }^{ 2 }{ \cos }^{ 2 }\theta +{ r }^{ 2 }{ \sin }^{ 2 }\theta ={ 0 }^{ 2 }+{ 1 }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }\left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) =1$

$\displaystyle \Rightarrow { r }^{ 2 }=1$

$\displaystyle \Rightarrow r=\sqrt { 1 } =1$ (Since,

$r>0$ )

$\displaystyle \therefore \cos { \theta } =0$ and

$\sin { \theta } =1$

$\displaystyle \therefore \theta =\frac { \pi }{ 2 }$
So, the polar form is

$\displaystyle \therefore i=r\cos { \theta } +ir\sin { \theta } =\cos { \frac { \pi }{ 2 } } +i\sin { \frac { \pi }{ 2 } }$

Convert the given complex number in polar form : $\displaystyle 1-i$

Given,

$\displaystyle z=1-i$
Let

$\displaystyle r\cos { \theta =1 } and\quad r\sin { \theta } =-1$
On squaring and adding, we obtain

$\displaystyle { r }^{ 2 }{ \cos }^{ 2 }\theta +{ r }^{ 2 }{ \sin }^{ 2 }\theta ={ 1 }^{ 2 }+{ (-1) }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }\left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) =2$

$\displaystyle \Rightarrow { r }^{ 2 }=2$

$\displaystyle \Rightarrow r=\sqrt { 2 }$ (since,r>0 )

$\displaystyle \therefore \sqrt{2}\cos { \theta } =1$ and

$\sqrt{2} \sin { \theta } =-1$

$\displaystyle \therefore \quad \theta =-\frac { \pi }{ 4 }$ (As

$\theta$ lies in fourth quadrant.)
So, the polar form is

$\displaystyle \therefore 1-i=r\cos { \theta } +ir\sin { \theta } =\sqrt{2}\cos { \left (\frac { -\pi }{ 4 } \right) } +i \sqrt{2}\sin { \left(\frac { -\pi }{ 4 } \right) }$

$\displaystyle =\sqrt { 2 } \left[\cos { \left( \frac { -\pi }{ 4 } \right) } +i\sin { \left( \frac { -\pi }{ 4 } \right) } \right]$

If $\displaystyle \alpha$ and $\displaystyle \beta$ are different complex numbers with $\displaystyle \left| \beta \right| = 1$ , then find $\displaystyle \left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right|$ .

Let

$\displaystyle a=a+ib$ and

$\displaystyle \beta =x+iy$
It is given that,

$\displaystyle \left| \beta \right| = 1\Rightarrow \sqrt { { x }^{ 2 }+{ y }^{ 2 }}=1$

$\displaystyle \Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1....(i)$

$\displaystyle\therefore \left| \frac { \beta -\alpha }{ 1-\bar { \alpha } \beta } \right| =\left| \frac { \left( x+iy \right) -\left( a+ib \right) }{ 1-\left( a-ib \right) \left( x+iy \right) } \right|$

$\displaystyle =\left| \frac { \left( x-a \right) +i\left( y-b \right) }{ 1-\left( ax-aiy-ibx+by \right) } \right|$

$\displaystyle =\left| \frac { \left( x-a \right) +i\left( y-b \right) }{ \left( 1-ax-by \right) +i\left( bx-ay \right) } \right|$

$\displaystyle =\frac { \left| \left( x-a \right) +i\left( y-b \right) \right| }{ \left| \left( 1-ax-by \right) +i\left( bx-ay \right) \right| }$

$=\displaystyle \frac { \sqrt { { \left( x-a \right) }^{ 2 }+{ \left( y-b \right) }^{ 2 } } }{ \sqrt { { \left( 1-ax-by \right) }^{ 2 }+{ \left( bx-ay \right) }^{ 2 } } }$

$\displaystyle = \frac { \sqrt { { x }^{ 2 }+{ a }^{ 2 }-2ax+{ y }^{ 2 }+{ b }^{ 2 }-2by } }{ \sqrt { 1+{ a }^{ 2 }{ x }^{ 2 }+{ b }^{ 2 }{ y }^{ 2 }-2ax+2abxy-2by+{ b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }-2abxy } }$

$\displaystyle =\frac { \sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) +{ a }^{ 2 }+{ b }^{ 2 }-2ax-2by } }{ \sqrt { 1+{ a }^{ 2 }\left( { x }^{ 2 }+{ y }^{ 2 } \right) +{ b }^{ 2 }\left( { y }^{ 2 }+{ x }^{ 2 } \right) -2ac-2by } }$

$\displaystyle =\frac { \sqrt { 1+{ a }^{ 2 }+{ b }^{ 2 }-2ax-2by } }{ \sqrt { 1+{ a }^{ 2 }+{ b }^{ 2 }-2ax-2by } }$ [Using (i) ]

$\displaystyle =1$

Find the modulus and argument of the complex number $\displaystyle \frac { 1+2i }{ 1-3i }$

Let,

$\displaystyle z=\frac { 1+2i }{ 1-3i }$

$\displaystyle z=\frac { 1+2i }{ 1-3i } \times \frac { 1+3i }{ 1+3i }$

$=\displaystyle \frac { 1+3i+2i+6{ i }^{ 2 } }{ { 1 }^{ 2 }+{ 3 }^{ 2 } }$

$=\displaystyle \frac { 1+5i+6\left( -1 \right) }{ 1+9 }$

$\displaystyle =\frac { -5+5i }{ 10 }$

$=\displaystyle \frac { -1 }{ 2 } +\frac { 1 }{ 2 } i$

Let

$\displaystyle z=r\cos { \theta } +i r\sin { \theta }$

$\displaystyle i.e.,r\cos { \theta } =\frac { -1 }{ 2 }$ and

$r\sin { \theta } =\displaystyle \frac { 1 }{ 2 }$

On squaring and adding, we obtain

$\displaystyle { r }^{ 2 }\left( { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta \right) ={ \left( \frac { -1 }{ 2 } \right) }^{ 2 }+{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }=\frac { 1 }{ 4 } +\frac { 1 }{ 4 } =\frac { 1 }{ 2 }$

$r>0$

$\displaystyle \Rightarrow r=\frac { 1 }{ \sqrt { 2 } }$

$\displaystyle \therefore \frac { 1 }{ \sqrt { 2 } } \cos { \theta } =\frac { -1 }{ 2 }$ and

$\dfrac { 1 }{ \sqrt { 2 } } \sin { \theta } =\dfrac { 1 }{ 2 }$

$\displaystyle \Rightarrow \cos { \theta } =\frac { -1 }{ \sqrt { 2 } }$

and

$\sin { \theta } =\dfrac { 1 }{ \sqrt { 2 } }$

$\displaystyle \therefore \theta =\pi -\frac { \pi }{ 4 } =\frac { 3\pi }{ 4 }$

[As

$\displaystyle \theta$ lies in the II quadrant ]

Therefore the modulus and argument of the given complex number are

$\displaystyle \frac { 1 }{ \sqrt { 2 } }$ and

$\displaystyle \frac { 3\pi }{ 4 }$ respectively.

Find the modulus and the argument of the complex number $\displaystyle z=-\sqrt { 3 } +i$

Given,

$\displaystyle z=-\sqrt { 3 } +i$
Let

$\displaystyle r\cos { \theta =-\sqrt { 3 } }$ and

$r\sin { \theta } =1$
On squaring and adding, we obtain

$\displaystyle { r }^{ 2 }{ \cos }^{ 2 }\theta +{ r }^{ 2 }{ \sin }^{ 2 }\theta ={ \left( -\sqrt { 3 } \right) }^{ 2 }+{ 1 }^{ 2 }$

$\displaystyle \Rightarrow { r }^{ 2 }=3+1=4$

$(\because { \cos }^{ 2 }\theta +{ \sin }^{ 2 }\theta =1 )$

$\displaystyle \Rightarrow \quad r=\sqrt { 4 } =2$ (Conventionally r>0)

$\displaystyle \therefore$ Modulus of

$z$ i.e.

$|z|=2$

$\displaystyle \therefore 2\cos { \theta } =-\sqrt { 3 }$ and

$2\sin { \theta } =1$

$\displaystyle \Rightarrow \quad \cos { \theta } =\frac { -\sqrt { 3 } }{ 2 } and\quad \sin { \theta } =\frac { 1 }{ 2 }$

$\displaystyle \therefore \quad \theta =\pi -\frac { \pi }{ 6 } =\frac { 5\pi }{ 6 }$ [As

$\displaystyle \theta$ lies in the II quadrant]
Thus, the modulus and argument of the complex number

$\displaystyle -\sqrt { 3 } +i$ are 2 and

$\displaystyle \frac { 5\pi }{ 6 }$

Find the modulus of : $\displaystyle \frac { 1+i }{ 1-i } -\frac { 1-i }{ 1+i }$

Let

$\displaystyle z= \frac { 1+i }{ 1-i } -\frac { 1-i }{ 1+i } =\frac { { \left( 1+i \right) }^{ 2 }-{ \left( 1-i \right) }^{ 2 } }{ \left( 1-i \right) \left( 1+i \right) }$

$\displaystyle =\frac { 1+{ i }^{ 2 }+2i-1-{ i }^{ 2 }+2i }{ { 1 }^{ 2 }+{ 1 }^{ 2 } }=\frac { 4i }{ 2 } =2i$

$\displaystyle \therefore \left| z \right| =\left| 2i \right| =\sqrt { { 2 }^{ 2 } } =2$

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(ii) $\displaystyle 3{ x }^{ 2 }-4\sqrt { 3 } x+4=0$
(iii) $\displaystyle 2{ x }^{ 2 }-6x+3=0$

${ x }^{ 2 }-3x+5=0\\ a=2,b=-3,c=5$

Discriminant,

$D={ b }^{ 2 }-4ac =9-40 =-31$

Since

$D<0,$ the roots are imaginary for the given equation.

ii)

$3{ x }^{ 2 }-4\sqrt { 3 } x+4=0\\ a=3,b=-4\sqrt { 3 } ,c=4$

Discriminant,

$D$

$={ b }^{ 2 }-4ac =48-48 =0$

Since

$D=0,$ the roots are real and equal for the given equation.

${ 3x }^{ 2 }-4\sqrt { 3 } x+4=0\\ a=3,b=-4\sqrt { 3 } ,c=4\\ x=\dfrac{ [-b\pm \sqrt { { b }^{ 2 }-4ac } ] }{ 2a }\\ x=\dfrac{ 4\sqrt { 3 } }{ 6 }=\dfrac{2}{\sqrt3}$

iii)

$2{ x }^{ 2 }-6x+3=0\\ a=2,b=-6,c=3$

Discriminant,

$D =$

${ b }^{ 2 }-4ac =36-24 =12$

Since

$D>0,$ roots are real but not equal.

$x=\dfrac{ [-b\pm \sqrt { { b }^{ 2 }-4ac } ] }{ 2a }\\ x=\dfrac{ 6\pm \sqrt { 12 } }{ 4 }\\ x=\dfrac{ 6\pm 2\sqrt { 3 } }{ 4 }\\ x=\dfrac{ 3\pm \sqrt { 3 } }{ 2 }$

Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) $\displaystyle 2{ x }^{ 2 }+kx+3=0$
(ii) $\displaystyle kx\left( x-2 \right) +6=0$

$2{ x }^{ 2 }+kx+3=0$ has equal roots, then

$b^{ 2 }-4ac=0$ .

Here,

$a=2, b=k, c=3$

${ x }^{ 2 }+kx+3=0\\ b^{ 2 }-4ac=0\\ { k }^{ 2 }-4(2)(3)=0\\ { k^{ 2 } }=24\\ k=\pm \sqrt { 24 } =\pm 2\sqrt { 6 }$

ii)

$k{ x }^{ 2 }-2kx+6=0$ has equal roots, then

$b^{ 2 }-4ac=0$ .

Here,

$a=k, b=-2k, c=6$

$kx(x-2)+6=0\\ k{ x }^{ 2 }-2kx+6=0\\ 4{ k }^{ 2 }-4(k)(6)=0\\ 4{ k }^{ 2 }-24k=0\\ 4k(k-6)=0$

For

$k=0,$ equation is not quadratic.

$\\ k=6$

Represent the complex number $2 + 3i$ in argand plane

The magnitude of

$2+3i$ is

$\sqrt{13}$

Let the angle made be the point be

$\theta$ , we have

$tan\theta=\frac{3}{2}$

Which implies

$\theta=\tan ^{ -1 }{ \frac { 3 }{ 2 } }$

The number in argand plane will be

$\sqrt{13}{e}^{(i\tan ^{ -1 }{ \frac { 3 }{ 2 } ) } }$

If ${ b }^{ 2 }-4ac> 0$ in $a{ x }^{ 2 }+bx+c=0$ , then what can you say about roots of the equation? ( $a\ne 0$ )

Given

${ b }^{ 2 } - 4ac > 0$ and

$a\neq 0$ ,

Roots of

$a{x}^{2} + bx + c = 0$ will be given by

$\dfrac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$ , as quantity under square-root is positive, roots will be rational and distinct.

Show that the roots of the equation ${x}^{2}-2px+{p}^{2}-{q}^{2}+2qr-{r}^{2}=0$ are rational.

The given equation is

$x^{2}-2px+p^{2}-q^{2}+2qr-r^{2}=0$

The roots of the given equation are rational only when the discriminant is a perfect square.

The discriminant of the given equation is

$b^{2}-4ac= (2p)^{2}-4(p^{2}-q^{2}+2qr-r^{2})$

$\Rightarrow b^{2}-4ac=4p^{2}-4p^{2}+4(q^{2}-2qr+r^{2})$

$\Rightarrow b^{2}-4ac=4(q-r)^{2}=(2q-2r)^{2}$

Here we can see that

$b^{2}-4ac$ is a perfect square.

Therefore the roots of given equation are rational.

Show that the equation $2{x}^{2}-6x+7=0$ cannot be satisfied by any real values of $x$ .

Given equation is

$2x^{2}-6x+7=0$

The equation is in the form of

$ax^{2}+bx+c=0$ where

$a=2, b=-6, c=7$

Now calculate the value of discriminant

$b^{2}-4ac$

$\Rightarrow b^{2}-4ac=(-6)^{2}-4(2)(7)=36-56=-20$

$\Rightarrow b^{2}-4ac=-20<0$

Since the value of

$b^{2}-4ac$ is less than zero, the roots of the given equation are imaginary.

So the given equation

$2x^{2}-6x+7=0$ cannot be satisfied by any real value of

$x$

Simplify equation: $\quad \cfrac { 1 }{ x } -\cfrac { 1 }{ (x+1) } =\cfrac { 1 }{ (x+2) } -\cfrac { 1 }{ (x+4) }$ to get a quadratic equation. Find the nature of roots. Solve the equation using the formula.

$\rightarrow$ The following eq can be simplified to,

$\rightarrow$

$\dfrac{1}{(x)(x+1)}=\dfrac{2}{(x+2)(x+4)}$
by cross multipilcation method we are getting

$\rightarrow$

${x}^2+8+6x=2{x}^2+2x$
further solving,

$\rightarrow$

${x}^2-4x-8=0$
now finding D of the given eq,

$\rightarrow$

$D={b}^2-4ac$

$\rightarrow$

$D=16-4\times(-8)\times1$

$\rightarrow$

$D=48$ Since the D is positive there The eq has the real and distinct roots now solving eq

$\rightarrow$

$x=(-b+{D}^{1/2})/2a$

$\rightarrow$

$x=(-b-{D}^{1/2})/2a$
solving these eq, we get

$\rightarrow$

$x=0.341$ , and

$x=0.09150$

Simplify $\dfrac { 2 }{ i } +\dfrac { 3 }{ { i }^{ 3 } } +\dfrac { 4 }{ { i }^{ 4 } } +\dfrac { 5 }{ { i }^{ 5 } }$ in the form of $a+ib$ & represent graphically & find modulus & amplitude.

Given

$\dfrac { 2 }{ i } +\dfrac { 3 }{ { i }^{ 3 } } +\dfrac { 4 }{ { i }^{ 4 } } +\dfrac { 5 }{ { i }^{ 5 } } =a+ib$

$\Rightarrow { i }^{ 2 }=-1$

${ i }^{ 3 }=--i$

${ i }^{ 4 }=-1$

${ i }^{ 5 }=-i$

$\Rightarrow \dfrac { -2\times -2 }{ i } +\dfrac { 3 }{ \left( -i \right) } +4+\dfrac { 5 }{ { i } } =a+ib$

$\Rightarrow -2i+3i+4-5i=a+ib$

$4-4i=a+ib$

$\Rightarrow a=4,$

$b=-4$

$\Rightarrow \left| a+ib \right| =\sqrt { { 4 }^{ 2 }+{ 4 }^{ 2 } } =4\sqrt { 2 }$

$\Rightarrow arg \left( a+ib \right) =-\tan { \left( \dfrac { 4 }{ 4 } \right) }$ ( as it is 4th quadrant )

$\Rightarrow arg \left( a+ib \right) =\dfrac { -\pi }{ 4 }$

Hence, the answer is

$\dfrac { -\pi }{ 4 }.$

Find the value of the principal argument of the complex number $z = \dfrac {(1 + i\sqrt {3})^{2}}{(1 - i)^{3}}$ .

Given

$z=\dfrac{(1+i\sqrt{3})^{2}}{(1+i)3}=\dfrac{(1+i\sqrt{3})(1+i\sqrt{3})}{(1-i)(1-i)(1-i)}$

$\therefore arg (z)=\dfrac{\tan^{-1}\sqrt{3}+\tan^{-1}\sqrt{3}}{\tan^{-1}(-1)+\tan^{-1}(-1)+\tan^{-1}(-1)}$

$=2\tan^{-1}\sqrt{3}-3\tan^{-1}(-1)$

$=120+3\tan^{-1}(1)$

$(\because (1-i)\ lie\ in\ IV\ quad)$

$arg (z)=255^{o}$

$\therefore \theta=-\tan^{-1}(1)$

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle z^2 \, + \, | \, z \, | \, = \, 0$

Let

$z=x+iy$

Then,

$z^2+|z|=0$

$\implies (x+iy)^2+\sqrt{x^2+y^2}=0$

$\implies x^2-y^2+2xyi+\sqrt{x^2+y^2}=0+0i$

By comparing real an imaginary parts:

$x^2-y^2+\sqrt{x^2+y^2}=0$ ..........[1] or

$2xy=0\implies x=0$ or

$y=0$

Case

$1$ : if

$y=0$ , from eq.

$1$

$x^2+\sqrt{x^2}=0\implies x^2+|x|=0$ ..........[2]

Now if

$x>0$ , then

$x^2+x$ will not be

$0$ (sum of two positive numbers can't be

$0$ ).

So, for

$x<0$ ;

$x^2-x=0\implies x=0,1$ . But

$x=1$ doesn't satisfy eq. 2.

So only point that satisfies is

$y=0,x=0$

Case

$2$ : if

$x=0$ , from eq.

$1$

$-y^2+\sqrt{y^2}=0\implies -y^2+|y|=0$ ..........[3]

Now if

$y<0$ , then

$-y^2-y=0\implies y=0,-1$

So, for

$y>0$ ;

$-y^2+y=0\implies y=0,1$ . But

$y=1,0$ both satisfies eq. 3.

So the points that satisfy case

$2$ are

$y=0,x=0$ and

$x=0, y=1$ and

$x=0, y=-1$

Hence all points which satisfies the given equation

$z^2+|z|=0$ is

$(0,0),(0,1)$ and

$(0,-1)$

So,

$z=0+0i=0$

$z=0+1i=i$

$z=0-1i=-i$

1376372_889265_ans_3955ff0176f04ca0afda24be9acb865d.png

Indicate the point of the complex plane $z$ which satisfy the following equation.
Im $z^2 = 0$

Let

$z=x+iy\\ \implies { z }^{ 2 }={ \left( x+iy \right) }^{ 2 }\\\implies={ x }^{ 2 }-{ y }^{ 2 }+2ixy\\ Im\left( { z }^{ 2 } \right) =0\\ \therefore \quad xy=0$

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle | \, z \, |^3 \, + \, z \, = \, 0$

Let

$z=x+iy$

$|z|=\sqrt{x^2+y^2}$

$|z|^3+z=0$

$({x^2+y^2})^{2/3}$

$+x+iy=0$

$({x^2+y^2})^{2/3}$

$+x=0$ (Real part) and

$y=0$ (imaginary part)

$\implies (x^2)^{3/2}+x=0$

$\implies x^3+x=0$

$\implies (x^2+1)x=0$

$\implies x=0$

Hence points are

$(0,0)$ .

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle Re \, z^2 \, = \, 0$

Let

$z=x+iy\\ \implies { z }^{ 2 }={ \left( x+iy \right) }^{ 2 }\\\implies={ x }^{ 2 }-{ y }^{ 2 }+2ixy\\ Re\left( { z }^{ 2 } \right) =0\\ \therefore \quad { x }^{ 2 }-{ y }^{ 2 }=0\\ \implies (x+ y)(x-y)=0$

Indicate the point of the complex plane
a) $1+2i$
b) $2-3i$
c) $-2-4i$

1391972_889316_ans_4bb10b5d91804c89930921eeca950b04.png

Calculate $\displaystyle \sqrt[3]{- \, 1}$

Let

$z=\sqrt{-1}=e^{i\pi/2}$

$z^{1/3}=i^{1/3}=e^{i\pi/6}$

$\implies (\cos \pi/6+i\sin \pi/6)$

$\implies \cfrac{\sqrt 3}{2 }+\cfrac{i}{ 2}$

Calculate, $\displaystyle \sqrt[4]{- \, 1 \, \frac{1}{2} \, - \, i \, \frac{\sqrt{3}}{2}}.$

$z=(\cfrac{-3}{2}-\cfrac{-i\sqrt 3}{2})^{1/4}$

$z^4=(\cfrac{-3}{2}-\cfrac{-i\sqrt 3}{2})=re^{i\theta}$

$r=\sqrt{\cfrac{9}{4}+\cfrac{3}{4}}=\sqrt 3$

$\theta=\tan{-1} (\cfrac{\sqrt 3}{2}\times \cfrac{2}{3})=\pi/6=7\pi/6$

$z^4 =\sqrt 3 e^{i7\pi/6}$

$\implies z= \sqrt 3 e^{i7\pi/24}$

$\implies \sqrt 3(\cos\cfrac{7\pi}{24}+i\sin \cfrac{7\pi}{24})$

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle | \, z \, | \, + \, z \, = \, 0$

Let

$z=x+iy$

$|z|=\sqrt{x^2+y^2}$

$|z|+z=0$

$\sqrt{x^2+y^2}$

$+x+iy=0$

$\sqrt{x^2+y^2}$

$+x=0$ and

$y=0$

$\implies \sqrt x^2+x=0$

$\implies 2x=0$

$\implies x=0$

1073655_889259_ans_5a6e5a0b6fb14dd59641599125c343f4.PNG

Find the values of $k$ for which the given quadratic equation has real and distinct roots
$kx^2 + 2x + 1 = 0$

General quadratic equation is

$ax^2+bx+c=0$

As per the question, the given equation has real and distinct root.

So,

$D>0$

Therefore

$D=b^2-4ac=(2)^2-4\times k\times 1$

$D=4-4k=4(1-k)$

Therefore

$D>0$

$4(1-k)>0=(1-k)>0$

Hence

$k<1$

Prove the following inequalities:
$\left| z-1 \right| \le \left| \left| z \right| -1 \right| +\left| z \right| \left| \arg\ z \right|$

$\left| z \right|$
Now apply

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| \le \left| { z }_{ 1 } \right| +\left| { z }_{ 2 } \right|$

$L.H.S$ is

$=\left| z \right| \left| \cfrac { z }{ \left| z \right| } -1 \right| +\left| \left| z \right| -1 \right|$

(a) For any two non-zero complex numbers ${z}_{1}$ and ${z}_{2}$ if $\left| { z }_{ 1 }+{ z }_{ 2 } \right| =\left| { z }_{ 1 } \right| +\left| { z }_{ 2 } \right|$ , then prove that $arg\ {z}_{1}-arg\ {z}_{2}$ is zero.
(b) Prove the above result if we have $\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| { z }_{ 1 } \right| -\left| { z }_{ 2 } \right|$

(a)

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| =\sqrt { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ 2r }_{ 1 }{ r }_{ 2 }\cos { ( { \theta }_{ 1 }-{ \theta }_{ 2 } }) }$
and

$\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\sqrt { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }-{ 2r }_{ 1 }{ r }_{ 2 }\cos { ({ \theta }_{ 1 }-{ \theta }_{ 2 } }) }$
If

$\cos { \left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) =1 }$ i.e.

${ \theta }_{ 1 }-{ \theta }_{ 2 }=0$
or arg

${z}_{1-}$ arg

${z}_{2}=0$ , then

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| =\sqrt { { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ 2r }_{ 1 }r_{ 2 } } ={ r }_{ 1 }+{ r }_{ 2 }$

$\left| { z }_{ 1 } \right| +\left| { z }_{ 2 } \right|$
and

$\left| { z }_{ 1 }-{ z }_{ 2 } \right| ={ r }_{ 1 }-{ r }_{ 2 }=\left| { z }_{ 1 } \right| -\left| { z }_{ 2 } \right|$

Put the following in the form A + iB :
$\dfrac{(1 \, + \, i)^2}{3 \, - \, i}$

$\dfrac{(1 \, + \, i)^2}{3 \, - \, i} \, = \, \dfrac{1 \, + \, i^2 \, + \, 2i}{3 \, - \, i} \, = \, \dfrac{2i(3 \, + \, i)}{3^2 \, - \, i^2}$

$\dfrac{2(3i \, + \, i^2)}{10} \, = \, \dfrac{1}{5}(3i \, - \, 1) \, = \, -\dfrac{1}{5} \, + \, \dfrac{3}{5}i$

Simplify the following :
$\dfrac{3}{1 \, + \, i} \, - \, \dfrac{2}{2 \, - \, i} \, + \, \dfrac{2}{1 \, - \, i}$

$\dfrac{3(1 \, - \, i)}{2} \, - \, \dfrac{2(2 \, + \, i)}{4 \, + \, 1} \, +\, \dfrac{2(1 \, + \, i)}{2}$

$= \, \dfrac{1}{10}[15(1 \, - \, i) \, - \, 4(2 \, + \, i) \, + \, 10(1 \, + \, i)] \, = \, \dfrac{1}{10}(17 \, - \, 9i)$

If a, b, c are real numbers such that ac $\neq$ 0, then show that at least one of the equations $ax^2$ + bx + c = 0 and - $ax^2$ + bx + c = 0 has real roots.

Given

$\left\{a, b, c\right\}\epsilon R$

and

$ac\neq 0$

Let First eqn is

$ax^{2}+bx+c=0$

and other is

$-ax^{2}+bx+c=0$

and

$D_{1}$ and

$D_{2}$ are the discriminant of these eqn resp

Now,

$=D_{1}=b^{2}-4ac$

$D_{2}=b^{2}+4ac$

case

$i) ac < 0$

then

$D_{1}=b^{2}-4ac > 0$

$D_{2}=b^{2}-4ac\rightarrow$ cant say

case

$ii) ac > 0$

then

$D_{1}=b^{2}-4ac\rightarrow$ cant say

$D_{2}=b^{2}+4ac > 0$

We can clearly observe that in both case atleast one equation have

$D > 0$ i.e. having real roots

Find the length of the segments connecting the points represented by the following pairs of numbers :
3 - 2i, 3 + 5i

First we note that the length of segment joining the points

$z_1$ and

$z_2$ is |

$z_1 \, - \, z_2$ |.

Therefore distance=

$|(3 - 2i) - (3 + 5i)| = |- 7i| = 7$

locate the point representing the complex numbers $z$ on the Argand diagram for which
$\left| z+i \right| =\left| z-2 \right|$

$\left| z+i \right| =\left| z-2 \right|$ or $$ \left| z+i
\right| ^{ 2 }=\left| z-2 \right| ^{ 2 }$$
or

$\left| x+iy+i \right| ^{ 2 }=\left| x+iy-2 \right| ^{ 2 }$
or

${ x }^{ 2 }+(y+1)^{ 2 }=(x-2)^{ 2 }+{ y }^{ 2 }$
or

$4x+2y-3=0$
Hence point

$z$ satisfying the given equation lie on the
straight line

$(1)$ .
Alternative. We know that $$\left| z+i \right| =\left| z-(-i)
\right|

$is the distance from the point$ z$$ to the point
representing the number

$-i$ and

$\left| z+2 \right|$ is the
distance from the point

$z$ to the point representing the
number

$2$ . It is required to find the point for which these
distances are equal. The solution will thus be a locus of a point
equidistant from tow fixed point representing the number

$-i$
and

$2$ i.e from the point

$(0,-1)$ and

$(2,0)$ .
From the geometry we know that this locus is a straight line
perpendicular to a line segment connecting the points

$(0,-1)$
and

$(2,0)$ and passing through its mid-point. Hence all point

$z$ satisfying

$\left| z+i \right| =\left| z-2 \right|$ lie on this line.

If ${z}_{1},{z}_{2},{z}_{3}$ are three complex numbers, prove that
${ z }_{ 1 }Im\left(\bar { { z }_{ 2 } } { z }_{ 3 } \right) +{ z }_{ 2 }Im\left(\bar { { z }_{ 3 } } { z }_{ 1 } \right) +{ z }_{ 3 }Im\left(\bar { { z }_{ 1 } } { z }_{ 2 } \right) =0$ where $Im(w)=$ imaginary part of $w,w$ being a complex number.

${ z }_{ 1 }={ x }_{ 1 }+{ iy }_{ 1 },{ z }_{ 2 }{ z }_{ 3 }=\left( { x }_{ 2 }+{ iy }_{ 2 } \right) \left( { x }_{ 3 }+{ iy }_{ 3 } \right)$

$\therefore Im\left( \bar { { z }_{ 2 } } { z }_{ 3 } \right) =\left( { x }_{ 2 }{ y }_{ 3 }-{ x }_{ 3 }{ y }_{ 2 } \right)$

$\therefore { { z }_{ 1 } }Im\left( \bar { { z }_{ 2 } } { z }_{ 3 } \right) =\left( { x }_{ 1 }+{ iy }_{ 1 } \right) \left( { x }_{ 2 }{ y }_{ 3 }-{ x }_{ 3 }{ y }_{ 2 } \right)$

$\qquad \qquad \qquad \ ={ x }_{ 1 }\left( { x }_{ 2 }{ y }_{ 3 }-{ x }_{ 3 }{ y }_{ 2 } \right) +{ iy }_{ 1 }\left( { x }_{ 2 }{ y }_{ 3 }-{ x }_{ 3 }{ y }_{ 2 } \right)$

$\therefore \sum { { z }_{ 1 }Im\left( \bar { { z }_{ 2 } } { z }_{ 3 } \right) =0 }$
Each sigma will have six terms, three

$+ive$ and three

$-ive$ which will cancel.

Solve the following equations:
x-1= $\sqrt{a-x^{2}}$

Right hand side is non-negative for all (permissible) x, that is, for all x for which a -x

$^{2}$

$\geq$ 0, and left hand side is non-negative for x

$\geq$ 1. Squaring,. we get:(x-1)

$^{2}$ =a-x

$^{2}$ or 2x

$^{2}$ - 2x + 1- a = 0or x =

$\dfrac{2 \pm \sqrt{4-8(1-a)}}{4}= \dfrac{1\pm\sqrt{2a-1}}{2}$

These values of x are real if 2a -1 > 0 i.e. a >

$\dfrac{1}{2}$
The root x =

$\dfrac{1-\sqrt{2a-1}}{2}$ does not satisfy the condition
x > 1 and so is not a root of the given equation.And the root x = [1 +

$\sqrt{2a-1}$ ] / 2 satisfies the equation for a

$\geq$ 1. Hence for a < 1, the equation has no root and for a

$\geq$ 1, it has the root x = [1 +

$\sqrt{2a-1}$ ] / 2.

Among the complex numbers $z$ which satisfy the condition $\left| z-25i \right| \le 15,$ find the number having the least positive and greatest positive argument.

The complex numbers

$z$ satisfying the condition

$\left|z-25i \right| \le 15$
are represented by the points inside and
on this circle of radius

$15$ and centre at the point

$C (0,25)$ . From the figure, it is clear that the complex number

$z$ satisfying

$(1)$ and having least positive argument correspond to the point

$P(x,y)$ which is the point of contact
ofa ray issuing from the origin and lying in the first quadrant
to the above circle. The positive argument of all other points
which and on the circle is greater than the argument of

$P$
From figure, we have

$OC=25,\ CP=radius=15$
and

$\angle\ CPO=90$
Hence

$OP=\sqrt { { OC }^{ 2 }-{ CP }^{ 2 } } =\sqrt { { 25 }^{ 2 }-{ 15 }^{ 2 } } =20$ .
If

$\angle\ PCO=\theta$ , then

$\angle\ PON=\theta$ , Also

$cos\theta =\dfrac { PC }{ OC } =\dfrac { 15 }{ 25 } =\dfrac { 3 }{ 5 } ,$

$\therefore x=ON=OP\cos \theta =20\times (3/5)=12,$
and

$y=PN=OP\sin \theta =20\times (4/5)=16,$ .
Hence

$P$ represents the complex number

$z=x+iy=12+16i$ which
is therefore the required value of

$z$ satisfying the given
conditions. The corresponding point

$Q$ on the other tangent
from

$O$ will have greatest

$+ive$ argument. It will be

$- 12+16i$ whose argument is

$\pi -\theta$ .
1037370_1001107_ans_8280101678af40c0ab51703716a3dbc4.png

Locate the complex numbers $z=x+iy$ such that
$\left| z-i \right| =1,\arg\dfrac { z }{ z+i } =\dfrac { \pi }{ 2 }.$

Here are two conditions to satisfied.

$\left| z-i \right| =1\Rightarrow { x }^{ 2 }+\left( y-1 \right) ^{ 2 }=1$ ......

$(1)$
It represents a circle with centre at

$(0,1)$ and radius

$1$ . It clearly touches

$x-$ axis as

$r=k=1$ . Hence all points on the boundary of this circle are the solutions of this.
Again

$\arg\dfrac { z }{ z+i } =\dfrac { \pi }{ 2 }$

$\therefore\quad \dfrac { z }{ z+i } =r\left( \cos { \dfrac { \pi }{ 2 } } +i\sin { \dfrac { \pi }{ 2 } } \right)$

$0+ir$

$\therefore\quad R.P.=0$ ,

$I.P.=r=+ive$ ......

$(2)$
Now

$\dfrac { z }{ z+i } =\dfrac { x+iy }{ x+i\left( y+1 \right) } .\dfrac { x-i\left( y+1 \right) }{ x-i\left( y+1 \right) }$

$=\dfrac { x^{ 2 }+{ y }^{ 2 }+y-ix }{ { x }^{ 2 }+\left( y+1 \right) ^{ 2 } }$
Condition

$\left( 2 \right) \Rightarrow { x }^{ 2 }+{ y }^{ 2 }+y=0$
It represents a circle with centre at

$\left( 0,-\dfrac { 1 }{ 2 } \right)$ and radius

$\dfrac { 1 }{ 2 }$ , i.e.

${ x }^{ 2 }+\left( y+\dfrac { 1 }{ 2 } \right) ^{ 2 }=\left( \dfrac { 1 }{ 2 } \right) ^{ 2 }$
Also I.P.

$=-x>0$ or

$x<0$
Thus we have two circles

$x^{ 2 }+(y-1)^{ 2 }=1^{ 2 }$
and

$x^{ 2 }+\left( y+\dfrac { 1 }{ 2 } \right) ^{ 2 }=\left( \dfrac { 1 }{ 2 } \right) ^{ 2 }$ with

$x<0$
Because of the condition

$x<0$ , only left hand portion of the second circle will be valid except the points

$P$ and

$Q$ as for both

$x=0$ . Also

$P(0,0)$ does not satisfy 1st circle.
1034840_999493_ans_32fe7469e4e447aa8fd9f2b1fad00954.png

locate the point representing the complex numbers $z$ on the Argand diagram for which
$\left| z-1 \right| =\left| z-3 \right| =\left| z-i \right| ,$

$\left| z-1 \right| ^{ 2 }=\left| z-3 \right| ^{ 2 }=\left| z-i \right| ^{ 2 }$ on squaring

$\therefore \left( x-1 \right) ^{ 2 }+y^{ 2 }=\left( x-3 \right) ^{ 2 }+y^{ 2 }=x^{ 2 }+\left( y-1 \right) ^{ 2 }$

$I$

$II$

$III$
From

$I$ and

$II$ ,

$2(2x-4)=0$

$\therefore x=2$
From

$I$ and

$III$ ,

$-2x=-2y$

$\therefore x=y=2$

$\therefore z=x+i\ y=2+2i$
Remarks: Geometric speaking, the equation

$(1)$ means that

$z$ is a point equidistant from the point representing the number

$1,3,i$ , that is, equidistant from the points

$A(1,0)$ ,

$B (3,0)$ and

$C(0,1)$ . It is therefore the circumcentre of

$\Delta ABC$ .
So it is the intersection of perpendicular bisectors

$FP$ and

$EP$ of segment

$AB$ and

$AC$ , that is, the intersection

$P$ of the lines represented by the equation

$x=2$ and

$y=x$ . So we get

$x=y=2$ as before. (See fig.).
1035222_999781_ans_998ff8f10bec447fa8210513335f5cff.png

If $z_1 \, , \, z_2 \, , \,z_3 \, , \,z_4$ be the vertices of rhombus in argand palne and $\angle CBA \, = \, \pi/3$ , then prove that
$2\, z_2 \, = \, z_1(1 \, + \, i\sqrt3) \, + \, z_3(1 \, - \, i\sqrt3)$
and $2\, z_4 \, = \, z_1(1 \, - \, i\sqrt3) \, + \, z_3(1 \, + \, i\sqrt3)$

Rotation about B in anti-clockwise sense

$z_1 - z_2 = (z_3 - z_2) e^{i\pi /3}$
or

$z_1 - z_2 = (z_3 - z_2) \dfrac{1+i\sqrt{3}}{2}$

$z_1 - z_3 .$

$\dfrac{1+i\sqrt{3}}{2} = z_2 \left (1-\,\dfrac{1+i\sqrt{3}}{2} \right )$
or

$z_1 - \dfrac{z_3}{2}\left ( 1 \,+\, i\sqrt{3} \right ) = \dfrac{z_2}{2}\left ( 1 \,-\, i\sqrt{3} \right )$

Multiplying both sides by

$1 \,+ \,i\sqrt{3}$ , we get

$\therefore$

$z_1 (1 \, + \,i\sqrt{3}) - \dfrac{z_3}{2} \, \left (-2 \,+ \,2i \,\sqrt{3} \right ) = 2z_2$
or

$z_1 (1 \, + \,i\sqrt{3}) + z_3 (1 \, - \,i\sqrt{3}) = 2z_2$
Similarly we get

$2z_4$ by rotation about D.

locate the point representing the complex numbers $z$ on the Argand diagram for which
$\left| z \right| -4=\left| z-i \right| -\left| z+5i \right| =0$

We have two equations

$z=x+iy$ , these equation become

$\left| x+iy \right| =4$ i.e

${ x }^{ 2 }+{ y }^{ 2 }=16$ .
And

$\left| x+iy-i \right| =\left| x+iy+5i \right|$
or

${ x }^{ 2 }+(y-1)^{ 2 }={ x }^{ 2 }+(y+5)^{ 2 }$
i.e,,

$y=-2$
Putting

$y=-2$ in

$(1)$ ,

${ x }^{ 2 }+4=16$ or

$${ x } = \pm 2\sqrt { 3 }$$.
Hence the complex number

$z$ satisfying the given equations are

${ z }_{ 1 }=(2\sqrt { 3 } ,-2)$
and

${ z }_{ 2 }=(-2\sqrt { 3 } ,-2)$
that is,

${ z }_{ 1 }=-2\sqrt { 3 } -2i$ ,

${ z }_{ 2 }=-2\sqrt { 3 } -2i$

Find the range of $k$ for which the equation ${ x }^{ 2 }-4x+k=0$ has distinct real roots.

The given equation is

${ x }^{ 2 }-4x+k=0$

Here,

$a=1,b=-4,c=k$

$\quad \therefore D={ b }^{ 2 }-4ac={ (-4) }^{ 2 }-4\times 1\times k=16-4k\quad$

The given equation will have real and distinct roots, if

$D>0\Rightarrow 16-4k>0\Rightarrow 4k<16\Rightarrow k<\cfrac { 16 }{ 4 } =4$

Hence, the given equation will have distinct roots, if

$k< 4$

Find the value(s) of $k$ for which the equation ${ x }^{ 2 }+5kx+16=0$ has no real roots.

The given equation is

${ x }^{ 2 }+5kx+16=0$

Here,

$a=1,b=5k,c=16$

$\therefore D={ b }^{ 2 }-4ac={ (5k) }^{ 2 }-4\times 1\times 16=25{ k }^{ 2 }-64$

The given equation will have no real roots, if

$D< 0$

$\Rightarrow 25{ k }^{ 2 }-64<0\Rightarrow 25\left( { k }^{ 2 }-\cfrac { 64 }{ 25 } \right) <0\Rightarrow { k }^{ 2 }-\cfrac { 64 }{ 25 } <0$ (

$\because ab<0$ and

$a> 0\Rightarrow b< 0$ )

$\Rightarrow -\cfrac { 8 }{ 5 } <k<\cfrac { 8 }{ 5 } \left[ \because { x }^{ 2 }-{ a }^{ 2 }< 0\Rightarrow -a< x< -a\right]$

Determine whether the given quadratic equation has real roots and if so, find the roots.
$2{ x }^{ 2 }+5\sqrt { 3 } x+6=0\quad$

The given equation is

$2{ x }^{ 2 }+5\sqrt { 3 } x+6=0\quad$

Here

$a=2,b=5\sqrt { 3 } ,c=6$

$\therefore D={ b }^{ 2 }-4ac={ (75) }^{ }-4\times 2\times 6=27>0$

So, the given equation has real roots given by

$\alpha =\cfrac { -b+\sqrt { D } }{ 2a } =\cfrac { -5\sqrt { 3 } +\sqrt { 27 } }{ 2\times 2 } =\cfrac { -\sqrt { 3 } }{ 2 }$

$\beta =\cfrac { -b-\sqrt { D } }{ 2a } =\cfrac { -5\sqrt { 3 } -\sqrt { 27 } }{ 2\times 2 } =-2\sqrt { 3 }$

If $p,q,r$ and $s$ are real numbers such that $pr=2(q+s)$ , then show that at least one of the equations ${ x }^{ 2 }+px+q=0$ and ${ x }^{ 2 }+rx+s=0$ has real roots.

We have

${ x }^{ 2 }+px+q=0....(i)$

${ x }^{ 2 }+rx+s=0....(ii)$

Let

${D}_{1}$ and

${D}_{2}$ be the discriminants of equations (i) and (ii). Then

${ D }_{ 1 }=b^2-4ac={ p }^{ 2 }-4q$

similarly,

${ D }_{ 2 }={ r }^{ 2 }-4s$

$\Rightarrow { D }_{ 1 }+{ D }_{ 2 }={ p }^{ 2 }-4q+{ r }^{ 2 }-4s=\left( { p }^{ 2 }+{ r }^{ 2 } \right) -4(q+s)\quad$

$\Rightarrow { D }_{ 1 }+{ D }_{ 2 }={ p }^{ 2 }+{ r }^{ 2 }-4\left( \cfrac { pr }{ 2 } \right) \left[ \because pr=2(q+s)\therefore q+s=\cfrac { pr }{ 2 } \right]$

$\Rightarrow { D }_{ 1 }+{ D }_{ 2 }={ p }^{ 2 }+{ r }^{ 2 }-2pr={ (p-r) }^{ 2 }\ge 0$

At least one of

${D}_{1}$ and

${D}_{2}$ is greater than or equal to zero
At least one of the two equations has real roots.

For each of the quadratic equations below, convert them into the standard form $ax^2+bx+c=0$ and determine the value of ' $b^2-4ac$ . Comment on the nature of roots from the value of ' $b^2-4ac$
(i) $5x^2=6x-7$
(ii) $4x^2+16=-16x$
(iii) $2x^2+6x+3=0$

$i)5{ x }^{ 2 }-6x+7=0$

${ b }^{ 2 }-4ac=6^{ 2 }-4\cdot 5\cdot 7$

$=36-140$

$=-104$ , No real roots

$ii)4{ x }^{ 2 }+16x+16=0$

${ b }^{ 2 }-4ac=16^{ 2 }-4\cdot 4\cdot 16$

$=256-256$

$=0$ , Equal real roots

$iii)2{ x }^{ 2 }+6x+3=0$

${ b }^{ 2 }-4ac=6^{ 2 }-4\cdot 2\cdot 3$

$=36-24$

$=12$ , Two distinct real roots.

Solve
$(\cfrac { 1 }{ 5 } +i\cfrac { 2 }{ 5 } )-(4+i\cfrac { 5 }{ 2 } )$

Given,

$\left(\dfrac{1}{5}+i\dfrac{2}{5}\right)-\left(4+i\dfrac{5}{2}\right)$

$=\dfrac{1}{5}-4+i\left (\dfrac{2}{5}-\dfrac{5}{2} \right )$

$=-\dfrac{19}{5}-i\dfrac{21}{10}$

Express the following complex numbers in the form $r\left( {\cos \theta + i\sin \theta } \right)$ :
1 + i tan $\alpha$

$1+i\tan \alpha$

$r=\sqrt{1^2+\tan ^2\alpha}=\sec \alpha$

$arg (\theta )=\tan^{-1}\left ( \dfrac{\tan \alpha}{1} \right )$

$=\alpha ,\alpha >0$

$=-\alpha ,\alpha <0$

The required form is

$(\sec\alpha ,\alpha ),(\sec \alpha ,-\alpha )$

What is the value of ${i^i}$
Where $i = \sqrt { - 1}$

Using Euler's formula

$e^{ix} = \cos x + i \sin x$
Using

$x= \cfrac{\pi}{2}$
we get

$e^{i\cfrac{\pi}{2}} = \cos \cfrac{\pi}{2} + i \sin \cfrac{\pi}{2}$
Now raising both sides by

$i^{th}$ power , we get

$e^{i^2(\cfrac{\pi}{2})} = i^i$

$i^2 = -1$

$i^i = e^{-\cfrac{\pi}{2}}$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
$5{x^2} - 6x + 2 = 0$

$5x^{2}-6x+2=0$

$D \equiv 6^{2}-4(2)(5)=36-40=-4<0$

Roots are imaginary & conjugate of each other.

Given that $iz^{2}$ = $1 + \frac{2}{z} + \frac{3}{z^{2}} +\frac{4}{z^{3}} +\frac{5}{z^{4}}+ ....$ and $z= n\pm\sqrt{-i}$ , find $\left \lfloor 100n \right \rfloor$

$i z^2 = 1 + \dfrac{2}{z} + \dfrac{3}{z^2} + \dfrac{4}{z^3} + .... \infty$ ....(1)

also

$iz = \dfrac{1}{2} + \dfrac{2}{z^2} + \dfrac{3}{z^3} + \dfrac{4}{z^4} .... \infty$ ....(2)

(1) - (2)

$iz(z - 1) = 1 + \dfrac{1}{z} + \dfrac{1}{z^2} + \dfrac{3}{z^3} .... \infty$

$iz (z - 1) = \dfrac{1}{1 - \dfrac{1}{z}}$

$iz (z - 1) = \dfrac{z}{z - 1}$

$z [iz - i - \dfrac{1}{z - 1}] = 0$

$\dfrac{z}{z - 1} [iz^2 - iz - iz + i - 1] = 0$

$\dfrac{z}{z - 1} [ iz^2 - 2iz + i - 1] = 0$

$\underset{(not \, possible)}{z = 0}$ or

$z = \dfrac{2i \pm \sqrt{-4 - 4i (i - 1)}}{2i}$

$= 1 \pm \dfrac{2}{2i} \sqrt{-1 + 1}$

$= 1 \pm \sqrt{-i}$

$n = \bot$ (comparing with

$n \pm \sqrt{-i}$ )

$\therefore [100 n] = 100$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
$x(x - 5) = 36$

$x(x-5)=36$

$\Rightarrow x^{2}-5x-36=0$

$D \equiv 5^{2}-4(-36)(1)=169=(13)^{2}$ (i.e. perfect square)

Roots are real, distinct & rational.

Find modulus of following
(i) $\pm \left( {4 + 3i} \right)$
(ii) $\pm \sqrt 2 + 0i$
(iii) $0 \pm \sqrt 2 i$

$\pm (4+3i)$

Modulus =

$\sqrt {4^2 + 3^2} = 5$

$\pm \sqrt2+0i$

Modulus =

$\sqrt {{\sqrt2}^2 + 0^2} = \sqrt2$

$0 \pm \sqrt2i$

Modulus =

$\sqrt {0^2 + {\sqrt2}^2} = \sqrt2$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
$5{x^2} - 4\sqrt 5 x + 4 = 0$

$5x^{2}-4\sqrt5x+4=0$

$D\equiv (4\sqrt5)^{2}-4(4)(5)=0$

Roots are equal & irrational real as b is irrational.

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
$3{x^2} - 18x + 27 = 0$

$3x^{2}-18x+27=0$

$D \equiv (18)^{2}-4(27)(3)$

$\equiv 324-320$

$\equiv 0$

$D=0 \Rightarrow$ Roots are equal & also integers as b is divisible by 4

$2a$ $

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
${x^2} + x - 3 = 0$

Given equation:

$x^3+x-3=0$

$a=1, b=1,c=-3$

$D=b^2-4ac$

$\implies D=(1)^2-4(1)(-3)$

$\implies D=1+12$

$\implies D=13$

Thus

$D>0$

$\therefore$ There are two distinct real zeros.

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
${x^2} + x + {1 \over 4} = 0$

Given equation:

$x^2+x+\cfrac{1}{4}=0$

$\Rightarrow 4x^2+4x+1=0$

$a=4, b=4,c=1$

$D=b^2-4ac$

$\implies D=(4)^2-4(4)(1)$

$\implies D=16-16$

$\implies D=0$

Thus

$D=0$

$\therefore$ Only one real zero.

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
$4{x^2} - 6x + 2 = 0$

Given equation:

$4x^2-6x+2=0$

$a=4, b=-6,c=2$

$D=b^2-4ac$

$\implies D=(-6)^2-4(4)(2)$

$\implies D=36-32$

$\implies D=4$

Thus

$D>0$

$\therefore$ Roots are real and distinct.

Find the value of $i^{i}$

$z = i^{i}$

$\Rightarrow$

$z^{4} = (i^{4})^{i} = 1^{i}$

$\Rightarrow$

$4 \ln z = i\ln 1 =0$

$\Rightarrow$

$z = e^{0} =1$

$\Rightarrow z=1$

Find the modulus of $-15-8i$ .

Complex number =

$-15-8i$

Modulus =

$\sqrt{15^2 + 8^2}$

$\sqrt{289}$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
${x^2} - 2x - 15 = 0$

Given equation:

$x^2-2x-15=0$

$a=1, b=-2,c=-15$

$D=b^2-4ac$

$\implies D=(-2)^2-4(1)(-15)$

$\implies D=4+60$

$\implies D=64$

Thus

$D>0$

$\therefore$ Roots are real and distinct.

Simplify ${ \left[ i^{17}+{ \left( \dfrac { 1 }{ i} \right) }^{ 25 } \right] }^{ 3 }$

${ i }^{ 17 }={ i }^{ 16 }.i=i$

${ i }^{ 25 }={ i }^{ 24 }\times i=i$

$\cfrac { 1 }{ i } =-i$

${ \left( { i }^{ 17 }+{ \left( \cfrac { 1 }{ i } \right) }^{ 25 } \right) }^{ 3 }={ \left( i+\cfrac { 1 }{ i } \right) }^{ 3 }={ \left( i-i \right) }^{ 3 }={ 0 }^{ 3 }=0$

Represent the following complex numbers by point in Aegand's dialogram.
$(i)$ $0-2+3i$
$(ii)$ $2i$

1195373_1039422_ans_c129b38b01b44858aa5ae6064f267acb.jpg

The number of integral values that $k$ can take if $x^{2}+kx-k^{2}+5k > 0$ for all $x \in R$ , is

$\Rightarrow$ The given quadratic equation is

$x^2+kx-k^2+5k>0$

$\Rightarrow$ Here,

$a=1,\,b=k,\,c=-k^2+5k$

Since,

$x$ belongs to real number.

$\Rightarrow$

$b^2-4ac=0$

$\Rightarrow$

$(k)^2-4(1)(-k^2+5k)=0$

$\Rightarrow$

$k^2+4k^2-20k=0$

$\Rightarrow$

$5k^2-20k=0$

$\Rightarrow$

$k(5k-20)=0$

$\Rightarrow$

$k=0$ and

$5k-20=0$

$\therefore$

$k=0,\,4$

Find the real values of $x$ and $y$ , if
$(x+iy)(2-3i)=4+i$

$\left( {x + iy} \right)\left( {2 - 3i} \right) = 4 + i$

$2x - 3ix + 2iy + 3y = 4 + i$

$\left( {2x + 3y} \right) + i\left( {2y - 3x} \right) = 4 + i$

Comparing both sides of the above equation,

$2x + 3y = 4$ (1)

And,

$2y - 3x = 1$ (2)

From equation (1) and (2), by elimination method,

$x = \dfrac{5}{{13}}$ and

$y = \dfrac{{14}}{{13}}$

If $Arg\left( {\frac{{z + 1}}{{z - 1}}} \right) = \frac{\pi }{6}$ , then find the locus of z.

Given,

$\dfrac{z+1}{z-1}$

$=\dfrac{x+iy+1}{x+iy-1}$

$=\dfrac{(x+1)+iy}{(x-1)+iy}\times \dfrac{(x-1)-iy}{(x-1)-iy}$

$=\dfrac{\left(x+1\right)\left(x-1\right)+y^2-2iy}{\left(x-1\right)^2+y^2}$

$\therefore z=\dfrac{x^2-1+y^2}{\left(x-1\right)^2+y^2}-i\dfrac{2y}{\left(x-1\right)^2+y^2}$

$\theta =\tan ^{-1}\left ( \dfrac{b}{a} \right )=\tan ^{-1}\left ( \dfrac{2y}{x^2+y^2-1} \right )$

$\tan \dfrac{\pi}{6}=\dfrac{1}{\sqrt 3}$

$\dfrac{1}{\sqrt 3}=\dfrac{2y}{x^2+y^2-1}$

$x^2+y^2-1=2\sqrt 3 y$

$x^2+y^2+0x-2\sqrt 3 y-1=0$

$g=0,f=-\sqrt 3,c=-1$

$center=(-g,-f)=(0,\sqrt 3)$

$radius=\sqrt{g^2+f^2-c}=\sqrt{0^2+(-\sqrt{3})^2-(-1)}=2$

Locus of z is a circle with center

$(0,\sqrt 3)$ and radius

$2$

Let $z=x+iy$ and $|(2z-3)|=|(z-6)|$ then prove that $x^2+y^2=3$ .

Given,

$z=x+iy$ and

$|(2z-3)|=|(z-6)|$

or,

$\sqrt{(2x-3)^2+(2y)^2}=\sqrt{(x-6)^2+(y)^2}$

or,

${(2x-3)^2+(2y)^2}={(x-6)^2+(y)^2}$

or,

$4x^2-12x+9+4y^2=x^2-12+36+y^2$

or,

$3x^2+3y^2=27$

or,

$x^2+y^2=9$ .

Let $z=x+iy$ . If $\dfrac{z-1}{z+1}$ is purely imaginary then prove that $|z|=1$ .

Now,

$\dfrac{z-1}{z+1}$

$=\dfrac{x-1+iy}{x+1+iy}$

$=\dfrac{(x-1+iy)(x+1-iy)}{(x+1)^2+y^2}$

$=\dfrac{x^2-1+y^2+i\{y(x+1)-y(x-1)\}}{(x+1)^2+y^2}$

Now if this complex number is purely imaginary then we must have,

$x^2-1+y^2=0$

or,

$x^2+y^2=1$

or,

$|z|=1$ .

Evaluate : $\sqrt { - 16} + 3\sqrt { - 25} + \sqrt { - 36} - \sqrt { - 625}$

$\sqrt {-16}+3\sqrt {-25}+\sqrt {-36}-\sqrt {-625}$

$=4i +15i +6i -25i$

$=25i -25i$

$=0$

Find the arguments of each of the complex numbers.
$z = -1 - i\sqrt{3}$
$z = -\sqrt{3} + i$
$z=1+i\sqrt{3}$

1. Argument of

$z = -1 - i\sqrt{3}$

$=-\pi+\tan^{-1}\left|\dfrac{-\sqrt{3}}{-1}\right|=-\pi+\dfrac{\pi}{3}=-\dfrac{2\pi}{3}$ [ Since the point represented by the complex number lies in the third quadrant]

2. Argument of

$z = -\sqrt{3} + i$

$=\pi-\tan^{-1}\left|\dfrac{1}{-\sqrt{3}}\right|=\dfrac{5\pi}{6}$ [ Since the point represented by the complex number lies in the second quadrant]

3.Argument of

$1+i\sqrt{3}$

$=\tan^{-1}\left|\dfrac{\sqrt{3}}{1}\right|=\dfrac{\pi}{3}$ [ Since the point represented by the complex number lies in the first quadrant]

Find the principal argument of the complex number $\sin \dfrac{6\pi}{5} + i(1 + \cos\dfrac{6\pi}{5})$

$\rightarrow \sin { \cfrac { 6\pi }{ 5 } } +i\left( 1+\cos { \cfrac { 6\pi }{ 5 } } \right)$

$\rightarrow 2\sin { \cfrac { 3\pi }{ 5 } } \cos { \cfrac { 3\pi }{ 5 } } +2i\cos ^{ 2 }{ \cfrac { 3\pi }{ 5 } }$

$\rightarrow 2\cos { \cfrac { 3\pi }{ 5 } } \left( \sin { \cfrac { 3\pi }{ 5 } } +i\cos { \cfrac { 3\pi }{ 5 } } \right)$

$\rightarrow 2\cos { \cfrac { 3\pi }{ 5 } } \left( \sin { \left( \cfrac { \pi }{ 2 } +\cfrac { \pi }{ 10 } \right) } +i\cos { \left( \cfrac { \pi }{ 2 } +\cfrac { \pi }{ 10 } \right) } \right)$

$2\cos { \cfrac { 3\pi }{ 5 } } \left( \cos { \left( \cfrac { -\pi }{ 10 } \right) } +i\sin { \left( \cfrac { -\pi }{ 10 } \right) } \right)$

argument

$=-\cfrac { \pi }{ 10 }$

Find the modulus and amplitude for each of the following complex numbers.
i) $7 - 5i$ vi) $\sqrt{3} - i$
ii) $\sqrt{3} + \sqrt{2 i}$ vii) $3$
iii) $-8 + 15 i$ viii) $1 + i$
iv) $-3 (l - i)$ ix) $1 + i \sqrt{3}$
v) $-4 - 4i$ x) $(1 + 2i)^2 (1 - i)$

Complex number:

$z=x+iy$

Modulus:

$z=\sqrt{x^2+y^2}$

Amplitude:

$\theta = \tan^{-1}\left | \dfrac{b}{a} \right |$

(i)

$z=7-5i$

$\Rightarrow\left | z \right |=\sqrt{7^2+(-5)^2}=8.6$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{-5}{7} \right |=35.53$

(ii)

$z=\sqrt{3}+\sqrt{2}i$

$\Rightarrow\left | z \right |=\sqrt{\sqrt{3}^2+\sqrt{2}^2}=2.23$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{\sqrt{2}}{\sqrt{3}} \right |=39.23$

(iii)

$z=-8+15i$

$\Rightarrow\left | z \right |=\sqrt{(-8)^2+15^2}=17$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(15)}{-8} \right |=61.92$

(iv)

$z=-3+3i$

$\Rightarrow\left | z \right |=\sqrt{(-3)^2+3^2}=4.24$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(-3)}{3} \right |=45$

(v)

$z=-4-4i$

$\Rightarrow\left | z \right |=\sqrt{(-4)^2+(-4)^2}=5.65$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(-4)}{(-4)} \right |=45$

(vi)

$z=\sqrt{3}-i$

$\Rightarrow\left | z \right |=\sqrt{(\sqrt{3})^2+(-1)^2}=2$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(-1)}{(\sqrt{3})} \right |=30$

(vii)

$z=3+i(0)$

$\Rightarrow\left | z \right |=\sqrt{(3)^2+(0)^2}=3$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(0)}{(3)} \right |=0$

(viii)

$z=1+i$

$\Rightarrow\left | z \right |=\sqrt{(1)^2+(1)^2}=1.414$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(1)}{(1)} \right |=45$

(ix)

$z=1+\sqrt{3}i$

$\Rightarrow\left | z \right |=\sqrt{(1)^2+(\sqrt{3})^2}=2$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(\sqrt{3})}{(1)} \right |=60$

(x)

$(1+2i)^2(1-i)=(-3+4i)(1-i)=-6+7i$

$z=-6+7i$

$\Rightarrow\left | z \right |=\sqrt{(-6)^2+({7})^2}=2$

$\Rightarrow \theta =\tan^{-1}\left | \dfrac{(\sqrt{7})}{(-6)} \right |=49.39$

Solve the equation ${z}^{2}+|z|=0$ , where $z$ is a complex number.

Simplifying ${z^2} + \left| z \right| = 0$

${z^2} = - \left| z \right|$

$\left| {{z^2}} \right| = \left| z \right|$

${\left| z \right|^2} - \left| z \right| = 0$

$\left| z \right|\left( {\left| z \right| - 1} \right) = 0$

$\left| z \right| = 0,1$

If $\left| z \right| = 0$ , then $z = 0$ and if $\left| z \right| = 1$ , then,

${z^2} = - 1$

$z = \pm i$

Therefore, the solutions are $0,i, - i$ .

Find the modulus, argument and the principal argument of the complex numbers.
$-2(\cos 30^o+i\sin 30^o)$

For

$z=r(\cos(\theta)+i\sin(\theta))$

$|z| = |r|$

$\arg(z)=\theta$
Principal argument is the argument that is in the interval

$(-180^{o}, 180^{o}]$
Here,

$|z| = 2$
Argument =

$30^{o}$
Principal argument =

$30^{o}$

Let $a=\dfrac{1+i}{2}$ then prove that $a^6+a^4+a^2+1=0$ .

Given

$a=\dfrac{1+i}{2}$ .

Now

$a^2=\dfrac{(1+i)^2}{2}=\dfrac{2i}{2}=i$ .

This gives

$a^4=-1$ and

$a^6=-i$ .

Then

$a^6+a^4+a^2+1$

$=(-i)+(-1)+i+1$ [ Using the values of

$a^2, a^4,a^6$ ]

$=0$

Find the modulus, argument and the principal argument of the complex numbers.
$6(\cos 310^o-i\sin 310^o)$

For

$z=r(\cos(\theta)+i\sin(\theta))$

$|z| = |r|$

$\arg(z)=\theta$
Principal argument is the argument that is in the interval

$(-180^{o}, 180^{o}]$
Here,

$|z| = 6$
Argument =

$-310^{o}$
Principal argument =

$-310^{o} + 360^{o} = 50^{o}$

If $i{z^3} + {z^2} - z + i = 0$ , then find $\left| z \right|$ .

$iz^3+z^2−z+i=0$

$\implies iz^3+i^2z+z^2+i=0$

$\implies iz(z^2+i)+(z^2+i)=0$

$\implies (z^2+i)(iz+1)=0$

$\implies (z^2+i)i(z−i)=0$

$⇒ z^2 =−i$ or

$z =i$

$z^2 =e^{i(3π/2)}$ or

$z =e^{i(π/2)}$

$z =±e^{i(3π/4)}$ or

$z =e^{i(π/2)}$

We can see that all the roots have modulus as

$1$ .

So, |z| =1

Find the number of $a$ for which given equation have equal roots $x^2+3x-(a^2+a-2)=0$

$x^2+3x-(c^2+a-2)=0$

For roots to be equal,

$b^2+4ac=0$

$9+4(a^2+a-2)=0$

$\implies a^2+a-2=\cfrac{-9}{4}$

$\implies a^2+a+\cfrac{1}{4}=0$

$\implies 4a^2+4a+1=0$

$\implies 4a^2+2a+2a+1=0$

$\implies 2a(2a+1)+1(2a+1)=0$

$\implies a=\cfrac{-1}{2}$

If $\dfrac {a+3l}{2+ib}=1-1$ , show that $(5a-7b)=0$

Simplify $\dfrac{{a + 3i}}{{2 + ib}} = 1 - i$

$a + 3i = \left( {1 - i} \right)\left( {2 + ib} \right)$

$a + 3i = \left( {2 + b} \right) + i\left( {b - 2} \right)$

By comparing both sides,

$a = 2 + b$

$3 = b - 2$

$b = 5$

Put these values of $a$ and $b$ in left hand side of $5a - 7b = 0$ .

$5\left( {2 + b} \right) - 7\left( 5 \right)$

$= 5\left( {2 + \left( 5 \right)} \right) - 7\left( 5 \right)$

$= 0$

$= {\rm{RHS}}$

Find the modulus of $\dfrac{{\left( {1 + 3i} \right)\left( {2 - 5i} \right)}}{{\left( {2 - \sqrt 6 i} \right)\left( { - 3 + 2\sqrt 5 i} \right)}}$

modulus of

$\dfrac{(1+3i)(2-5i)}{(2-\sqrt6 i)(-3+2\sqrt5 i)}=\dfrac{|1+3i||2-5i|}{|2-\sqrt6 i||-3+2\sqrt5 i|}$

$=\dfrac{\sqrt10\sqrt29}{\sqrt10\sqrt29}$

$=\dfrac{1}{1}\\=1$

Convert the complex number $z=\dfrac { i-1 }{ \cos { \dfrac { \pi }{ 3 } +i } \sin { \dfrac { \pi }{ 3 } } }$ in the polar from and hence find the modulus and the argument of $x$ .

$i$ - 1 can be written as

$\sqrt2(i/\sqrt2 - 1/\sqrt2)$

$\sqrt2 (cos3\pi/4 - isin3\pi/4$ ) =

$\sqrt2 e^{i3\pi/4}$ -(i)

$cos\pi/3 + isin\pi/3$ in polar form can be written as

$e^{i\pi/3}$ -(ii)

Combining (i) and (ii),

z =

$\sqrt2(e^{\dfrac{5\pi}{12}})$

$\Rightarrow$ Modulus of z , |z| =

$\sqrt2$

and Principal argument of z =

$\dfrac{5\pi}{12}$

Find the modulus of $\dfrac{1+2i}{1-3i}$

Let

$z=\dfrac{1+2i}{1-3i}$

Rationalising, we get

$=\dfrac{1+2i}{1-3i}\times\dfrac{1+3i}{1+3i}$

$=\dfrac{1+3i+2i+6{i}^{2}}{1-9{i}^{2}}$

$=\dfrac{1+5i-6}{1+9}$

$=\dfrac{-5+5i}{10}$

$=\dfrac{-1}{2}+\dfrac{i}{2}$

$\therefore z=\dfrac{-1}{2}+\dfrac{i}{2}$

Modulus of

$z=\left|z\right|=\sqrt{{\left(\dfrac{-1}{2}\right)}^{2}+{\left(\dfrac{1}{2}\right)}^{2}}=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}=\dfrac{1}{\sqrt{2}}$

$\therefore$ Modulus of

$z=\dfrac{1}{\sqrt{2}}$

Solve: $\left(\dfrac{1}{1 - 2i} + \dfrac{3}{1 + i}\right)\left(\dfrac{3 + 4i}{2 - 4i}\right)$ .

$\left ( \dfrac{1}{1-2i}+\dfrac{3}{1+i} \right )\left ( \dfrac{3+4i}{2-4i} \right )$

$=\left ( \dfrac{1+i+3(1-2i)}{(1-2i)}\left ( \dfrac{3+4i}{2-4i} \right ) \right )$

$=\left ( \dfrac{1+i+3-6i}{1+i-2i-2i^{2}} \right )\left ( \dfrac{3+4i}{2-4i} \right )$

$=\left ( \dfrac{4-5i}{2-i} \right )\left ( \dfrac{3+4i}{2-4i} \right )$

$=\left ( \dfrac{12+16i-15i-20i^{2}}{4-8i-2i+4i^{2}} \right )$

$=\dfrac{i+32}{-10i}= \dfrac{-1}{10} \dfrac{-32}{10}$

$=\dfrac{-1}{10}+\dfrac{32i^{2}}{10i}$

$=\dfrac{-1}{10}+\dfrac{32i}{10}=\dfrac{32i-1}{10}$

If $Z=\cos \theta+i\sin \theta$ find the complex representation of $\dfrac {Z}{1-2Z{}}$ .

Solution:-

$Z = \cos{\theta} + i \sin{\theta}$

$\therefore \; \cfrac{Z}{1 - 2Z} = \cfrac{\cos{\theta} + i \sin{\theta}}{1 - 2 \left( \cos{\theta} + i \sin{\theta} \right)}$

$\Rightarrow \; = \cfrac{\cos{\theta} + i \sin{\theta}}{\left( 1 - 2 \cos{\theta} \right) - 2 i \sin{\theta}}$

Multiplying and dividing by the conjugate of denominator, we have

$\cfrac{Z}{1 - 2Z} = \cfrac{\cos{\theta} + i \sin{\theta}}{\left( 1 - 2 \cos{\theta} \right) - 2 i \sin{\theta}} \times \cfrac{\left( 1 - 2 \cos{\theta} \right) + 2 i \sin{\theta}}{\left( 1 - 2 \cos{\theta} \right) + 2 i \sin{\theta}}$

$\Rightarrow \; = \cfrac{ \left( \cos{\theta} + i \sin{\theta} \right) \left( \left( 1 - 2 \cos{\theta} \right) + 2 i \sin{\theta} \right)}{{\left( 1 - 2 \cos{\theta} \right)}^{2} - {\left( 2 i \sin{\theta} \right)}^{2}}$

$\Rightarrow \; = \cfrac{\cos{\theta} \left( 1 - 2 \cos{\theta} \right) - 2 \sin^{2}{\theta} + i \left( \sin{\theta} \left( 1 - 2 \cos{\theta} \right) + 2 \sin{\theta}\cos{\theta} \right)}{1 + 4 \cos^{2}{\theta} - 4 \cos{\theta} - \left( -4 \sin^{2}{\theta} \right)}$

$\Rightarrow \; = \cfrac{\cos{\theta} - 2 \cos^{2}{\theta} - 2 \sin^{2}{\theta} + i \left( \sin{\theta} - 2 \sin{\theta} \cos{\theta} + 2 \sin{\theta} \cos{\theta} \right)}{1 - 4 \cos{\theta} + 4 \cos^{2}{\theta} + 4 \sin^{2}{\theta}}$

$\Rightarrow \; = \cfrac{\cos{\theta} - 2 \left( \cos^{2}{\theta} + \sin^{2}{\theta} \right) + i \sin{\theta}}{1 - 4 \cos{\theta} + 4 \left( \cos^{2}{\theta} + \sin^{2}{\theta} \right)}$

$\Rightarrow \; = \cfrac{\left( \cos{\theta} - 2 \right) + i \sin{\theta}}{5 - 4 \cos{\theta}}$

Hence, the complex representation of

$\cfrac{Z}{1 - 2Z}$ will be

$\cfrac{\left( \cos{\theta} - 2 \right) + i \sin{\theta}}{5 - 4 \cos{\theta}}$ .

Solve:
$\sqrt[4]{-81}$ .

We are given ,

$a$ imaginary number.

Let

$Z = \sqrt [4] {-81}$

$Z= (-81)^{1/4}$

$Z^{4}= (-81)$

now, as we know,

$i^{2}= -1$

$Z^{4}= 81\ i^{2}$

$Z^{4}= \dfrac{81}{4} (2i)^{2}$

$Z^{4} = \dfrac{81}{4} (1+2 i+ i^{2})^{2}$

here

$1+ i^{2} =0$

$Z^{4} =\dfrac{81}{4} ((1+i)^{2})^{2}$

$Z = \sqrt [4] {\dfrac{81}{4} \times (1+i)^{4}}$

$Z= \left(\dfrac{3^{4}}{2^{4}} \times (1+i)^{4} \right)^{1/4}$

$Z= \dfrac{3}{\sqrt{2}} (1+i)$

So,

$\sqrt [4] {-81} = \dfrac{3}{\sqrt{2}} (1+i)$

For $z=x+iy$ find the real and imaginary part of $e^z$ .

For

$z=x+iy$ we have,

$e^z$

$=e^{x+iy}$

$=e^x(\cos y+i\sin y)$ .

So the real part and the imaginary parts are

$e^x \cos y$ and

$e^x \sin y$ .

Let ${x_1}$ and ${x_2}$ be two complex numbers such that $\left| {{x_1} + {x_2}} \right| = \left| {{x_1}} \right| + \left| {{x_2}} \right|$ . show that arg $\left( {{x_1}} \right) - \arg \left( {{x_2}} \right) = 0$

Let

$x_{1}=a_{1}+ib_{1}$

$x_{2}=a_{2}+ib_{2}$

$\left | x_{1} \right |+\left | x_{2} \right |=\left | x_{1}+x_{2} \right |$

$\sqrt{a_{1}^{2}+b_{1}^{2}}+\sqrt{a_{2}^{2}+b_{2}^{2}}=\sqrt{(a_{1}+a_{2})^{2}+(b_{1}+b_{2})^{2}}$

square both sides

$2\sqrt{(a_{1}^{2}+b_{1}^{2})(a_{2}^{2}+b_{2}^{2})}=2a_{1}a_{2}+2b_{1}b_{2}$

$\dfrac{b_{1}}{a_{1}}=\dfrac{b_{2}}{a_{2}}$

$arg(x_{1})=arg(x_{2})=\dfrac{b_{1}}{a_{1}}=\dfrac{b_{2}}{a_{2}}$

$arg(x_{1})-arg(x_{2})=0$

1112371_1072455_ans_1890deae41a54965b99e2586238fdddb.jpg

find the modulus and argument of the complex numbers.
$\frac{{2 + 14i}}{{{{\left( {2 - i} \right)}^2}}}$

Consider the given complex expression ,

$\dfrac{2+14i}{{{\left( 2-i \right)}^{2}}}$

$\dfrac{2+14i}{{{\left( 2-i \right)}^{2}}}=\dfrac{2+14i}{\left( 4+{{i}^{2}}-2i \right)}$

$\because {{i}^{2}}=-1$

$=\dfrac{2+14i}{4-1-4i}$

$=\dfrac{2+14i}{3-4i}$

Rationalize

$\dfrac{2+14i}{3-4i}\times \dfrac{3+4i}{3+4i}$

Now , $\dfrac{2+14i}{3-4i}\times \dfrac{3+4i}{3+4i}=\dfrac{6+50i+56{{i}^{2}}}{{{\left( 3 \right)}^{2}}-{{\left( 4i \right)}^{2}}}$

$\dfrac{6+50i+56{{i}^{2}}}{{{\left( 3 \right)}^{2}}-{{\left( 4i \right)}^{2}}}=\dfrac{6+50i+56{{i}^{2}}}{{{3}^{2}}-16{{i}^{2}}}$

$\dfrac{6+50i+56\left( -1 \right)}{9-16\left( -1 \right)}=\dfrac{6+50i-56}{9+16}$

$\dfrac{50i-50}{25}=i-1=-1+i$

Now,

$-1+i=$ $r(\cos \theta + i \sin \theta)$ …..(1)

$-1=r\cos \theta$ ….(2)

$1=r\sin \theta$ ……(3)

Now taking square and add equations (2) and (3), we get

$2={{r}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )$

$2={{r}^{2}}(1)$

$r=\sqrt{2}$

Now divided equation (3) by equation (2), we get

$\dfrac{\sin \theta }{\cos \theta }=-1$

$\tan \theta =-\tan \dfrac{\pi }{4}$

$\tan \theta =\tan \left( \pi -\dfrac{\pi }{4} \right)=\tan \dfrac{3\pi }{4}$

$\tan \theta =\tan \dfrac{3\pi }{4}$

$\theta =\dfrac{3\pi }{4}$

Put the value of $r$ and $\theta$ in equation (1), we get

$-1+i=\sqrt{2}\left( \cos \dfrac{3\pi }{4}i\sin \dfrac{3\pi }{4} \right)$

Hence, this is the required question.

For what values of $a$ , that the roots of the equation $x^2 + 2 (3a + 5) x + 2 (9a^2 + 25) = 0$ are complex

Roots of the quadratic equation

$Ax^{2}+Bx+C=0$ are complex if

$B^{2}-4AC<0$

For the given equation

$x^{2}+2(3a+5)x+2(9a^{2}+25)=0$

$B^{2}-4AC=4(3a+5)^{2}-4(2)(9a^{2}+25)$

$=4(9a^{2}+25+30a)-8(9a^{2}+25)$

$=-4(9a^{2}+25-30a)$

$=-4(3a-5)^{2}$

which will be zero if

$a=\dfrac{5}{3}$

or will be negative for any other value of

$a$

So the roots are real and equal if

$a=\dfrac{5}{3}$

and the roots are complex for any other value of

$a$

Find the modulus and argument of the complex numbers.
$\dfrac{{5 - i}}{{2 - 3i}}$

Consider the given complex number, $\dfrac{5-i}{2-3i}$ .

By rationalisation of given number .

$\dfrac{5-i}{2-3i}=\dfrac{5-i}{2-3i}\times \dfrac{2+3i}{2+3i}=\dfrac{10-13i-3{{i}^{2}}}{4-9{{i}^{2}}}$

$\because {{i}^{2}}=-1$

$\dfrac{10-13i-3{{i}^{2}}}{4-9{{i}^{2}}}=\dfrac{10-13i-3\left( -1 \right)}{4-9\left( -1 \right)}=\dfrac{13-13i}{13}=1-i$

$\dfrac{5-i}{2-3i}=1-i$

Now,

$1-i=r\left( \cos \theta -i\sin \theta \right)$ ……(1)

$1=\cos \theta$ …..(2)

$1=\sin \theta$ …..(3)

Taking square and add equation (1) and (2),we get

${{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{1}^{2}}+{{1}^{2}}$

${{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=2$

${{r}^{2}}=2$

$r=\sqrt{2}$

Equation (3) divided by equation (2), we get

$\tan \theta =1$

$\tan \theta =\tan \dfrac{\pi }{4}$

$\theta =\dfrac{\pi }{4}$

The value of $\theta$ and $r$ put in equation (1), we get

$1-i=\sqrt{2}\left( \cos \theta -i\sin \theta \right)$

This is the required answer.

If $\left| {\dfrac{{z - 5i}}{{z + 5i}} } \right|=1$ , prove that $z$ is real.

Let

$z=x+iy$ , where x and y are real numbers.

Given,

$\left| \dfrac { z-5i }{ z+5i } \right| =1 \Rightarrow |z-5i|=|z+5i|$

$\Rightarrow |x+iy-5i|=|x+iy+5i|\Rightarrow |x+i(y-5)|=|x+i(y+5)|$

$\Rightarrow\sqrt{x^2+(y-5)^2}=\sqrt{x^2+(y+5)^2}$

Squaring both sides,

$x^2+(y-5)^2=x^2+(y+5)^2\Rightarrow (y-5)^2=(y+5)^2$

This is true only if y = 0, as

$(0-5)^2=(0+5)^2$

$, z=x+iy=x+i\times 0=x$

Thus,

$z$ is real.

Find the modulus of the complex number $\dfrac{i}{1-i}$ .

Given the complex number is

$\dfrac{i}{1-i}$

$=\dfrac{i(1+i)}{1-i^2}$

$=\dfrac{-1+i}{2}$ .

Now the modulus of the complex number is

$\sqrt{\left(\dfrac{-1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2}$

$=\sqrt{\dfrac{1}{2}}$

$=\dfrac{1}{\sqrt{2}}$ .

Number of real roots of $2\sqrt{2x+1}=2x-1$

$2\sqrt { 2x+1 } =2x-1\\ \Rightarrow 4(2x+1)=4{ x }^{ 2 }-4x+1\quad (\text{squaring both sides})\\ \Rightarrow 4{ x }^{ 2 }-12x-3=0\\ \because \quad { b }^{ 2 }-4ac={ (-12) }^{ 2 }-4.4.(-3)=192>0\\ \therefore \text{ given equation has 2 distinct real roots}$

Comment on the nature of the roots of the Quadratic equation $3\sqrt{3}x^2+10x+\sqrt{3}=0$ .

$3\sqrt 3 x^2+10x+\sqrt 3=0$

$a=3\sqrt 3,b=10,c=\sqrt 3$

$\triangle=b^2-4ac\\ =100-4(3\sqrt 3)(\sqrt 3)\\ =100-36\\ =64\ge 0$

$\therefore\ \triangle\ge 0$

Hence, roots are real and distinct.

Find the conjugate and modulus of the following complex numbers.
$\overline{(2+3i)}+\overline{(5i-4i^3)}+\overline{(6i-7i^4)}$ .

1332381_1081784_ans_577543a1dba74ce984d6b17440ea6c1f.jpg

Show that the roots of the equation ${x}^{2}+2(3a+5)x+2(9{a}^{2}+25)=0$ are complex unless $a=\cfrac{5}{3}$

For Real roots

$\Rightarrow D \geq 0$

$\Rightarrow (2(3a+5))^{2}-4(1)(2(a_{a}^{2}+25))\geq 0$

$\Rightarrow 4(a_{a}^{2}+25+30a)-4(18a^{2}+50)\geq 0$

$\Rightarrow -9a^{2}+30a-25 \geq 0$

$\Rightarrow 9a^{2}-30a+25\leq 0$

$\Rightarrow (3a-5)^{2} \leq 0$

$\Rightarrow (3a-5)^{2}=0$

$\Rightarrow$ Only possible if

$a=\dfrac{5}{3}$

If $\left| Z \right| =2$ and $arg(Z)=\cfrac{\pi}{4}$ then write $Z$ .

$Z=\left | Z \right |\left ( \cos\theta +i\sin\theta \right )$

$=2\left ( \cos\dfrac{\pi }{4}+i \sin\dfrac{\pi }{4} \right )$

$=2\left ( \dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}} \right )$

$Z=\sqrt{2}\left ( 1+i \right )$

What is the nature of roots of quadratic equation $5{x^2} - 7x + 2 = 0?$

$5x^2-7x+2=0$

$a=5$

$b-7$

$c=2$

$b^2-4ac$

$=49-4\times 5\times 2$

$b^2-4ac=9$

$b^2-4ac > 0$

$\therefore$ The roots are real and distinct.

Find the modulus are arguments $z = -\sqrt{3} + i$

$z=-\sqrt{3}+i$

Modulus

$\left | z \right |=\sqrt{(-\sqrt{3})^2+1^2}$

$=\sqrt{3+1}$

$=2$

$\left | z \right |=2$

Argument

$\theta =\tan^{-1}\dfrac{b}{a}$

$=\tan^{-1}\dfrac{1}{-\sqrt{3}}$

$=-30$

$\therefore \theta = 180-30=150$

$\theta =150 \times \dfrac{\pi}{180}$

$\theta =\dfrac{5\pi}{6}$

If $z_1=6+i$ $z_2=3-4i$ then find $z_1z_2$

$z_1=6+i\\z_2=3-4i\\z_1z_2=(6+i)(3-4i)\\z_1z_2=18+3i-24i+4=22-21i$

Find the Modulas and argument of $\dfrac{1+i}{1-i}$

Writing both numerator and denominator in Euler notation:

$1+i = \sqrt{2}\Bigg(\dfrac{1}{\sqrt{2}} + i\dfrac{1}{\sqrt{2}}\Bigg)=\sqrt{2}e^{i\frac{\pi}{4}}$
Similarly,

$1-i=\sqrt{2}e^{-i\frac{\pi}{4}}$

$\therefore\dfrac{1+i}{1-i}=e^{i\pi/2}$

Modulus = 1, Argument =

$\pi/2$

If $z+m=88$ & $z-m=z$ , then $m=______$

1362423_1127647_ans_a4b30b16034c41a9b061b05f1c04dece.jpg

Which terms of the $8\vec {i},\ 7-4\vec {i},\ 6-2\vec {i}$ ,.... is purely imaginary

1356706_1128485_ans_db7a81c1dcc04f49a5a0bef841ae807e.jpg

If z = x + yi and $\dfrac{\left| z - 1 -i\right| + 4}{3 \left| z - 1 -i\right| -2} = 1$ , show that $x^2 + y^2 - 2x - 2y - 7 = 0$ .

$\dfrac{\left| z - 1 -i\right| + 4}{3 \left| z - 1 -i\right| -2} = 1$

Put

$z = x + yi$

$\dfrac{|x+iy-1-i|+4}{3|x+iy-1-i|-2} = 1$

$\dfrac{|(x-1)+i(y-1)|+4}{3|(x-1)+i(y-1)-2} = 1$

$\dfrac{\sqrt{(x-1)^{2}+(y-1)^{2}}+4}{3\sqrt{(x-1)^{2}+(y-1)^{2}}-2} = 1$

$\sqrt{(x-1)^{2}+(y+1)^{2}+4} = 3\sqrt{(x-1)^{2}+(y-1)^{2}}-2$

$2\sqrt{(x-1)^{2}+(y-1)^{2}} = 6$

$\sqrt{(x-1)^{2}+(y-1)^{2}} = 3$

$\sqrt{(x^{2}+1-2x+y^{2}+1-2y)} = 9$

$x^{2}+y^{2}-2x-2y -7 = 0$ Hence proved

1097434_1135114_ans_5cc4aa497a954170b9fa161e09922d8a.jpg

Find the values of $m$ for which the quadrilateral equation ${x}^{2}-m\left(2x-8\right)-15=0$ has
(i) equal roots
(ii) both roots positive.

Given eq. is;

$f(x)=x^2-2mx+8m-15=0$

1). For equal roots,

$D=0$

$\Rightarrow$

$D=b^2-4ac=0$

$\Rightarrow$

$4m^2-32m+60=0$

$\Rightarrow$

$m^2-8m+15=0$

$\Rightarrow$

$m=3,5$

2). For both roots positive,

$D>=0$ &

$f(0)>0$

$f(0)=8m-15>0$

$\Rightarrow$

$m>\dfrac{15}{8}$ ........(1).

$D>=0$

$\Rightarrow$

$m^2-8m+15>=0$

$\Rightarrow$

$(m-3)(m-5)>=0$ ..........(2).

From (1) & (2),

$\Rightarrow$

$m>=5$

$\Rightarrow$

$m\epsilon [5, \infty)$

if $z=r \cos \theta$ then find the value $|e^{iz}|.$

Given :

$z = r \cos \theta$

To find :

$|e^{iz}|$

Answer :

$z = r \cos \theta$

$iz = ir \cos \theta$

$e^{iz} = e^{ir \cos \theta}$

Taking modulus :

$|e^{iz} = |e^{ir \cos \theta} |$

$= 1$

(

$\because$ any term in the form

$|e^{i\alpha}| = 1$ )

If $a > 0,|z| =a$ , then find the real part of $\left ( \dfrac{z-a}{z+a} \right )$ .

Given

$|z| = a$ let

$z = x+iy \Rightarrow \sqrt{x^{2}+y^{2}} = a...(1)$

$\dfrac{z-a}{z+a} = \dfrac{(x-a)+iy}{(x+a)+iy} = \dfrac{[(x-a)+iy][(x+a)+iy(-1)]}{[(x+a)+iy][(x+a)-iy]}$ [rationalizing]

$= \dfrac{(x^{2}-a^{2})-iy(x-a)+iy(x+a)+y^{2}}{(x+a)^{2}+y^{2}}$

$= \dfrac{iy[(x+a)-(x-a)]}{y^{2}+(a+x)^{2}}$

$[\because x^{2}+y^{2} = a^{2}]$

$= \dfrac{2ayi}{y^{2}+(x+a)^{2}}$

$\therefore Re(\dfrac{z-a}{z+a}) = 0$

In the Argand plane ,the vector $OP$ ,where $O$ is the origin and $P$ represents the complex number $z=4-3i$ ,is turned in the clockwise sense through ${180^ \circ }$ and streched $3$ times. the complex number represented by the new vector is -----.

$|z|=5$ Let

$z_1$ be the new vector obtained by rotating z in the clockwise sense through 180^., therefore

$z_1=e^{−iπz}=(\cosπ−i\sinπ)$ ,i.e., z=−4+3i The unit vector in the direction of

$z_1$ is

$\rightarrow\dfrac{4}{5}+\dfrac{3}{5i}$ . Therefore required vector

$=3|z|(\dfrac{-4}{5}+\dfrac{3}{5i})=15(\dfrac{-4}{5}+\dfrac{3}{5i})=−12+9i$

Solve the equation $z^{2}=\overline {z}$

$z^{2}=z$

$(x+iy)^{2}=(x+iy)$

$x^{2}-y^{2}+2ixy=x+iy$

$x^{2}-y^{2}=x$

$x=\dfrac{1}{2}$

$\left(\dfrac{1}{4}\right)-y^{2}=\dfrac{1}{2}$

$-y^{2}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2+1}{4}=\dfrac{3}{4}\Rightarrow y=\dfrac{+\sqrt{3}i}{2}$

$z=\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2}$

If $g(x)=x^{4}-x^{3}\div x^{2}+3x-5$ , find $g(2+3i)$ ?

$g\left( x \right) =\dfrac { { x }^{ 4 }-{ x }^{ 3 } }{ { x }^{ 2 }+3x-5 } =\dfrac { { x }^{ 3 }\left( x-1 \right) }{ { x }^{ 2 }+3x-5 }$

${ \left( 2+3i \right) }^{ 3 }=8-27i+36i-54=-46+9i$

${ \left( 2+3i \right) }^{ 2 }=4-9+12i=-5+12i$

Now,

$g\left( 2+3i \right) =\dfrac { { \left( 2+3i \right) }^{ 3 }\left( \left( 2+3i \right) -1 \right) }{ { \left( 2+3i \right) }^{ 2 }+3\left( 2+3i \right) -5 }$

$=\dfrac { \left( -46+9i \right) \left( 1+3i \right) }{ \left( -5+12i \right) +\left( 6+9i \right) -5 } =\dfrac { -46-138i+9i-27 }{ -4+21i }$

$=\dfrac { -73-129i }{ -4+21i }$

If $1 + \dfrac{1}{i} = A + iB,$ then find the value of B.

1100456_1175933_ans_520c154431fc40eaaae73779d6a99ccd.jpg

Represent follow complex no. in polar form.
$z=-1+\sqrt3i$

$z=-1+\sqrt3i$

$z = r(cos\theta +i sin\theta ) \rightarrow$ polar from

$z = 2[-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i]$

on comparing,

$r=2.$

$cos\theta = -1/2 \Rightarrow \theta = \dfrac{2\pi }{3}$

$sin\theta = \dfrac{\sqrt{3}}{2} \Rightarrow \theta = \dfrac{\pi }{3}\Rightarrow \dfrac{2\pi }{3}$

$\therefore \theta = \dfrac{2\pi }{3}$

$\therefore r = 2$ and

$\theta = \dfrac{2\pi }{3}$

$\therefore z = 2[cos(\dfrac{2\pi }{3})+isin(\dfrac{2\pi }{3})]\rightarrow$ polar form.

Find the real and imaginary part of $\dfrac{{3 - 2i}}{{7 + 4i}}$

Let

$z_1=\dfrac{3-2i}{7+4i}$

Multiplying and dividing

$z_1$ by

$7-4i$

We get,

$z_1=\dfrac{\left(3-2i \right) \left(7-4i \right)}{ \left(7+4i \right) \left(7-4i \right)}$

$=\dfrac{21-12i-14i-8}{{7}^{2}+{4}^{2}}$

$=\dfrac{13-26i}{65}$

$z_1=\dfrac{1-2i}{5}$

$z_1=\dfrac{1}{5}+\dfrac{-2i}{5}$

Comparing with

$z=x+iy$

We get,

Real part

$=\dfrac{1}{5}$

Imaginary part

$=\dfrac{-2}{5}$

If the quadratic equation $(b^2+c^2)x^2-2(a+b)c x+(c^2+a^2)=0$ has equal roots. Then find a relation between $a,b,c$ .

Given, the quadratic equation

$(b^2+c^2)x^2-2(a+b)c x+(c^2+a^2)=0$ has equal roots.

Then we've the discriminant of the given quadratic equation must be

$0$ .

Then,

$4(a+b)^2 c^2-4(b^2+c^2)(a^2+c^2)=0$

or,

$(a^2+2ab+b^2)c^2-(b^2c^2+a^2b^2+c^4+a^2c^2)=0$

or,

$c^4-2abc^2+a^2b^2=0$

or,

$(c^2-ab)^2=0$

or,

$c^2=ab$ .

$a,c,b$ are in G.P.

Find the modulus of the complex number, $\sqrt{2}i+\sqrt{-2}i$ .

$\sqrt{2}i+\sqrt{2}i$

$=\sqrt{2}i+\sqrt{-1}\times \sqrt{2}\times i$

$=\sqrt{2}i+i\times \sqrt{2}\times i$

$=\sqrt{2}i-\sqrt{2}$

$=\sqrt{2}(-1+i)$ .

Modulus is

$\sqrt{2}\cdot \sqrt{(-1)^2+i^2}=\sqrt{2}\cdot \sqrt{2}=2$ .

1181589_1196702_ans_4dade393acdc4e208c877fc1791d20e3.jpg

Find the value of $\left(\dfrac{2i}{1+i}\right)^2$

Now,

$\left(\dfrac{2i}{1+i}\right)^2$

$=\dfrac{-4}{(1+i)^2}$

$[\because (1+i)^2=1+2i-1=2i]$

$=\dfrac{-4}{2i}=\dfrac{-2}{i}=(2i);$

$[\because i^2=-1\Rightarrow i=-yi]$

1181594_1196523_ans_6d3e6614d2664ef1875e432cccd60db4.jpg

The equation $(\cos\:p - 1) x^2 +(\cos \: p)x+\sin \:p=0$ has real roots, then 'P' can take any value in the interval

$\left(\cos{p}-1\right){x}^{2}+\left(\cos{p}\right)x+\left(\sin{p}\right)=0$

As roots are real, discriminant

$\ge 0$

$\triangle\ge 0\Rightarrow {b}^{2}-4ac\ge 0$

$\Rightarrow {\cos}^{2}{p}-4\left(\cos{p}-1\right)\sin{p}\ge 0$

$\Rightarrow {\cos}^{2}{p}-4\cos{p}\sin{p}+4\sin{p}\ge 0$

$\Rightarrow {\cos}^{2}{p}-4\cos{p}\sin{p}+4{\sin}^{2}{p}-4{\sin}^{2}{p}+4\sin{p}$ by adding and subtracting

$4{\sin}^{2}{p}$

$\Rightarrow {\left(\cos{p}-2\sin{p}\right)}^{2}+4\sin{p}\left(1-\sin{p}\right)\ge 0$

$\Rightarrow \cos{p}-2\sin{p}$ is always positive.

$-1\le \sin{p}\le 1$

$\Rightarrow 1\ge -\sin{p}\ge -1$

$\Rightarrow 1+1\ge 1-\sin{p}\ge 0$ by adding

$1$

$\Rightarrow 2\ge 1-\sin{p}\ge 0$

$\sin{p}\ge 0$

$\sin{p}$ is positive and is in first and third quadrant.

$\therefore p\in\left(0,\pi\right)$ in first and second quadrant in anticlockwise direction.

If $z = 2 - i \sqrt 7$ , then show that $3z^3 - 4z^2 + z + 88 = 0.$

1303637_1183968_ans_bc311f3b1b7f479ebef8345acd95083c.JPG

Represent the complex number $2+5i,2-5i,-2+5i$ and $-2-5i$ in Argand's diagram.

1233819_1187047_ans_969b8e0fff7d4755baa96de2372b853a.png

If $z=\sqrt{20i-21}+\sqrt{21+20i}$ then principal value of arg z can not be

$\sqrt{20i+21}=\sqrt{21+20i} =\sqrt{25-4+20i}$

$=\sqrt{{\left(5\right)}^{2}+{\left(2i\right)}^{2}+2\times5\times 2i}$

$=\sqrt{{\left(5+2i\right)}^{2}}$

$\Rightarrow 20i+21 = {\left(5+2i\right)}^{2}$

$\sqrt{20i-21}=\sqrt{4-25+20i}$

$=\sqrt{{\left(2\right)}^{2}+{\left(5i\right)}^{2}+20i}$

$=\sqrt{{\left(2-5i\right)}^{2}}$

$\Rightarrow 20i-21={\left(2-5i\right)}^{2}$

$\Rightarrow z=\pm \left(5+2i\right) \pm \left(2-5i\right)$

$\Rightarrow z=5+2i+25i,5+2i-2-5i,-5-2i+2+5i,-5-2i-2-5i$

$\therefore z=7+7i,3-3i,-3+3i, -7-7i$

$=7\left(1+i\right),3\left(1-i\right) 3\left(-1+i\right),7\left(-1-i\right)$

$\Rightarrow arg\left(z\right)=\dfrac{\pi}{4}, \dfrac{-\pi}{4}, \dfrac{3\pi}{4}, \dfrac{-3\pi}{4}$

$\therefore \arg{z}$ cannot be more than

$\dfrac{3\pi}{4}$ and less than

$\dfrac{-\pi}{4}$

If $\alpha , \beta$ are the roots of $x ^ { 2 } - 8 x + A = 0$ and $\gamma , \delta$ are the roots of $x ^ { 2 } - 72 x + B = 0 ,$ if $\alpha < \beta < \gamma < \delta$ are in GP, then find the value of $A + B .$

Given eq. is

$x^2-8x +A=0$ ...(1)

Since

$\alpha ,\beta$ are roots of eq. (1)

sum of roots

$\alpha +\beta =\frac {8}{1}=8$

Product of roots

$\alpha \cdot \beta =\frac {A}{1}=A$

and

$x^2 - 72 x +B =0$ ...(2)

Since

$\gamma ,\delta$ are roots of eq. (2)

$\gamma +\delta =72, \,\,\,\,\,\,\,\,\gamma .\delta =B$

Since

$\alpha ,\beta ,\gamma ,\delta$ are in G.P.

Let common ratio be

$\pi$

$\therefore \beta = d\cdot \pi, \gamma =d\cdot \pi^2, \delta = d\cdot \pi^3$

$\alpha +\beta =8 \Rightarrow \alpha +d\cdot \pi=8\Rightarrow \alpha (1+\pi)=8\Rightarrow 1+\pi=\frac {8}{\pi}$

$\alpha \cdot\beta = A \Rightarrow \alpha \cdot \alpha \pi = A \Rightarrow \alpha ^2\cdot \pi = A$

$\gamma +\delta = 72 \Rightarrow \alpha \cdot \pi^2+alpha \cdot \pi^3=72\Rightarrow alpha \cdot \pi^2 (1+\pi)=72$

$\Rightarrow alpha \cdot \pi^2 \frac {8}{2}=72\Rightarrow \pi =3$

$\gamma \cdot \delta =B \Rightarrow \alpha \cdot \pi^2\cdot \alpha \cdot \pi^3=B$

$= \alpha^2 \cdot \pi \cdot \pi^4 =B$

$=A\cdot 81= B$

$\therefore A:B = 81:1$

$\therefore A+B = 82$

Let $x-iy=\sqrt{\dfrac{a-ib}{c-id}}$ then prove that $(x^2+y^2)^2=\dfrac{a^2+b^2}{c^2+d^2}$ .

Given,

$x-iy=\sqrt{\dfrac{a-ib}{c-id}}$

Now squaring both sides we get,

$(x-iy)^2=\dfrac{a-ib}{c-id}$

Now taking modulus both sides we get,

$|x-iy|^2=\left|\dfrac{a-ib}{c-id}\right|$

or,

$(x^2+y^2)=\sqrt{\dfrac{a^2+b^2}{c^2+d^2}}$

or,

$(x^2+y^2)^2=\dfrac{a^2+b^2}{c^2+d^2}$ .

If $a,b,c\ \in\ R^{+}$ and $2b=a+c$ , then check the nature of roots of equation $ax^{2}+2bx+c=0$ .

$a,b,c \, \in\, R^+$

and

$2b=a+c ------(1)$

equation,

$ax^{2}+2bx+c=0$

Nature of roots

$D=B^{2}-4AC$

$=(2b)^{2}-4ac$

$=4b^{2}-4ac$

$=(a+c)^{2}-4ac$

$=a^{2}+c^{2}+2ax-4ac$

$=a^{2}+c^{2}-2ac$

$=(a-c)^{2} > 0$

then, real roots

Hence, this is the answer.

If $\pi/2\ and\ \pi/4$ are respectively the arguments nof $Z_1 and\ \overline{Z_2}$ , what is the value of arg $(z_1/z_2)$

$Arg(z_1)=\dfrac{\pi}{2}$

$Arg(\bar{z_2})=\dfrac{\pi}{4}$

$Arg(z_2)=-Arg(\bar{z_2})=-\dfrac{\pi}{4}$

$Arg(z_1/z_2)=Arg(z_1)-Arg(z_2)$

$=\dfrac{\pi}{2}-\left(-\pi/4\right)$

$=3\pi/4$ .

1172957_1270264_ans_8cb6dc2e2e544dfe8836350b66b9b061.jpeg

$x^{2}+6x+9=0$
$x=?$

$x^2+6x+9=0$

$x^2+3x+3x+9=0$

$x(x+3)+3(x+3)=0$

$(x+3)(x+3)=0$

$\therefore \ x=-3, -3$

If the discriminant of $3x^2+2x+a=0$ is double the discriminant of $x^2-4x+2$ , then find the value of a.

Consider the given equations.

$3x^2+2x+a=0$

$....... (1)$

$x^2-4x+2=0$

$........ (2)$

Since,

$D_1=2D_2$

Therefore,

$b_1^2-4a_1 c_1=2(b_2^2-4a_2 c_2)$

$2^2-4\times 3\times a=2((-4)^2-4\times 1 \times 2)$

$4-12a=2(16-8)$

$4-12a=16$

$-12a=12$

$a=-1$

Hence, this is the answer.

Discriminant of $-x^{2}+\frac{1}{2}x+\frac{1}{2}=0$ is

$-{x}^{2}+\dfrac{1}{2}x+\dfrac{1}{2}=0$

$a=1 , b=\dfrac{1}{2}, c=\dfrac{1}{2}$

$D={ b }^{ 2 }-4ac={ \left( { \dfrac { 1 }{ 2 } } \right) }^{ 2 }-4(-1)\left( \dfrac { 1 }{ 2 } \right)$

$=\dfrac{1}{4}+2=\dfrac{1+8}{4}=\dfrac{9}{4}$

$\therefore$ The diccriminant is

$\dfrac{9}{4}$

$\sqrt { D } =\sqrt { \dfrac { 9 }{ 4 }} =\pm \dfrac{3}{2}$

If $z _ { 1 } = 9 + 5 i$ and $z _ { 2 } = 3 + 5 i$ and if $arg \left( \dfrac { z - z _ { 1 } } { z - z _ { 2 } } \right) = \dfrac { \pi } { 4 }$ then $| z - 6 - 8 i | = 3 \sqrt { 2 }.$

Given :

${ z }_{ 1 }=9+5i$

${ z }_{ 2 }=3+5i$

arg

$\left( \dfrac { z-{ z }_{ 1 } }{ z-{ z }_{ 2 } } \right) =\dfrac { \pi }{ 4 }$

let

$z=x+iy$

$\therefore$

$\dfrac { z-{ z }_{ 1 } }{ z-{ z }_{ 2 } } =\dfrac { \left( x-9 \right) +i\left( y-5 \right) }{ \left( x-3 \right) +i\left( y-5 \right) } =\dfrac { \left[ \left( x-9 \right) +i\left( y-5 \right) \right] \left[ \left( x-3 \right) -i\left( y-5 \right) \right] }{ { \left( x-3 \right) }^{ 2 }+{ \left( y-5 \right) }^{ 2 } }$

$=\dfrac { \left( x-9 \right) \left( x-3 \right) +\left( y-9 \right) \left( y-5 \right) +i\left( x-3 \right) \left( y-5 \right) -i\left( x-9 \right) \left( y-5 \right) }{ { \left( x-3 \right) }^{ 2 }+{ \left( y-5 \right) }^{ 2 } }$

$\therefore$ arg

$\left( \dfrac { z-{ z }_{ 1 } }{ z-{ z }_{ 2 } } \right) ={ \tan }^{ -1 }\dfrac { Im\left( \dfrac { z-{ z }_{ 1 } }{ z-{ z }_{ 2 } } \right) }{ Re\left( \dfrac { z-{ z }_{ 1 } }{ z-{ z }_{ 2 } } \right) } =\dfrac { \pi }{ 4 }$

$\Rightarrow \dfrac { \left( x-3 \right) \left( y-5 \right) -\left( x-9 \right) \left( y-5 \right) }{ \left( x-9 \right) \left( x-3 \right) +\left( y-5 \right) \left( y-5 \right) }$

$=\tan\dfrac { \pi }{ 4 }$

$\Rightarrow \dfrac { \left( xy-5x-3y+15 \right) -\left( xy-5x-9y+45 \right) }{ \left( { x }^{ 2 }-12x+27 \right) +\left( { y }^{ 2 }-10y+25 \right) } =1$

$\Rightarrow \dfrac { 6y-30 }{ { x }^{ 2 }+{ y }^{ 2 }-12x-10y+52 } =1$

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }-12x-16y+82=0$

$\Rightarrow { \left( x-6 \right) }^{ 2 }+{ \left( y-8 \right) }^{ 2 }=18$

$\Rightarrow { \left| z-6-8i \right| }^{ 2 }={ \left( 3\sqrt { 2 } \right) }^{ 2 }$

$\therefore$

$\left| z-6-8i \right| =3\sqrt { 2 }$

Hence proved

Express: $Z=\dfrac {i-1}{\cos \dfrac {\pi}{3}+i\sin \dfrac {\pi}{3}}$ in polar form.

$Z = \frac{i-1}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}$

$= \frac{i -1}{(\frac{1}{2}+i\frac{\sqrt{3}}{2})} \times \frac{(\frac{1}{2}-\frac{i\sqrt{3}}{2})}{(\frac{1}{2}-\frac{\sqrt{3}i}{2}}$

$= \frac{\frac{i}{2}+\frac{\sqrt{3}}{2}-\frac{1}{2}+\frac{\sqrt{3}i}{2}}{\frac{1}{4}+\frac{3}{4}}$

$= (\frac{\sqrt{3}}{2}-\frac{1}{2})+i(\frac{1}{2}+\frac{\sqrt{3}}{2})$

polor from

$z = r (cos\theta +isin\theta )$

$rcos\theta = \frac{\sqrt{3}-1}{2}, rsin\theta = \frac{\sqrt{3}+1}{2}.$

squaring and adding

$\therefore r^{2} = \frac{3-2\sqrt{3}+1+3+2\sqrt{3}+1}{4} = 2$

$\therefore r =\sqrt{2}$

$cos\theta = \frac{\sqrt{3}-1}{2\sqrt{2}} = \frac{\sqrt{3}}{2}\times \frac{1}{\sqrt{2}}-\frac{1}{2}\times \frac{1}{\sqrt{2}}$

$=cos \frac{\pi }{6}. cos(\frac{\pi }{4}) - sin\frac{\pi }{6}.sin\frac{\pi }{4}$

$cos\theta = cos(\frac{\pi }{6}+\frac{\pi }{4})$

$\therefore \theta = \frac{5\pi }{12} (cos(A+B)) =cosA.cosB - sinA sinB)$

Hence, required poloar from

$z = \sqrt{2} (cos \frac{5\pi }{12}+isin \frac{5\pi }{12} )$

1222731_1297779_ans_36f72e74579d4befbf635fea2dad7f66.jpg

Simplify:
$\left(\dfrac{2i}{1+i}\right)^2$

Now,

$\left(\dfrac{2i}{1+i}\right)^2$

$=\left(\dfrac{-4}{2i}\right)$ [ Since

$i^2=-1$ and

$(1+i)^2=2i$ ]

$=2i$ .

If $a+ib=\dfrac{p+i}{2q-i}$ , prove that $a^2+b^2=\dfrac{(p^2+1)}{4q^2+1}$

$a+ib = \dfrac{p+i}{2q-i}$

Taking modulus on both sides

$\left | a+ib \right | = \left | \dfrac{p+i}{2q-i} \right |$

$a^{2}+b^{2} = \dfrac{p^{2}+1}{4q^{2}+1}$

Simplify and hence find the modulus and argument of $\frac{{3 + 2i}}{{2 - 5i}} + \frac{{3 - 2i}}{{2 + 5i}}$ and $\frac{{(3 + i)(3 - i)}}{{2 + i}}$ .

1212249_1293641_ans_392e7306d1e24646b6508b9b0d9a487a.JPG

Simplify : $\dfrac {(\cos \theta+i\sin \theta)^{4}}{(\cos \theta+i\sin \theta)^{5}}$

$\dfrac{(\cos\theta +i\sin\theta)4}{(\cos\theta +i\sin\theta)5}=\dfrac{1}{(\cos\theta +i\sin\theta)}$

$\Rightarrow \dfrac{1(\cos\theta -i\sin\theta)}{(\cos\theta +i\sin\theta)(\cos\theta -i\sin\theta)}$

$\Rightarrow\dfrac{\cos\theta -i\sin\theta}{(\cos^2\theta -i\sin^2\theta)}=\dfrac{\cos\theta -i\sin\theta}{\cos^2\theta +\sin^2\theta)}$

$=\cos\theta -i\sin\theta$ .

1202697_1301693_ans_0e2988f529d24fe3a1c619b8718ffb05.jpg

Represent the complex number $Z = 1 + i$ , $Z = -1 + i$ in the Argand's diagram and find their arguments.

1212244_1293580_ans_66f03f2961154bf2bcd9f66c0e7c9e3c.JPG

For what value of $k,\left( 4-k \right) { x }^{ 2 }+\left( 2k+4 \right) x+\left( 8k+1 \right) =0$ is a perfect square:

$\left( 4 - k \right) {x}^{2} + \left( 2k + 4 \right) x + 8k + 1 = 0$

Here,

$a = 4 + k, \quad b = 2k + 4, \quad c = 8k + 1$

The given equation will be a perfect square, i.e., real and equal roots if

$D = 0$

${b}^{2} - 4ac = 0$

${\left( 2k + 4 \right)}^{2} - 4 \left( 4 - k \right) \left( 8k + 1 \right) = 0$

$4 {k}^{2} + 16 + 16k - 4 \left( 32k - 8 {k}^{2} + 4 - k \right) = 0$

${k}^{2} + 4 + 4k - 32k + 8 {k}^{2} - 4 + k = 0$

$9 {k}^{2} - 27k = 0$

${k}^{2} - 3k = 0$

$k \left( k - 3 \right) = 0$

$k = 0$ or

$k = 3$

Hence for

$k = 0$ or

$k = 3$ , the given equation will be a perfect square.

Locate the complex number $z$ such that $\log_{\cos\dfrac{\pi}{6}}\dfrac{|z-2|+5}{4|z-2|-4}<2$

Let

$\left| z-2 \right| =x$

So,

$\log _{ \cfrac { \sqrt { 3 } }{ 2 } }{ \left( \cfrac { x+5 }{ 4x-4 } \right) } <2$

So,

$\cfrac { x+5 }{ 4x-4 } <{ \left( \cfrac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }$

$\cfrac { x+5 }{ 4x-4 } <\cfrac { 3 }{ 4 }$

$\cfrac { x+5 }{ x-1 } <\cfrac { 3 }{ 1 }$

$x+5<3x-3$

$8<2x$

So,

$x>4$

So,

$\left| z-2 \right| >4$

Thus, complex number z is region outside of a circle with radius

$4$ and center

$\left( 2,0 \right)$

Express each of the complex number given in the Exercises $1$ to $10$ in the form $a+ib$ .
$(5i)\left(-\dfrac{3}{5}i\right)$

$i=\sqrt{-1}\Rightarrow i^2=-1$

$(5i)\left ( -\dfrac{3}{5}i \right )$

$=5\times \left ( -\dfrac{3}{5} \right )\times i^2$

$=-3(-1)$

$=3+i\times 0$

Convert the given complex number in polar form: $- 1 - i$ .

Given complex number is

$-1-i$

Let

$r\cos \theta =-1$ and

$r\sin \theta =-1$

On squaring and adding, we obtain

$r^2\cos^2 \theta +r^2 \sin^2 \theta =(-1)^2+(-1)^2$

$\Rightarrow r^2 (\cos^2 \theta +\sin^2 \theta )=1+1$

$\Rightarrow r^2=2$

$\Rightarrow r=\sqrt 2$ [ Conventionally,

$r > 0$ ]

$\therefore \sqrt 2 \cos \theta =-1$ and

$\sqrt 2 \sin \theta =-1$

$\Rightarrow \cos \theta =-\dfrac {1}{\sqrt 2}$ and

$\sin \theta =-\dfrac {1}{\sqrt 2}$

$\theta =-\left( \pi -\dfrac {\pi}{4}\right) =-\dfrac {3\pi}{4}$ [ As

$\theta$ lies in the III quadrant ]

$\therefore -1-i=r\cos \theta +ir \sin \theta \\=\sqrt 2 \cos \dfrac {-3\pi}{4}+i\sqrt 2 \sin \dfrac {-3\pi}{4}$

$=\sqrt 2 \left( \cos \dfrac {-3\pi}{4}+i\sin \dfrac {-3\pi}{4}\right)$

This is the required polar form.

Express :
$\dfrac{(a+ib)^{2}}{a-ib}-\dfrac{(a-ib)^{2}}{a+ib}$
in the forum $x+iy$

$\cfrac{{\left( a + ib \right)}^{2}}{a - ib} - \cfrac{{\left( a - ib \right)}^{2}}{a + ib}$

$= \cfrac{{\left( a + ib \right)}^{3} - {\left( a - ib \right)}^{3}}{{a}^{2} + {b}^{2}}$

$= \cfrac{{a}^{3} - {b}^{3}i + 3aib \left( a + ib \right) - {a}^{3} + {b}^{3} i + 3abi \left( a - ib \right)}{{a}^{2} + {b}^{2}}$

$= \cfrac{3abi \left( a + ib + a - ib \right)}{{a}^{2} + {b}^{2}}$

$= \cfrac{6{a}^{2}bi}{{a}^{2} + {b}^{2}}$

$= 0 + \cfrac{6 {a}^{2}b}{{a}^{2} + {b}^{2}} i$

Here

$x = 0, \quad y = \cfrac{6 {a}^{2} b}{{a}^{2} + {b}^{2}}$

Find the value of P so that the quadratic equation px(x-3)+9=0 has two equal roots.

Px(x-3) + 9 = 0

px2-3px+9=0

a=p , b=-3p , c=9

b2-4ac= (-3p)2-4(p)(9)

=9p2-36p

for having equal roots

b2-4ac = 9p2-36p = 0

9p2-36p = 0

9p(p-4) = 0

9p=0 (or) p-4=0

p=0(rejected) or p=4

If $y=\log { \left( \dfrac { \sqrt { \left( x+1 \right) } -1 }{ \sqrt { \left( x+1 \right) } +1 } \right) } +\dfrac { \sqrt { x } }{ \sqrt { \left( x+1 \right) } }$ the by using substitution $x=\tan^{2}\theta,y$ reduces to

$y=\log{\left(\dfrac{\sqrt{x+1}-1}{\sqrt{x+1}-1}\right)}+\dfrac{\sqrt{x}}{\sqrt{x+1}}$

Given:

$x={\tan}^{2}{\theta}$

$y=\log{\left(\dfrac{\sqrt{{\tan}^{2}{\theta}+1}-1}{\sqrt{{\tan}^{2}{\theta}+1}-1}\right)}+\dfrac{\sqrt{{\tan}^{2}{\theta}}}{\sqrt{{\tan}^{2}{\theta}+1}}$

We know that

$1+{\tan}^{2}{\theta}={\sec}^{2}{\theta}$

$y=\log{\left(\dfrac{\sqrt{{\sec}^{2}{\theta}}-1}{\sqrt{{\sec}^{2}{\theta}}-1}\right)}+\dfrac{\sqrt{{\tan}^{2}{\theta}}}{\sqrt{{\sec}^{2}{\theta}}}$

$y=\log{\left(\dfrac{\sec{\theta}-1}{\sec{\theta}+1}\right)}+\dfrac{\tan{\theta}}{\sec{\theta}}$

$y=\log{\left(\dfrac{1-\cos{\theta}}{1+\cos{\theta}}\right)}+\dfrac{\dfrac{\sin{\theta}}{\cos{\theta}}}{\dfrac{1}{\cos{\theta}}}$

We know that

$1+\cos{\theta}=2{\cos}^{2}{\dfrac{\theta}{2}}$ and

$1-\cos{\theta}=2{\sin}^{2}{\dfrac{\theta}{2}}$

$y=\log{\dfrac{{\sin}^{2}{\dfrac{\theta}{2}}}{{\cos}^{2}{\dfrac{\theta}{2}}}}+\sin{\theta}$

$\therefore y=2\log{\tan{\dfrac{\theta}{2}}}+\sin{\theta}$

Find the arguments of ${z}_{1}=5+5i,\,{z}_{2}=-4+4i,\,{z}_{3}=-3-3i$ and ${z}_{4}=2-2i,$ where $i=\sqrt{-1}$

Since

${z}_{1},\,{z}_{2},\,{z}_{3}$ and

${z}_{4}$ lies in

$I,\,II,\,III$ and

$IV$ quadrants respectively.The arguments are given by

1] arg

${\left({z}_{1}\right)}={\tan}^{-1}{\left|\dfrac{5}{5}\right|}={\tan}^{-1}{1}=\dfrac{\pi}{4}$

2] arg

${\left({z}_{2}\right)}=\pi-{\tan}^{-1}{\left|\dfrac{4}{-4}\right|}$

$=\pi-{\tan}^{-1}{1}=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$

3] arg

${\left({z}_{3}\right)}=\pi+{\tan}^{-1}{\left|\dfrac{-3}{-3}\right|}$

$=\pi+{\tan}^{-1}{1}=\pi+\dfrac{\pi}{4}=\dfrac{5\pi}{4}$

and

4] arg

${\left({z}_{4}\right)}=2\pi-{\tan}^{-1}{\left|\dfrac{-2}{2}\right|}$

$=2\pi-{\tan}^{-1}{1}=2\pi-\dfrac{\pi}{4}=\dfrac{7\pi}{4}$

Find the condition on the complex constants $\alpha ,\beta$ if ${ z }^{ 2 }+\alpha z+\beta =0$ has two distinct roots on the line Re(z)= 1.

Let the roots be

$x+iy$

$\&$

$x-iy$

$\Rightarrow Re\left( z\right) =1$

$\Rightarrow x=1$

Sum of roots

$=-\alpha$

$=2x$

$\Rightarrow \alpha =-2$

Product of roots

$=x^2+y^2$

$=1+y^2=\beta$

$\Rightarrow \beta \in \left( 1,\infty \right)$

Hence, solved.

Find the nature of the roots of the quadratic equation $4x^{2}-5x+3=0$ .

$4x^{2}-5x+3=0$

Comparing with

$ax^{2}+bx+c=0$ , we get,

$a=4, b=-5, c=3$ .

Therefore,

$D=b^{2}-4ac=25-4(4)(3)=25-48=-23 < 0$ .

Hence, the given equation has no real roots.

Let $z$ and $\omega$ be the complex numbers.If $Rs\left(z\right)=\left|z-2\right|$ , $Re\left(\omega\right)=\left|\omega-2\right|$ and $arg\left(z-\omega\right)=\dfrac{\pi}{3},$ find the value of $Im\left(z+\omega\right)$

Let

$z=x+iy,\,\,x,y\in \,R$ and

$i=\sqrt{-1}$

$\because\,Re\left(z\right)=\left|z-2\right|$

$\Rightarrow\,x=\sqrt{{\left(x-2\right)}^{2}+{y}^{2}}$

$\Rightarrow\,{x}^{2}={\left(x-2\right)}^{2}+{y}^{2}$

$\Rightarrow\,{x}^{2}={x}^{2}-4x+4+{y}^{2}$

$\Rightarrow\,{y}^{2}=4\left(x-1\right)$

$\therefore\,z=1+{t}^{2}+2ti$ , parametric form and let

$\omega=p+iq$

Similarly,

$\omega=1+{s}^{2}+2si$

$\therefore\,z-\omega=\left({t}^{2}-{s}^{2}\right)+2i\left(t-s\right)$

$\Rightarrow\,arg{\left(z-\omega\right)}=\dfrac{\pi}{3}$

$\therefore\,{\tan}^{-1}{\left(\dfrac{2}{t+s}\right)}=\dfrac{\pi}{3}$

$\Rightarrow\,\dfrac{2}{t+s}=\sqrt{3}$

Now,

$z+\omega=2+{t}^{2}+{s}^{2}+2i\left(t+s\right)$

$\therefore\,Im{\left(z+\omega\right)}=2\left(t+s\right)=\dfrac{4}{\sqrt{3}}$

Explain the nature of roots of a quadratic equation $x^2-x-2=0$ .

$\textbf{Find the nature of the roots of the given quadratic equation by using discriminant condition.}$

$\text{Given,}$

$\Rightarrow x^{2}-x-2 = 0$

$\text{The give equation is in the form of, }ax^{2}+bx+c = 0$

$\text{So,}$

$\Rightarrow a = 1$

$\Rightarrow b = -1$

$\Rightarrow c = -2$

$\text{Finding the discriminant of the equation,}$

$\Rightarrow D = b^{2} - 4ac$

$\text{Putting the values of a, b and c,}$

$\Rightarrow D = (-1)^{2}-4(1)(-2)$

$\Rightarrow D = 9$

$\Rightarrow D>0$

$\text{Since, D is positive, so, the equation has two distinct real roots.}$

$\textbf{Hence, the roots of the quadratic equation, }\mathbf{x^{2}-x-2 = 0}\textbf{ are real in nature.}$

Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational ? Justify your answer.

Yes, for this Discriminant should not be a perfect square.

Consider the quadratic equation

$2x^{2}+x-4=0$ , which has rational coefficients.

The roots of the given quadratic equation are

$\dfrac{-1+\sqrt{33}}{4} and \dfrac{-1-\sqrt{33}}{4}$ , which are irrational.

Find the value of k for which the given quadratic equation $x^{2}-4 x+k=0$ distinct real roots.

Equation is

$x^2-4x+k=0$

Condition for distinct real roots

$b^2>4ac\\4^2>4k\\k<4$

Find the nature of the roots of the quadratic equation $(b-c)x^2+2(c-a)x+(a-b)=0$ .

$(b-c)x^2+2(c-a)x+(a-b)=0$

$D=b^2-4ac=(c-a)^2-(b-c)(a-b)$

$c^2+a^2-2ac-ab+b^2+ac-bc$

$=a^2+b^2+c^2-ac-bc-ab$

$=2(a^2+b^2+c^2-ab-bc-ca)$

So, the roots are real and may or may not be equal.

Simplify: $\dfrac{1}{i+1} + \dfrac{1}{i-1}$

Let

$Z=\dfrac{1}{i+1}+\dfrac{1}{i-1}$

$=\dfrac{i-1+i+1}{{i}^{2}-1}$

$=\dfrac{2i}{-1-1}$

$=\dfrac{-2i}{2}$

$Z=-i$

Simplify: $\dfrac{\sqrt{2+3i}}{\sqrt{2-3i}}$

Let

$Z=\dfrac{\sqrt{2+3i}}{\sqrt{2-3i}}$

$=\dfrac{\sqrt{2+3i}}{\sqrt{2-3i}}\times \dfrac{\sqrt{2+3i}}{\sqrt{2+3i}}$

$=\dfrac{2+3i}{\sqrt{4-9{i}^{2}}}$

$=\dfrac{2+3i}{\sqrt{4+9}}$

$\therefore Z=\dfrac{2}{\sqrt{13}}+i\dfrac{3}{\sqrt{13}}$

Compare the quadratic equation $\sqrt 3 {x^2} + 2\sqrt 3 = 0\,\,to\,\,a{x^2} + bx + c = 0$ and find the value of discriminant and hence write the nature of the roots.

If we compare the quadratic equation

$\sqrt 3 {x^2} + 2\sqrt 3 = 0\,\,to\,\,a{x^2} + bx + c = 0$ .

Then we get,

$a=\sqrt{3}, b=0, c=2\sqrt{3}$ .

Now discriminant

$=b^2-4ac=0^2-4\times \sqrt{3}\times 2\sqrt{3}=-24.$

Since the discriminant is

$<0$ , so the roots of the given equation are imaginary.

For what value of $k$ are the roots of the quadratic equation $kx^{2}+1=-4x$ equal and real?

Given the quadratic equation

$kx^{2}+1=-4x$ .

or,

$kx^2+4x+1=0$ .

Now if the roots of the given equation be real and equal then we'll have,

$4^2-4k=0$ [ Discriminant equal to

$0$ ]

or,

$k=4$ .

If $a=\dfrac{1+i}{\sqrt{2}}$ then show that $a^6+a^4+a^2+1=0$ .

Given,

$a=\dfrac{1+i}{\sqrt{2}}$ .

Now

$a^2=\dfrac{(1+i)^2}{{2}}$

or,

$a^2=\dfrac{(2i)}{{2}}$

or,

$a^2 =i$ .

Then

$a^4=-1, a^6=-i$ .

Then

$a^6+a^4+a^2+1=-i-1+i+1=0$ .

If the roots of the quadratic equation $kx\left(x+2\right)+6=0$ are real and equal.then find the value of $k$

$kx\left(x+2\right)+6=0$

$\Rightarrow k{x}^{2}+2kx+6=0$ is quadratic in

$x$

Discriminant

$\Delta={b}^{2}-4ac$ where

$a=k,b=2k$ and

$c=6$

Roots are real and equal

$\Rightarrow {b}^{2}-4ac=0$

$\Rightarrow 4{k}^{2}-4\times k\times 6=0$

$\Rightarrow 4k\left(k-6\right)=0$

$\Rightarrow k\neq 0,k=6$ if

$k=0$ , then

$0\left(0+2\right)+6\neq 0$ from the equation

$\therefore k=6$

Find the value of $k$ for the quadratic equation $3{y}^{2}+ky+12=0$ has two equal roots.

$3{y}^{2}+ky+12=0$

Comparing equation with

$a{x}^{2}+bx+c=0$

$a=3$ ,

$x=y$ ,

$b=k$ and

$c=12$

Since the equation has

$2$ equal roots,

$D=0$

$\Rightarrow {b}^{2}-4ac=0$

$\Rightarrow {k}^{2}-4\times 3\times 12=0$

$\Rightarrow {k}^{2}=144$

$\therefore k=\pm 12$

$\dfrac { 3\sqrt { -2 } +2\left( -5 \right) }{ 3\sqrt { -2 } -2\sqrt { -2 } }$

$\dfrac{3\sqrt{-2}+2\left(-5\right)}{3\sqrt{-2}-2\sqrt{-2}}$

$=\dfrac{3i\sqrt{2}-10}{\sqrt{-2}}$

$=\dfrac{3i\sqrt{2}-10}{i\sqrt{2}}$

$=\dfrac{3i\sqrt{2}}{i\sqrt{2}}-\dfrac{10}{i\sqrt{2}}$

$=3-\dfrac{10\sqrt{2}}{i\sqrt{2}\times \sqrt{2}}$

$=3-\dfrac{10\sqrt{2}}{2i}$

$=3+5i\sqrt{2}$ because

$\dfrac{1}{i}=\dfrac{i}{i\times i}=-i$

Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(1+i)^{3}$

$(1+i)^3\\1+3(1)^2(i)+3(1)(i)^2+i^3\\1+3i-3-i\\-2+2i$

If $a=\dfrac { -1+i\sqrt { 3} }{ 2 } ,b=\dfrac { -1-i\sqrt { 3 } }{ 2 }$ then show that ${ a }^{ 2 }=b$ and ${ b }^{ 2 }=a.$

$a=\dfrac{-1+i\sqrt{3}}{2}$

$\Rightarrow {a}^{2}={\left(\dfrac{-1+i\sqrt{3}}{2}\right)}^{2}$

$=\dfrac{1+3{i}^{2}-2\sqrt{3}i}{4}$

$=\dfrac{1-3-2i\sqrt{3}}{4}$

$=\dfrac{-2-2i\sqrt{3}}{4}$

$=\dfrac{-1-i\sqrt{3}}{2}=b$

$b=\dfrac{-1-i\sqrt{3}}{2}$

$\Rightarrow {b}^{2}={\left(\dfrac{-1-i\sqrt{3}}{2}\right)}^{2}$

$=\dfrac{1+3{i}^{2}+2\sqrt{3}i}{4}$

$=\dfrac{1-3+2i\sqrt{3}}{4}$

$=\dfrac{-2+2i\sqrt{3}}{4}$

$=\dfrac{-1+i\sqrt{3}}{2}=a$

Write the real and imaginary part of ${ (i-\sqrt { 3 } ) }^{ 3 }$ .

$(i-\sqrt 3)^3$

$=-i-3(-1)\sqrt 3+9i-3\sqrt 3$

$(i-\sqrt 3)^3$

$=8i$

Real part is

$0$

Imaginary part is

$8i$

1) Express the following in the form of a = ib, a,b $\epsilon$ R $i = \sqrt{-1}$ . State the values of a and b.
$(2+3i)(2-3i)$

$(2+3i)(2-3i)\\4-(-9)=13\\a=13,b=0$

Show that $\dfrac{\sqrt 5 + i\sqrt 2}{\sqrt 5 - i\sqrt 2} + \dfrac{\sqrt 5 - i\sqrt 2}{\sqrt 5 + i\sqrt 2}$ is real.

Let

$Z=\dfrac{\sqrt 5 + i\sqrt 2}{\sqrt 5 - i\sqrt 2 }+ \dfrac{\sqrt 5 - i\sqrt 2}{\sqrt 5 + i\sqrt 2}$

$=\dfrac{(\sqrt 5+i\sqrt 2)^2+(\sqrt 5-i\sqrt 2)^2}{(\sqrt 5-i\sqrt 2)(\sqrt 5+i\sqrt 2)}$

$=\dfrac {5-2+2i\sqrt {10}+5-2-2i\sqrt {10}}{5+2}$

$Z=\dfrac {6}{7}$

Show that $\dfrac{\sqrt 7 + i\sqrt 3}{\sqrt 7 - i\sqrt 3} + \dfrac{\sqrt 7 - i\sqrt 3}{\sqrt 7 + i\sqrt 3}$ is real.

$\dfrac{\sqrt 7 + i\sqrt 3}{\sqrt 7 - i\sqrt 3} + \dfrac{\sqrt 7 - i\sqrt 3}{\sqrt 7 + i\sqrt 3}$

$\dfrac{(\sqrt 7+i\sqrt 3)^2+(\sqrt 7-i\sqrt 3)^2}{(\sqrt 7-i\sqrt 3)(\sqrt 7+i\sqrt 3)}$

$\dfrac {7-3+2i\sqrt {21}+7-3-2i\sqrt {21}}{7+3}$

$\dfrac {8}{10} =\dfrac 45$

Simplify: $\left( -\sqrt { 3 } +\sqrt { -2 } \right) \left( 2\sqrt { 3 } -i \right)$

Given

Let

$Z=\left( -\sqrt { 3 } +\sqrt { -2 } \right) \left( 2\sqrt { 3 } -i \right)$

$=\left( -\sqrt { 3 } +i\sqrt { 2 } \right) \left( 2\sqrt { 3 } -i \right)$

$=-6+i\sqrt { 3 } +2\sqrt { 6 } i+\sqrt { 2 }$

$Z=\left( \sqrt { 2 } -6 \right) +i\left( \sqrt { 3 } +2\sqrt { 6 } \right)$

Find the value of $p$ if the following quadratic equation has equal roots :
${ 4x }^{ 2 }-(p-2)x+1=0$

1323517_1491533_ans_f1f5249748ad494f95fb16a33acfc68f.JPG

Find the modulus of $\dfrac{{i + 1}}{{1 - i}}$ .

First we solve

$\frac{1+i}{1-i}$

Let

$z=\frac{1+i}{1-i}$

Rationalizing the same

$=\frac{1+i}{1-i}\times \frac{1+i}{1+i}$

$=\frac{(1+i)(1+i)}{(1-i)(1+i)}$

Using

$(a+b)(a-b)=a^{2}-b^{2}$

$=\frac{(1+i)^{2}}{1^{2}-i^{2}}$

Using

$(a+b)^{2}=a^{2}+2ab+b^{2}$

$=\frac{(1)^{2}+(i)^{2}+2.1.i}{(1)^{2}-i^{2}}$

$=\frac{1+i+2i}{(1)^{2}-(i)^{2}}$

1201509_1416201_ans_185b299ca7474995aa3136f3525f2884.jpg

$\left( 3k+1 \right) { x }^{ 2 }-2\left( k+1 \right) x+1=0$ has equal and real roots.

$(3k+1)x^{2}-2(k+1)x+1=0$

when D=0,then the equation real equal roots.

$D=b^{2}-4ac=0$

$\Rightarrow (-2(k+1))^{2}-4(3k+1)(1)=0$

$\Rightarrow 4(k+1)^{2}-4(3k+1)=0$

$\Rightarrow k^{2}+1+2k-3k-1=0$

$\Rightarrow k^{2}-k=0$

$\Rightarrow k(k-1)=0$

so k=0 or k=1

1199789_1441428_ans_ce3241c2c7e841efbf8d937bb93168dd.jpg

If arg $\left(z\right)<0$ then find arg $\left(-z\right)-$ arg $\left(z\right)$

Let

$z=r\left(\cos{\theta}+i\sin{\theta}\right)$

arg

$\left(z\right)=\theta<0$

$\Rightarrow -z=-r\left(\cos{\theta}+i\sin{\theta}\right)$

$=r\left(\cos{\left(\pi+\theta\right)}+i\sin{\left(\pi+\theta\right)}\right)$

$\Rightarrow$ arg

$\left(-z\right)=\pi+\theta$

$\therefore$ arg

$\left(-z\right)-$ arg

$\left(z\right)=\pi+\theta-\theta=\pi$

The modulus and amplitude of $(1+i\sqrt { 3 } )^{ 8 }$ are respectively.

1307410_1516639_ans_f3b876167b5b44bf98b42214fda614c1.jpeg

Express the following complex numbers in the standard from $a+ib$ :
$\dfrac{2+3i}{4+5i}$

1429686_1666055_ans_6d69c8ec4f5f4bdc8e27612651de30ba.jpg

Evaluate the following:
$\dfrac{1}{i^{58}}$

1408404_1665938_ans_480240f7526e486aa7682a198786ae20.png

Find the modulus of $(1-i)^{10}$

As we know that

$|a+i b|=\sqrt{a^2+b^2}$

$\implies |1-i|=\sqrt{1^2+(-1)^2}=\sqrt{2}$

$|(1-i)^{10}|=(|1-i|)^{10}=(\sqrt{2})^{10}=2^{5}=32$

Find the least positive integral value of n for which
$\left ( \dfrac{1+i}{1-i} \right )^{n}$ is real.

1409221_1666277_ans_496dcda851d7420bb815f3df2da001f8.jpg

Give three examples from your environment of Points.

Three examples of points are:

Full stop, capital cities indicator on map, moon from longer distance.

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\dfrac{-16}{1+i\sqrt{3}}$

Let

$z=\dfrac{-16}{1+i\sqrt{3}}=\dfrac{-16(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}=\dfrac{-16(1-i\sqrt{3})}{1+3}=-4+4i\sqrt{3}$ .

Then,

$\left | z \right |=\sqrt{(-4)^{2}+(4\sqrt{3})^{2}}=\sqrt{16+48}=8$

Let

$\alpha$ the acute angle given by

$tan \,\alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}$ .

Then,

$tan\, \alpha =\dfrac{\left | 4\sqrt{3} \right |}{\left | -4 \right |}=\sqrt{3}\Rightarrow \alpha =\dfrac{\pi }{3}$

Clearly, z lies in the second quadrant. Therefore,

$arg(z)=\pi -\alpha =\pi -\dfrac{\pi }{3}=\dfrac{2\pi }{3}$ .

Find the real values of $\theta$ for which the complex number $\dfrac{1+i\, cos \,\theta}{1-2\, i\, cos \, \theta }$ is purely real.

1409223_1666281_ans_210651901f824879b7d239f1f55e7e03.jpg

Find the modulus of the following complex number
$\sin\, 120^{\circ}-i\, \cos\, 120^{\circ}$

$Z=\sin\, 120^{\circ}-i\, \cos\, 120^{\circ}$

$120^{\circ}=\dfrac{\pi}{2}+\dfrac{\pi}{6}$

$|Z|=\sqrt{sin^2{120^{\circ}}+cos^2{120^{\circ}}}$

$|Z|=\sqrt{1}=1$

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\sqrt{3}+i$

Let

$z=\sqrt{3}+i$ . Then,

$\left | z \right |=\sqrt{(\sqrt{3})^{2}+(1)^{2}}=2$ .

Let

$\theta$ be the argument of z and

$\alpha$ be the acute angle given by

$tan\, \alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}$ .

Then,

$tan\, \alpha =\dfrac{1}{\sqrt{3}}\Rightarrow \alpha =\dfrac{\pi }{6}$

Clearly, z lies in the first quadrant. So,

$arg(z)= \alpha =\dfrac{\pi }{6}$ .

Evaluate : $(\sqrt{-1})^{30}$

$(\sqrt{-1})^{30}$

$=i^{30}$

$=(i^4)^{7} i^2$

$=-1$ ........ As

$i^2=-1,i^4=1$

Evaluate : $i^{62}$

$i^{62}$

$=(i^4)^{15} i^2$

$=1^{15}(-1)$ ........... as

$i^2=-1, i^4=1$

$=-1$

Evaluate : $(\sqrt{-1})^{192}$

$(\sqrt{-1})^{192}$

$=i^{192}$ as

$\sqrt{-1}=i$

$=(i^4)^{48}$

$=1^{48}$ as

$i^4=1$

$=1$

Evaluate $i^{19}$

$i^{19}$

$=(i^4)^4i^3$

$=1\times -i$

$=-i$

Note :

$i^2=-1,i^3=-i,i^4=1$

Evaluate : $i^{-131}$

$i^{-131}$

$=\dfrac{1}{i^{131}}$

$=\dfrac{1}{(i^4)^{32}i^3}\\$

$=\dfrac{1}{-i}\\$

$=\dfrac{-1}{i}\times \dfrac{i}{i}\\$

$=\dfrac{-i}{i^2}\\$

$=i$

Note :

$i^2=-1,i^3=-i,i^4=1$

Evaluate : $\left(i^{41}+\dfrac{1}{i^{71}}\right)$

$i^{41}+\dfrac{1}{i^{71}}$

$=(i^4)^{10} i +\dfrac{1}{(i^4)^{17} i^3}$

$=i+\dfrac{1}{-i}$

$=i+i$

$=2i$

Note :

$i^2=-1,i^3=-i,i^4=1$

Evaluate : $i^{-9}$

$\textbf{Step-1: Rearranging the terms}$

$i^{-9}=\dfrac{1}{i^9}$

$=\dfrac{1}{i\times (i^4)^2}$

$\textbf{Step-2: Solving the obtained form}$

$=\dfrac{1}{i\times (i^4)^2}$

$=\dfrac{1}{i}$

$\boldsymbol{[\because i^2=-1,i^4=1)]}$

$=\dfrac{1}{i}\times \dfrac{i}{i}$

$=\dfrac{i}{i^2}$

$=\dfrac{i}{-1}$

$=-i$

$\textbf{Hence , the simplified form is }\mathbf{ -i.}$

Evaluate : $(\sqrt{-1})^{93}$

$(\sqrt{-1})^{93}$

$=i^{93}$ ...... as

$\sqrt{-1}=i$

$=(i^4)^{23} i$

$=1\times i$ ....... as

$i^4=1$

$=i$

Evaluate : $\left(i^{53}+\dfrac{1}{i^{53}}\right)$

$\left(i^{53}+\dfrac{1}{i^{53}}\right)$

$=(i^4)^{13}i+\dfrac{1}{(i^4)^{13}i}$

$=i+\dfrac{1}{i}$

$=\dfrac{i^2+1}{i}\\$

$=\dfrac{-1+1}{i}\\$

$=0$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$2i$

Given

$z=2i=r(\cos\theta+i \sin\theta)$

Now, separating real and complex part, we get

$0=r\cos\theta$ ---- ( 1 )

$2=r\sin\theta$ ---- ( 2 )

Squaring and adding equation ( 1 ) and ( 2 ), we get

$\Rightarrow$

$4=r^2$

Since,

$r$ is always positive number

$\Rightarrow$

$r=2$

Hence, modulus is

$2$ .

Now, dividing equation ( 2 ) by equation ( 1 ), we get

$\Rightarrow$

$\dfrac{r\sin\theta}{r\cos\theta}=\dfrac{2}{0}$

$\Rightarrow$

$\tan\theta=\infty$

$\therefore$

$\theta=\dfrac{\pi}{2}$

$\theta$ lies in first quadrant.

Argument of

$z=\theta=\dfrac{\pi}{2}$

Representing the complex number in its polar form will be

$z=2\left\{\cos\left(\dfrac{\pi}{2}\right)+i\sin\left(\dfrac{\pi}{2}\right)\right\}$

Prove that:
$1+i^{2}+i^{4}+i^{6}=0$

1859802_1708726_ans_397f809d52334786857158c4253c432c.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\sqrt 3+i$

$z=\sqrt3+i$

$|z|=\sqrt {(\sqrt 3)^2+1^2}=2$

$arg(z)=\tan^{-1} \dfrac yx=\tan^{-1} \dfrac {1}{\sqrt3}=\dfrac {\pi}{6}$

polar form=

$2\left(\cos \dfrac{\pi}{6}+i\sin \dfrac{\pi}{6}\right)$

Prove that:
$(1+i^{10}+i^{20}+i^{30})$ is a real number.

$(1+i^{10}+i^{20}+i^{30})=\left\{1+i^{4\times 2+2}+i^{4\times 5}+i^{4\times 7+2}\right\}$

$=\left\{1+(i^{4})^{2}\times i^{2}+(i^{4})^{5}+(i^{4})^{7}\times i^{2}\right\}$

$=(1+i^{2}+1+i^{2})=(1-1+1-1)=0$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$4$

let

$z=4+0i$

For

$z=x+iy$

$|z|=\sqrt {x^2+y^2}=\sqrt {4^2+0^2}=4$

For

$z=x+iy$

$arg (z)=\tan^{-1} \dfrac yx=\tan^{-1} \dfrac 04=0$

polar form

$=4(\cos 0+i\sin 0)$

$4, 0, 4(\cos 0+i\sin 0)$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$1-i$

$z=1-i=0-i \Rightarrow |z|=\sqrt {1^2+(-1)^2}=\sqrt 2$

$z=x+iy=1-i$

$arg z=\tan^{-1}\dfrac yx=\tan^{-1} -1=\dfrac{-\pi}{4}$

polar form

$=\sqrt 2 \left\{ \cos \left(\dfrac{-\pi}{4}\right)+i\sin \left(\dfrac{-\pi}{4}\right)\right\}$

Prove that:
$\dfrac{1}{i}-\dfrac{1}{i^{2}}+\dfrac{1}{i^{3}}-\dfrac{1}{i^{4}}=0$

$\dfrac{1}{i}=\dfrac{1}{i}\times \dfrac{i^{3}}{i^{3}}=i^{3}=-i, \dfrac{1}{i^{2}}-\dfrac{1}{-1}=-1, \dfrac{1}{i^{3}}\times \dfrac{i}{i}=i$ and

$\dfrac{1}{i^{4}}=1$

$\therefore\left\{\dfrac{1}{i}-\dfrac{1}{i^{2}}+\dfrac{1}{i^{3}}-\dfrac{1}{i^{4}}\right\}=(-i+1+i-1)=0$

Prove that:
$6i^{50}+5i^{33}-2i^{15}+6i^{48}=7i$

1972896_1708729_ans_0fa2e1482e6041a1874368962f2299d5.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$-1+\sqrt 3 i$

1972889_1708895_ans_27af74d3986e42b5bf4831ef40a7915c.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$-2$

$z=-2=-2+0i \Rightarrow |z|=\sqrt {(-2)^2+(0)^2}=2$

$z=x+iy=-2+0i$

$arg z=\tan^{-1}\dfrac yx=\tan^{-1}0=\pi$

polar form

$=2(\cos \pi+i\sin \pi)$

Find the modulus of the following :
$(3+\sqrt {-5})$

$(3+\sqrt {-5})$

$=(3+\sqrt {5}i)$ as

$\sqrt{-1}=i$

Modulus

$=\sqrt{3^2+\sqrt5^2}$

$=\sqrt{9+5}$

$=\sqrt{14}$

Find the modulus of the following :
$(7+24i)$

$(7+24i)$

Modulus

$=\sqrt{7^2+24^2}$

$=\sqrt{625}$

$=25$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\dfrac{1+i}{1-i}$

1972988_1708915_ans_cf7464551e6a42d08648cc1d21fcfd08.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\dfrac{1-3i}{1+2i}$

1860576_1708952_ans_56d6321a30364d8dad93c20de8a25f23.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$-4+4\sqrt 3 i$

1972899_1708906_ans_09bf4936704c4ed48d18422e74f54913.jpeg

Find the modulus of the following :
$(-3-4i)$

$(-3-4i)$

$=(-3)+(-4)i$

Modulus

$=\sqrt{(-3)^2+(-4)^2}$

$=\sqrt {25}$

$=5$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\dfrac{1+3i}{1-2i}$

1972881_1708922_ans_1d5884a124d94fcfb7005861c9417e93.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\dfrac{5-i}{2-3i}$

1860524_1708956_ans_23d2e307ce8e44899edb7ebee4c6262e.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$2-2i$

1972893_1708898_ans_e61ad2d374254f58af09104cfae7c1c0.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$-3\sqrt 2+3\sqrt 2 i$

1972897_1708909_ans_be322fbc360a4d0bbda5448d6e7734dc.jpeg

Find the modulus of the following :
$5$

$z=5+0i$

Modulus

$|z|=\sqrt{5^2+0^2}$

$=5$

Evaluate $(i^{57}+i^{70}+i^{91}+i^{101}+i^{108})$

Given expression

$=(i^{56}\times i+i^{68}\times i^{2}+i^{88}\times i^{3}+i^{100}\times i+i^{104})$

$=(i^{}+i^{2}+i^{3}+i+1)=(+i-1-i+i+1)=-i$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$-\sqrt 3-i$

Let

$z=-\sqrt 3-i\Rightarrow r^2=|z|^2=(-\sqrt 3)^2+(-1)^2=4\Rightarrow r=|z|=2$ .

$\tan \alpha =\left| \dfrac{-1}{-\sqrt 3}\right|=\dfrac{1}{\sqrt 3}\Rightarrow \alpha =\dfrac{\pi}{6}$ .

The given number represents the point

$P(-\sqrt 3, -1)$ which lies in the third quadrant.

$\therefore arg(z)=\theta=-(\pi -\alpha)=-\left(\pi -\dfrac{\pi}{6}\right)=\dfrac{-5\pi}{6}$ .

$\therefore z=2\left[\cos \left(\dfrac{-5\pi}{6}\right)+i\sin \left(\dfrac{-5\pi}{6}\right)\right]$ .

Find the modulus of the following :
$\dfrac {(3+2i)^2}{(4-3i)}$

$z=\dfrac {(3+2i)}{(4-3i)}=\dfrac {(9+4i^2+12i)}{(4-3i)}=\dfrac {(5+12i)}{(4-3i)}\times \dfrac {(4+3i)}{(4+3i)}=\dfrac {-16+63i}{25}$

$\Rightarrow \ |z|^2 = \left\{\left(\dfrac {-16}{25}\right)^2 +\left(\dfrac {63}{25}\right)^2\right\}\\=\left(\dfrac {256}{625}+\dfrac {3969}{625}\right)\\=\dfrac {4225}{625}$

$\Rightarrow \ |z|=\sqrt {\dfrac {4225}{625}}=\dfrac {65}{25}=\dfrac {13}{5}$

Find the modulus of the following :
$3i$

$3i=0+3i$

Modulus

$=\sqrt{0^2+3^2}$

$=3$

Find the modulus of each of the following :
$(1+2i) (i-1)$

Let

$z=(1+2i)(i-1)$

$=i-1+2i^2-2i$

$=-i-3$

$\therefore |z|=\sqrt {(-1)^2+(-3)^2}=\sqrt {10}$

Find the modulus of the following :
$\dfrac {(2-i)(1+i)}{(1+i)}$

Given,

$z=\dfrac {(2-i)(1+i)}{(1-i)}\\=\dfrac {(3+i)}{(1-i)}\times \dfrac {(1+i)}{(1+i)}\\=\dfrac {(2+4i)}{2}\\=(1+2i)$

Modulus

$|z|=\sqrt{1^2+2^2}=5$

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$(\sin 120^o-i\cos 120^o)$

1902351_1708980_ans_75ca519b2bfd45768193bc11e6ba66df.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$\dfrac{2+6\sqrt 3 i}{5+\sqrt 3 i}$

1972882_1708963_ans_9ce3bad7d29c492dad4385e03b3a588a.jpeg

Find the modulus and argument of the following complex numbers and hence express each of them in polar form:
$(i^{25})^3$

1947813_1708972_ans_bdad4bf2aa264288bc1ebe74b56a60cd.jpg

Write the principal argument of $(1+i\sqrt{3})^{2}$

1853444_1709106_ans_5e7e8b228b264ef9be51d5853d8772cd.jpg

Evaluate $\left(\dfrac{i^{180}+i^{178}+i^{176}+i^{174}+i^{172}}{i^{170}+i^{168}+i^{166}+i^{164}+i^{162}}\right)$

Given expression

$=\dfrac{i^{172}(i^{8}+i^{6}+i^{4}+i^{2}+1)}{i^{162}(i^{8}+i^{6}+i^{4}+i^{2}+1)}=i^{(172-162)}=i^{10}=i^{8}\times i^{2}=-1$

Find the sum $(i^{n}+i^{n+1}+i^{n+2}+i^{n+3})$ , where $n\in N$

Given sum

$=i^{n}\times (1+i+i^{2}+i^{3})=i^{n}(1+i-1-i)=0$

Evaluate $(i^{40+1}-i^{4n-1})$

Given expression

$=(i^{40}\times i-i^{4n}\times i^{-1})=\left(i-\dfrac{1}{i}\right)=\dfrac{(i^{2}-1)}{i}\times \dfrac{i}{i}=\dfrac{-2i}{-1}=2i$

Find the principal argument of $(-2i)$

$z=6\left(\cos\dfrac{3\pi}{4}+i\sin\dfrac{3\pi}{4}\right)=6\left\{\cos \left(\pi-\dfrac{\pi}{4}\right)+i\sin \left(\pi-\dfrac{\pi}{4}\right)\right\}$

$\Rightarrow z=6\left(-\cos\dfrac{3\pi}{4}+i\sin\dfrac{\pi}{4}\right)=6\left(\dfrac{-1}{\sqrt{2}}+i.\dfrac{1}{\sqrt{2}}\right)=\dfrac{6}{\sqrt{2}}(-1+i)$

$\Rightarrow z=3\sqrt{3}(-1+i)$

Find the value of $p$ for which the quadratic equation
$\left( 2p+1 \right) { x }^{ 2 }.\left( 7p+2 \right) x+\left( 7p-3 \right) =0$ has equal roots. Also find these roots.

since the quadratic equation has equal roots

$D=0$
ie.,

${ b }^{ 2 }-4ac=0$
In

$\left( 2p+1 \right) { x }^{ 2 }.\left( 7p+2 \right) x+\left( 7p-3 \right) =0$
Here

$a=\left( 2p+1 \right) ,b=-\left( 7p+2 \right) ,c=\left( 7p-3 \right)$

$\therefore { \left( 7p+2 \right) }^{ 2 }-4\left( 2p+1 \right) \left( 7p-3 \right) =0\quad$

$\Rightarrow 49{ p }^{ 2 }+4+28p-(8p+4)(7p-3)=0\quad \quad$

$\Rightarrow$

$49{ p }^{ 2 }+4+28p-56{ p }^{ 2 }+24p-28p+12=0$

$\Rightarrow$

$7{ p }^{ 2 }+24p+16=0\Rightarrow 7{ p }^{ 2 }-24p-16=0$

$\Rightarrow$

$7{ p }^{ 2 }-28p+4p-16=0\Rightarrow 7p(p-4)+4(p-4)=0$

$\Rightarrow$

$(7p+4)(p-4)=0\Rightarrow p=-\cfrac{4}{7}$ or

$p=4$
For

$p=\cfrac{-4}{7}$

$\left( 2\times \cfrac { -4 }{ 7 } +1 \right) { x }^{ 2 }-\left( 7\times \cfrac { -4 }{ 7 } +2 \right) x+\left( 7\times \cfrac { -4 }{ 7 } -3 \right) =0$

$\Rightarrow \cfrac { -1 }{ 7 } { x }^{ 2 }+2x-7=0\Rightarrow { x }^{ 2 }-14x+49=0\Rightarrow { x }^{ 2 }-7x-7x+49=0\Rightarrow x(x-7)-7(x-7)=0$

$\Rightarrow { (x-7) }^{ 2 }=0\Rightarrow x=7,7$
for

$p=4$

$\left( 2\times 4+1 \right) { x }^{ 2 }-\left( 7\times 4+2 \right) x+\left( 7\times 4-3 \right) =0\Rightarrow 9{ x }^{ 2 }-30x+25=0\Rightarrow 9{ x }^{ 2 }-15x-15x+25=0$

$0\Rightarrow 3x(3x-5)-5(3x-5)=0\Rightarrow (3x-5)(3x-5)=0\Rightarrow x=\cfrac { 5 }{ 3 } ,\cfrac { 5 }{ 3 }$

Find the value(s) of p for which the equation $2 x^{2}+3 x+p=0$ has real roots.

Given:

$2 x^{2}+3 x+p=0\\$
Let us consider,

$a=2, b=3, c=p\\$
By using the formula,

$\begin{aligned}D &=b^{2}-4 a c \\&=(3)^{2}-4(2)(p) \\&=9-8 p\end{aligned}\\$
since, roots are real.

$9-8 p \geq 0$

$\begin{array}{l}9 \geq 8 \mathrm{p} \\8 \mathrm{p} \leq 9 \\\mathrm{p} \leq 9 / 8\end{array}\\$

If 'a' is a complex number such that $\mid a \mid$ = 1, find arg(a), so that equation $az^2 + z + 1 = 0$ has one purely imaginary root.

$az^2 + z + 1 = 0$

Taking conjugate of both sides,

$\overline{az^2 + z + 1} = \bar{0}$

$\Rightarrow\bar{a}(\bar{z})^2 + \bar{z} + \bar{1}$

$\bar{a}z^2 - z + 1 = 0$

(since

$\bar{z} = -z$ as z is purely imaginary)

Eliminating z from both the equations, we get

$(\bar{a} - a)^2 + 2(a + \bar{a}) = 0$

Let,

$a = cos\theta + i sin \theta\space (\because\mid{a}\mid = 1)$

$\Rightarrow(-2isin\theta)^2 + 2(2cos\theta) = 0$

$\Rightarrow cos\theta = \dfrac{-1 \pm\sqrt{1 + 4}}{2}$

$\dfrac{\sqrt{5} - 1}{2}$

Hence,

$\alpha = cos\theta + i sin\theta, \space where\space \theta = cos^{-1}\dfrac{\sqrt{5}-1}{2}$

State whether the following quadratic equation has two distinct real roots. Justify your answer.
$2x^2 - 6x + \dfrac{9}{2}= 0$

Given equation is

$2x^2 - 6x + \dfrac{9}{2}= 0$

$\therefore D = b^2 - 4ac$

$\Rightarrow D = (-6)^2 - 4(2)\left ( \dfrac{9}{2} \right )$

$\Rightarrow D = 36 - 36$

$\therefore D = 0$

Hence the given quadratic equation has two real and equal roots.

Convert the following in the polar form:
$\dfrac{1+7i}{(2-i)^{2}}$

Here,

$z=\dfrac{1+7i}{(2-i)^{2}}$

$=\dfrac{1+7i}{(2-i)^{2}}=\dfrac{1+7i}{4+i^{2}-4i}=\dfrac{1+7i}{4-1-4i}\\$

$=\dfrac{1+7i}{3-4i}\times \dfrac{3+4i}{3+4i}=\dfrac{3+4i+12i+28i^{2}}{3^{2}+4^{2}}\\$

$=\dfrac{3+4i+21i-28}{3^{2}+4^{2}}=\dfrac{-25+25i}{25}$

$=-1+i$

Let

$r\cos \theta=-1$ and

$r\sin \theta=1$

On squaring and adding, we obtain

$r^{2}(\cos^{2}\theta+\sin^{2}\theta)=1$

$\Rightarrow r^{2}(\cos^{2}\theta+\sin^{2}\theta)=2$

$\Rightarrow r^{2}=2$

$[\cos^{2}\theta+\sin^{2}\theta=1]$

$\Rightarrow r=\sqrt{2}$ [Conventionally

$r >0$ ]

$\therefore \sqrt{2}\cos\theta=-1$ and

$\sqrt{2}\sin \theta=1$

$\Rightarrow \cos\theta=\dfrac{-1}{\sqrt2}$ and

$\sin \theta=\dfrac{1}{\sqrt{2}}$

$\therefore \theta=\pi -\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ ........... [As

$\theta$ lies in

$II$ quadrant]

$\therefore z=r\cos \theta+i\ r\sin\theta$

$=\sqrt{2}\cos\dfrac{3\pi}{4}+i\sqrt{2}\sin \dfrac{3\pi}{4}=\sqrt{2}\left(\cos\dfrac{3\pi}{4}+i\sin \dfrac{3\pi}{4}\right)$

This is the required polar form.

Convert the following in the polar form:
$\dfrac{1+3i}{1-2i}$

Here,

$z=\dfrac{1+3i}{1-2i}$

$=\dfrac{1+3i}{1-2i}\times \dfrac{1+2i}{1+2i}$

$=\dfrac{1+2i+3i-6}{1+4}$

$=\dfrac{-5+5i}{5}=1+i$

Let

$r\cos \theta=-1$ and

$r\sin \theta$

$=1$

on squaring and adding. we obtain

$r^{2}(\cos^{2}\theta+\sin^{2}\theta)$

$=1+1$

$\Rightarrow r^{2}(\cos^{2}\theta+\sin^{2}\theta)=2$

$\Rightarrow r^{2}=2$

$[\cos^{2}\theta+\sin^{2}\theta=1]$

$\Rightarrow r=\sqrt{2}$ [Conventionally,

$r > 0$ ]

$\therefore \sqrt{2}\cos\theta=-1$ and

$\sqrt{2}\sin \theta=1$

$\Rightarrow \cos\theta=\dfrac{-1}{\sqrt{2}}$ and

$\sin \theta=\dfrac{1}{\sqrt{2}}$

$\therefore \theta=\pi -\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ [As

$\theta$ lies in

$II$ quadrant]

$\therefore z=r\cos \theta+i\ r\sin \theta$

$=\sqrt{2}\cos \dfrac{3\pi}{4}+i\sqrt{2}\sin \dfrac{3\pi}{4}=\sqrt{2}\left(\cos \dfrac{3\pi}{4}+i\sin \dfrac{3\pi}{4}\right)$

This is the required polar form.

Convert the given complex number in polar form : $\sqrt 3+i$

Given complex number is

$\sqrt 3+i$

Let

$r\cos \theta =\sqrt 3$ and

$r\sin \theta =1$

On squaring and adding, we obtain

$r^2\cos^2 \theta +r^2 \sin^2 \theta =(\sqrt 3)^2+1^2$

$\Rightarrow r^2 (\cos^2 \theta +\sin^2 \theta )=3+1$

$\Rightarrow r^2=4$

$\Rightarrow r=\sqrt 4=2$ [ Conventionally,

$r > 0$ ]

$\therefore 2\cos \theta =\sqrt 3$ and

$2\sin \theta =1$

$\Rightarrow \cos \theta =\dfrac {\sqrt 3}{2}$ and

$\sin \theta =\dfrac 12$

$\therefore \theta =\dfrac {\pi}{6}$ [ As

$\theta$ lies in the I quadrant ]

$\therefore \sqrt 3+i =r\cos \theta +ir\sin \theta =2\cos \dfrac{\pi}{6}+i2\sin \dfrac {\pi}{6}=2\left( \cos \dfrac {\pi}{6}+i\sin \dfrac {\pi}{6}\right)$

This is the required polar form.

Find the modulus and argument of the complex number $\dfrac {1 + 2i}{1 - 3i}$ .

Let

$z=\dfrac{1+2i}{1-3i}$ , then

$z=\dfrac{1+2i}{1-3i}\times \dfrac{1+3i}{1+3i}=\dfrac{1+3i+2i+6i^{2}}{1^{2}+3^{2}}=\dfrac{1+5i+6(-1)}{1+9}$

$=\dfrac{-5+5i}{10}=\dfrac{-5}{10}+\dfrac{5i}{10}=\dfrac{-1}{2}+\dfrac{1}{2}i$

Let

$z=r\cos \theta+ir\sin \theta$

i.e.,

$z=r\cos \theta=\dfrac{-1}{2}$ and

$r\sin \theta=\dfrac{1}{2}$

On squaring and adding we obtain

$r^{2}(\cos^{2}\theta+\sin^{2}\theta)=\left(\dfrac{-1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}\\$

$\Rightarrow r^{2}=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\\$

$\Rightarrow r=\dfrac{1}{\sqrt{2}}$ [Conventionally

$r > 0$ ]

$\therefore \dfrac{1}{\sqrt{2}}\cos\theta=\dfrac{-1}{2}$ and

$\dfrac{1}{\sqrt{2}}\sin \theta=\dfrac{1}{2}$

$\Rightarrow \cos\theta=\dfrac{-1}{\sqrt{2}}$ and

$\sin \theta=\dfrac{1}{\sqrt{2}}$

$\therefore \theta=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$ [As

$\theta$ lies in the

$II$ quadrant]

Therefore, the modulus and argument of the given complex number are

$\dfrac{1}{\sqrt{2}}$ and

$\dfrac{3\pi}{4}$ respectively.

State whether the following quadratic equation has two distinct real roots. Justify your answer.
$2x^2 + x - 1 = 0$

Given equation is

$2x^2 + x - 1 = 0$

$\therefore D = b^2 - 4ac$

$\Rightarrow D =(1)^2 - 4(2) (-1)$

$\Rightarrow D = 1 + 8$

$\Rightarrow D = 9$

$\therefore D > 0$

Hence the given quadratic equation has two distinct real roots.

Find the modulus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

$\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\\=\dfrac{(1+i)^{2}-1(1-i)^{2}}{(1-i)(1+i)}\\$

$=\dfrac{1+i^{2}+2i-1-i^{2}+2i}{1^{2}+1^{2}}\\$

$=\dfrac{4i}{2}=2i$

$\therefore \left|\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}\right|=|2i|=\sqrt{2^{2}}=2$

Find whether the following quadratic equations have a repeated root: $16y^2-40y+25=0$

REPEATED ROOTS MEAN d=0.

$d=b^2-4ac$

$d=(-40)^2-4(16)(25)$

$d=1600-1600$

$d=0=0$

Yes, roots are repeated.

Find whether the following quadratic equations have a repeated root: $x^2+6x+9=0$

REPEATED ROOTS MEAN d=0.

$d=b^2-4ac$

$d=(6)^2-4(1)(9)$

$d=36-36$

$d=0$
Yes, roots are repeated.

Find whether the following quadratic equations have a repeated root: $y^2-6y+6=0$

REPEATED ROOTS MEAN d=0.

$d=b^2-4ac$

$d=(-6)^2-4(1)(6)$

$d=36-24$

$d=12\neq 0$
Yes, roots are not repeated.

Find whether the following quadratic equations have a repeated root: $9x^2+4x+6=0$

REPEATED ROOTS MEAN d=0.

$d=b^2-4ac$

$d=(4)^2-4(9)(6)$

$d=16-216$

$d=-200\neq 0$

Yes, roots are not repeated.

Examine whether the following quadratic equations have real roots or not: $x^2-10x+2=0$

TO CHECK REAL ROOTS , 'd ' SHOULD BE GREATER THAN 0.

$d=b^2-4ac$

$d=(-10)^2-4(1)(2)$

$d=100-8$

$d=92$
Yes, roots are real.

Find whether the following quadratic equations have a repeated root: $9x^2-12x+4=0$

REPEATED ROOTS MEAN d=0.

$d=b^2-4ac$

$d=(-12)^2-4(9)(4)$

$d=144-144$

$d=0=0$
Yes, roots are repeated.

Examine whether the following quadratic equations have real roots or not: $x^2+x-1=0$

TO CHECK REAL ROOTS , 'd ' SHOULD BE GREATER THAN 0.

$d=b^2-4ac$

$d=(1)^2-4(1)(-1)$

$d=1+4$

$d=5$

Yes, roots are real.

Examine whether the following quadratic equations have real roots or not: $7x^2+8x-1=0$

TO CHECK REAL ROOTS , 'd ' SHOULD BE GREATER THAN 0.

$d=b^2-4ac$

$d=(8)^2 2-4(7)(-1)$

$d=64+28$

$d=92>0$

Yes,roots are real.

Examine whether the following quadratic equations have real roots or not: $2x^2+3x+4=0$

TO CHECK REAL ROOTS , 'd ' SHOULD BE GREATER THAN 0.

$d=b^2-4ac$

$d=(3)^2-4(2)(4)$

$d=9-32$

$d=-23$

No, roots aren't real.

Examine whether the following quadratic equations have real roots or not: $x^2-12x-16=0$

TO CHECK REAL ROOTS , 'd ' SHOULD BE GREATER THAN 0.

$d=b^2-4ac$

$d=(-12)^2-4(1)(-16)$

$d=144-64$

$d=208$

Yes, roots are real.

Without solving, determine whether the following equations have real roots or not.If yes,find them: $2x^2-4x+3=0$

$d=b^2-4ac$

$d=(-4)^2-4(2)(3)$

$d=16-24$

$d=-8$
Since,d < 0, no real roots exists for the given equation.

Without solving, determine whether the following equations have real roots or not.If yes,find them: $y^2-\dfrac{2}{3}y+\dfrac{1}{9}=0$

$d=b^2-4ac$

$d=(\dfrac{-2}{3})^2-4(1)(\dfrac{1}{9})$

$d=\dfrac{4}{9}-\dfrac{4}{9}$

$d=0$
Since,d > 0, roots are real and equal for the given equation.

$x=\dfrac{-b \sqrt{d}}{2a}$

$x=\dfrac{-(-\dfrac{2}{3})\pm \sqrt{0}}{2 \times 1}$

$x=\dfrac{2}{3\times \dfrac{1}{3}}$

$x=\dfrac{1}{3}$

Comment upon the nature of roots of the following equations: $x^2+10x+39=0$

$d=b^2-4ac$

$d=(10)^2-4(1)(39)$

$d=100-156$

$d=-56$

Since,d < 0, no real roots exists.

Without finding the roots,comment upon the nature of roots of each of the following quadratic equations : $2x^2-5x-3=0$

$d=^2-4ac$

$d=(-5)^2-4(2)(-3)$

$d=25+24$

$d=49$

$\therefore$ Roots are real and unique.

Comment upon the nature of roots of the following equations: $4x^2+7x+2=0$

$d=b^2-4ac$

$d=(7)^2-4(4)(2)$

$d=49-32$

$d=17$

Since,d > 0, roots are unique and real.

Without finding the roots,comment upon the nature of roots of each of the following quadratic equations : $2x^2-6x+3=0$

$d=^2-4ac$

$d=(-6)^2-4(2)(3)$

$d=36-24$

$d=12$

$\therefore$ Roots are real and unique.

Find the value of k for which the following quadratic equation has equal roots.
$9x^2+8kx+16=0$

Since roots are equal.

$\therefore d=0$ ....(1)

$9x^2+8kx+16=0$

$d=b^2-4ac$

$d=(8k)^2-4(9)(16)$

$(\because (a+b)^2=a^2+b^2+2ab)$

$d=64k^2-576$

From (1),d=0

$\therefore$ Equation will be

$0=64k^2-576$

$k^2=\dfrac{576}{64}$

$k^2=9$

$k=\pm \sqrt{9}$

$k=3,-3$

$\therefore$ Values of k are-3,3.

Find the value of k for which the quadratic equation $4x^2-2(k+1)x+(k+4)=0$ has equal roots.

Since roots are equal.

$\therefore d=0$ ..........(1)

$4x^2-2(k+1)x+(k+4)=0$

$d=b^2-4ac$

$d=(-2(k+1)^2)-4(4)(k+4)$

$d=(-2k-2)^2-16k-64$

$(\because (a-b)^2=a^2+b^2-2ab)$

$d=4k^2-8k-60$
From (1),d=0

$\therefore$ Equation will be

$4k^2-8k-60=0$
Dividing by 4

$k^2-2k-15=0$

$4k^2-5k+3k-15=0$

$k(k-5)+3(k-5)=0$

$(k-5)(k+3)-0$

$k-5=0 \ \ \ \ \ \ k+3=0$

$k=5 \ \ \ \ \ \ \ \ \ k=-3$

If -5 is root of the quadratic equation $2x^2+2px=15=0$ and the quadratic equation $p(x^2+x)+k=0$ has equal roots,find the value of k.

$2x^2+2px-15=0$

Put x=-5

$2(-5)^2+2p(-5)-15=0$

$50-10p-15=0$

$35=10p$

$p=3.5$

Equation

$p(x^2+x)+k=0$ has equal roots i.e d=0

$p(x^2+x)+k=0$

$p=3.5$

$3.5(x^2+x)+k=0$

$3.5 x^2+3.5x+k=0$

$d=b^2-4ac$

$d=(3.5)^2-4(3.5)(k)$

$d=12.25-14k$

Putting d=0

$\therefore$ Equation will be:

$0=12.25-14k$

$14k=12.25$

$k=\dfrac{12.25}{14}$

$k=0.875$

Define the addition of two complex numbers.

Let

$z_{1}=a+i b$ and

$z_{2}=c+i d$ be any two complex numbers. Then addition of

$z_{1}$ and

$z_{2}$ defined as

$z_{1}+z_{2}=(a+c)+i(b+d)$ which is again a complex number. Properties of addition of complex numbers:
(i) Closure property: The sum of two complex numbers is always a complex number, i.e., addition is closed in

$\mathrm{C}$ set of complex numbers.
(ii) Commutative property: For any two complex numbers

$z_{1}$ and

$z_{2}$ we have

$z_{1}+z_{2}=z_{2}+z_{1}$
(iii) Associative properly: For any complex numbers

$z_{1}, z_{2}$ and

$Z_{3},$ we have

$z_{1}+\left(z_{2}+z_{3}\right)=\left(z_{1}+z_{2}\right)+z_{3}$
(iv) Existence of additive identity: For any complex number

$z,$ we have

$z+0=0+z=z$ Thus, 0 is the additive identity for complex numbers.
(v) Existence of additive inverse: Every complex number

$z$ -a

$+$ ib has

$-z=(-a)+i(-b)$ as its additive inverse, as

$z+(-z)=(-z)+z=0$

Determine the nature of roots of the following quadratic equation.
$2y^{2} - 7y + 2 = 0$

$2y^{2} - 7y + 2 = 0$

Comparing the given equation with the quadratic equation

$ax^{2} + bx + c = 0$

$a = 2, b = -7$ and

$c = 2$

Discriminant,

$\Delta = b^{2} - 4ac = \left(- 7\right)^{2} - 4 \times 2 \times 2 = 49 - 16 = 33$

Since the discriminant

$> 0$

so, the roots of the given quadratic equation are real and unequal.

Determine the nature of roots of the following quadratic equation.
$m^{2} + 2m + 9 = 0$

$m^{2} + 2m + 9 = 0$
Comparing the given equation with the quadratic equation

$ax^{2} + bx + c = 0$

$a = 1, b = 2, c = 9$

Discriminant,

$\Delta = b^{2} - 4ac = \left(2\right)^{2} - 4 \times 1 \times 9 = 4 - 36 = -32$
Since the discriminant

$< 0$

so, the roots of the given quadratic equation are not real.

Determine the nature of roots of the following quadratic equation.
$x^{2} - 4x + 4 = 0$

$x^{2} - 4x + 4 = 0$
Comparing the given equation with the quadratic equation

$ax^{2} + bx + c = 0$

$a = 1, b = -4$ and

$c = 4$

Discriminant,

$\Delta = b^{2} - 4ac = \left(- 4\right)^{2} - 4 \times 1 \times 4 = 16 - 16 = 0$
Since the discriminant

$= 0$

so, the roots of the given quadratic equation are real and equal.

Choose the correct answer for the following question.
$\sqrt{5}m^{2} - \sqrt{5}m + \sqrt{5}$ which of the following statement is true for this given equation?
Real and unequal roots
Real and equal roots
Roots are not real
Three roots

1815251_1851032_ans_aa005d97a14046b7960f7b65abcf29b3.PNG

Define a complex number.

A number of the form

$a + ib$ , where a and b are real numbers, is defined to be a complex number, where

$\mathrm{i}^{2}=-1.$
Here 'a' is called the real part, denoted by Re (z) and 'b' is called the imaginary part, denoted by

$\mathrm{Im}(\mathrm{z}),$ of the complex number

$z=\mathrm{a}+\mathrm{ib}$

Determine the nature of root of the quadratic equation.
$3x^{2} - 5x + 7 = 0$

1809914_1851064_ans_a9cc05767e3a40d1964f5ae36a213f70.PNG

Determine the nature of root of the quadratic equation.
$\sqrt{3}x^{2} + \sqrt{2}x - 2\sqrt{3}= 0$

1809911_1851067_ans_1875fe22b677498fb1cab815a1fef3e1.PNG

Find the values of k,for which the given equation has real roots: $kx^2+4x+1=0$

Roots are equal

$\therefore d=0$

$d=b^2-4ac$

$d=(4)^2-4(1)(k)$

$d=16-4k$
Put d=0

$0=16-4k$

$4k=16$

$k=4$

Define purely real and purely imaginary numbers.

A complex number z is said to be:

$\bullet$ Purely real, if

$\operatorname{Im}(z)=0$

$\bullet$ Purely imaginary, if

$Re(z) = 0$

Define multiplication numbers.

Let

$z_{1}=a+i b$ and

$z_{2}=c+i d$ be any two complex numbers. Then, the product

$z_{1} z_{2}$ is defined as follows:

$z_{1} z_{2}=(a c-b d)+i(a d+b c)$
Properties of product of complex numbers:

$\bullet$ Closure property: The product of two complex numbers is always a complex number.

$\bullet$ Commutative law: For any two complex numbers

$z_{1}$ and

$z_{2}$ we have

$z_{1} z_{2}=z_{2} z_{1}$

$\bullet$ Associative law: For any three complex numbers

$z_{1} z_{2}$ and

$z_{3}$ we have

$\left(z_{1} z_{2}\right) z_{3}=z_{1}\left(z_{2} z_{3}\right)$

$\bullet$ Existence of multiplicative identity: For every complex number

$z$ , we have

$z \cdot 1=1$ .

$z=z$ . Thus, 1 is the multiplicative identity.

$\bullet$ Existence of multiplicative inverse: Let

$z=a+i b .$ Then

$z^{-1}=\cfrac{1}{z}=\cfrac{1}{a+i b}=\cfrac{1}{a+i b} \cdot \cfrac{a-i b}{a-i b}=\cfrac{a-i b}{a^{2}+b^{2}}$
Clearly,

$z \cdot \cfrac{1}{z}=\cfrac{1}{z} \cdot z=1$
Thus, every z=a+ib has its multiplicative inverse, given

$z^{-1}=\cfrac{1}{z}=\cfrac{a}{a^{2}+b^{2}}-\cfrac{a}{a^{2}+b^{2}} i \quad(z \neq 0)$

Define the division of two complex numbers.

Let

$z_{1}$ and

$z_{2}$ be two complex numbers and

$z_{2} \neq 0 .$ Then

$\dfrac{z_{1}}{z_{2}}$ is defined by

$\dfrac{z_{1}}{z_{2}}=z_{1}\left(\dfrac{1}{z_{2}}\right)$
Power of i:
We have

$i^{2}=-1 \Rightarrow i^{3}=i^{2} \cdot i=-i$

$\Rightarrow i^{4}=i^{2} \cdot i^{2}=(-1) \cdot(-1)=1$

$\Rightarrow i^{5}=i^{4} \cdot i=i$

$\Rightarrow i^{6}=i^{5} \cdot i=i \cdot i=-1$

$\Rightarrow i^{7}=i^{6} \cdot i=-i$
In general, for any integer

$k, i^{4 k}=1, i^{4 k+1}=i$

$i^{4 k+2}=-1, i^{4 k+3}=-i$

Define the modulus of a complex number.

Let.

$z=a+i b$ be a complex number. Then, the modulus of

$z$ , denoted by

$|z|$ to be the nonnegative real number

$\sqrt{a^{2}+b^{2}}$

Write down the modulus of $(-1-i)^3$

Let

$z=(-1-i)^{3}=(-1)^3-i^3-3(-1)^2(i)+3(-1)(i)^2=-1-3 i+3+i=2-2 i$
Then,

$|z|=\sqrt{2^{2}+(-2)^{2}}=\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}$

Find the modulus and the arguments of the following:
1

Let

$z=1 = 1+i 0$
Modulus

$=|z|=\sqrt{x^{2}+y^{2}}=\sqrt{1+0}=1$

$r \cos \theta=1$ and

$r \sin \theta=0$

$\Rightarrow \cos \theta=1$ and

$\sin \theta=0$

$\therefore \theta=0^o \quad$

Write down the modulus of $3+\sqrt{-3}$

$3+\sqrt{-3}$

Let

$z=3+i \sqrt{3}$
Then,

$|z|=\sqrt{3^{2}+(\sqrt{3})^{2}}=\sqrt{12}=2 \sqrt{3}$

Define the difference of two complex numbers.

Let

$z_{1}$ and

$z_{2}$ be any two complex numbers. The different

$z_{1}-z_{2}$ is defined as follows:

$z_{1}-z_{2}=z_{1}+\left(-z_{2}\right)$

Find the modulus and the arguments of the following:
-3

Let

$z=-3=-3+i 0$
Modulus

$=|z|=\sqrt{x^{2}+y^{2}}=\sqrt{(-3)^{2}+0^{2}}=3$

$r \cos \theta=-3$ and

$r \sin \theta=0$

$\Rightarrow \cos \theta=-1$ and

$\sin \theta=0$

$\therefore \theta=\pi$

Write down the modulus of i

Let

$z=i=0+i$
Then,

$|z|=\sqrt{0^{2}+1^{2}}=1$

Find the modulus of $z=\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$

$\begin{aligned}\text { Let } z=& \dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}=\dfrac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)} \\&=\dfrac{(1+2 i-1)-(1-2 i-1)}{1+1}=\dfrac{4 i}{2}=2 i=0+2 i \\\therefore|z|=\sqrt{0^{2}+2^{2}}=2 \end{aligned}$

Find the modulus and the arguments of the following:
$\dfrac{1+2i}{1-3i}$

Let

$z=\dfrac{1+2 i}{1-3 i}$

$=\dfrac{1+2 i}{1-3 i} \times \dfrac{1+3 i}{1+3 i}=\dfrac{1+2 i+3 i-6}{1+9}$

$=\dfrac{-5+5 i}{10}=\dfrac{-1}{2}+\dfrac{1}{2} i$
Modulus

$=r=\sqrt{x^2+y^2}=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}=\dfrac{1}{\sqrt{2}}$

$r \cos \theta=-\dfrac{1}{2}$ and

$r \sin \theta=\dfrac{1}{2}$

$\Rightarrow \cos \theta=-\dfrac{1}{\sqrt{2}}$ and

$\sin \theta=\dfrac{1}{\sqrt{2}}$

$\therefore \theta=\pi-\dfrac{\pi}{4}=\dfrac{3 \pi}{4}$

Find the modulus and the arguments of the following:
i

Let

$z=i=0+1 i$
Modulus

$=|z|=r=\sqrt{x^{2}+y^{2}}=\sqrt{0+1}=1$

$r \cos \theta=0$ and

$r \sin \theta=1$

$\Rightarrow \cos \theta=0$ and

$\sin \theta=1$

$\therefore \theta=\dfrac{\pi}{2}$

Find the modulus and the arguments of the following:
$\dfrac{1+i}{1-i}$

Let

$z=\dfrac{1+i}{1-i} \quad$

$=\dfrac{1+i}{1-i} \times \dfrac{1+i}{1+i}=\dfrac{1+2 i-1}{1+1}=i=0+i$

$\therefore$ Modulus

$=|{z}|=\sqrt{x^{2}+y^{2}}$

$\therefore r=\sqrt{0+1^{2}}=1$

$r \cos \theta=0$ and

$r \sin \theta=1$

$\Rightarrow \cos \theta=0$ and

$\sin \theta=1$

$\therefore \theta=\dfrac{\pi}{2}$

Find the modulus and the arguments of the following:
$-8 i$

Let

$z=-8 i=0+(-8) i$
Modulus

$=|z| =r=\sqrt{x^2+y^2}=\sqrt{0+64}=8$

$r \cos \theta=0$ and

$r \sin \theta=-8$

$\Rightarrow \cos \theta=0$ and

$\sin \theta=-1$

$\Rightarrow \theta=-\frac{\pi}{2}$

Write the formula to find the nature of roots?

$discriminat = b^2 – 4ac$

Find the modulus and the arguments of the following:
$-\sqrt{3}+i$

Let

$z=-\sqrt{3}+i$
Modulus

$=|z|=r=\sqrt{x^2+y^2}=\sqrt{3+1}=2$

$r \cos \theta=-\sqrt{3}$ and

$r \sin \theta=1$

$\Rightarrow \cos \theta=-\dfrac{\sqrt{3}}{2}$ and

$\sin \theta=\dfrac{1}{2}$

$\therefore \theta=\dfrac{2 \pi}{3}$

Find the modulus and the arguments of the following:
$1+\mathrm{i}$

Let

$z=1+i$
Modulus

$= |z|=r=\sqrt{x^2+y^2}=\sqrt{1+1}=\sqrt{2}$

$r \cos \theta=1$ and

$r \sin \theta=1$

$\cos \theta=\dfrac{1}{\sqrt{2}}$ and

$\sin \theta=\dfrac{1}{\sqrt{2}}$

$\Rightarrow \theta=\dfrac{\pi}{4}$

Find the modulus and the arguments of the following:
$\dfrac{1}{1+i}$

Let

$z=\dfrac{1}{1+i}$

$=\dfrac{1}{1+i} \dfrac{1-i}{1-i}=\dfrac{1-i}{1+1}=\dfrac{1}{2}-\dfrac{1}{2} i$
Modulus

$=|z|=r=\sqrt{x^{2}+y^{2}}$

$=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}=\dfrac{1}{\sqrt{2}}$

$r \cos \theta=\dfrac{1}{2}$ and

$r \sin \theta=-\dfrac{1}{2}$

$\Rightarrow \cos \theta=\dfrac{1}{\sqrt{2}}$ and

$\sin \theta=\dfrac{-1}{\sqrt{2}}$

$\therefore \theta=-\dfrac{\pi}{4}$

Find the modulus and the arguments of the following:
$\sqrt{3}-i$

Let

$\sqrt{3}-i=z$
Modulus

$=|z|=r\sqrt{x^2+y^2}=\sqrt{3+1}=2$

$r \cos \theta=\sqrt{3}$ and

$r \sin \theta=-1$

$\Rightarrow \cos \theta=\dfrac{\sqrt{3}}{2}$ and

$\sin \theta=-\dfrac{1}{2}$

$\therefore \theta=-\dfrac{\pi}{6} \quad$

Find the modulus and the arguments of the following:
$-1-i \sqrt{3}$

Let

$z=-1-i \sqrt{3}$
Modulus

$=|z|=r=\sqrt{x^2+y^2}=\sqrt{1+3}=2$

$r \cos \theta=-1$ and

$r \sin \theta=-\sqrt{3}$

$\Rightarrow \cos \theta=\dfrac{-1}{2}$ and

$\sin \theta=-\dfrac{\sqrt{3}}{2}$

$\therefore \theta=\dfrac{\pi}{3}-\pi=\dfrac{-2 \pi}{3}$

Find the modulus of the following ;
$\dfrac1{(3-2i)}$

Let

$z = \frac {1}{ 3-2i}$

$\Rightarrow z = \frac { 1}{ 3 -2i} \times \frac { 3+2i}{ 3+2i} = \frac { 3+2i}{9-4i^2}$

$= \frac { 3 +2 i}{9+4} = \frac { 3 +2i}{13} = \frac {3}{13} + \frac {2}{13} i$
Modulus of

$z = \sqrt {( \frac {3}{13})^2 +( \frac {2}{13})^2 }$

$= \sqrt { \frac {9}{169} + \frac {4}{169} }$

$= \sqrt {\frac{13}{169}} = \sqrt {\frac{1}{13}} = \frac {1}{\sqrt {13} }$
Hence

$|z | = \frac {1}{\sqrt {13}}$

Find the arguments of the following numbers :
$\frac { 1 + i}{ 1 - i}$

Let

$X = \dfrac { 1+ i}{ 1- i} = a+ bi ...(i)$

$z = \dfrac { 1 + i}{ 1- i} \times \dfrac { 1+ i}{ 1+ i }$

$\Rightarrow z = \dfrac {( 1+i)^2}{1 - i^2} = \dfrac { 1 +i^2 + 2 i }{ 1+ 1 }$

$\Rightarrow z = \dfrac { 1- 1 +2 i}{2} = i .....(ii)$
From equation (i) and (ii) we get

$a= 0 , b = 1$

$\because z = D + i ( 0 , 1 )$ lies in 1st quadrant

$\therefore$ Argument

$\theta = tan^{-1} ( \frac {b}{a}) = tan^{-1}( \frac {1}{0})$

$= tan^{-1} ( \infty) = \frac { \pi}{2}$

Write the principle argument of $-2$

$-2 = -2 + 0 i$
let

$-2 + 0 i = r ( cos \theta + i sin \theta )$

$\Rightarrow r cos \theta = -2 ....(i)$

$r sin \theta = 0....(ii)$

$\Rightarrow tan \theta = \frac {0}{-2} = 0$
Hence principle argument of

$-z$ is

$\pi$

Find the arguments of the following numbers :
$\frac { 5 + i \sqrt {3} }{4 - i 2 \sqrt {3} }$

let

$z= \frac { 5+ i \sqrt {3} }{ 4- i2 \sqrt {3} }= a + bi$ ...(i)

$\Rightarrow x = \frac { 5 + i \sqrt {3}}{ 4 - i2 \sqrt {3} } \times \frac { 4 + i 2 \sqrt {3}}{ 4 + i 2 \sqrt {3} }$

$=\Rightarrow z = \frac { 20 + i4 \sqrt {3} + i 10 \sqrt {3} + i^2 6 }{ 4^2 -( i2\sqrt {3})^2 }$

$\Rightarrow z = \frac {20+14 \sqrt i - 6}{ 16 - i^2 4 \times 3}$

$\Rightarrow z = \frac { 14 + 14 \sqrt {3} i }{ 16+12}$

$\Rightarrow z = \frac { 14(1+ \sqrt {3} i)}{28}$

$\Rightarrow z= \frac {1}{2} + \frac { \sqrt {3}}{2} i ...(ii)$
Comparing equation (i) and (ii) we get

$a= \frac {1}{2} , b = \frac { \sqrt {3} }{2}$

$\because z = \frac {1}{2} + \frac {\sqrt {3}}{2} i = ( \frac {1}{2}, \frac { \sqrt {3}}{2})$ lies in Ist quadrant.

$\therefore$ Argument

$\theta = tan^{-1} \frac {b}{a} = tan^{-1 } \frac { \sqrt {3} / 2}{1/2}$

$\Rightarrow \theta = tan^{-1} \sqrt {3} = \frac {\pi}{3}$

Find the arguments of the following numbers :
$-1+ \sqrt {3} i$

Let

$z = -1 + \sqrt {3} i =- a+ ib$
On comparing

$a= 1 , b= \sqrt {3}$

$\because z = -1 + \sqrt {3} i =(-1, \sqrt {3})$ lies in II nd quadrant

$\therefore$ Argument

$\theta = \pi - tan^{-1} | \frac {b}{a} |$

$= \pi - tan^{-1} | \frac { \sqrt {3}}{1}|$

$= \pi - tan^{-1} ( \sqrt {3})$

$= \pi - \frac { \pi}{3} = \frac {2 \pi}{3}$

Let $z = 9 + bi$ , where $b$ is nonzero real and $i^2 = -1$ . If the imaginary part of $z^2$ and $z^3$ are equal, then find the value of $\dfrac{b}{3}$ .

Given that, $z=9+bi$

Since, imaginary parts of ${ z }^{ 2 }$ & ${ z }^{ 3 }$ are equal

Therefore, $\dfrac { { z }^{ 2 }-{ \bar { z } }^{ 2 } }{ 2i } =\dfrac { { z }^{ 3 }-{ \bar { z } }^{ 3 } }{ 2i }$

$\Rightarrow \left( z-\bar { z } \right) \left( z+\bar { z } \right) =\left( z-\bar { z } \right) \left( { z }^{ 2 }+{ \bar { z } }^{ 2 }+z\bar { z } \right)$

$\Rightarrow z+\bar { z } ={ \left( z+\bar { z } \right) }^{ 2 }-z\bar { z }$

$\Rightarrow z\bar { z } ={ 18 }^{ 2 }-18=306$

$\Rightarrow { \left| z \right| }^{ 2 }=306$

$\Rightarrow 81+{ b }^{ 2 }=306$

$\Rightarrow b=15$

$\Rightarrow \dfrac { b }{ 3 } =5$

Ans: 5

Solve the following equation for x ${ \left( 15+4\sqrt { 14 } \right) }^{ t }+{ \left( 15-4\sqrt { 14 } \right) }^{ t }=30$ where $t={ x }^{ 2 }-2\left| x \right|$
find the number of roots of the equation.

${ \left( 15+4\sqrt { 14 } \right) }^{ t }+{ \left( 15-4\sqrt { 14 } \right) }^{ t }=30$ ....(1)
Here,

$\left( 15+4\sqrt { 14 } \right) \left( 15-4\sqrt { 14 } \right) =1$

$\left( 15-4\sqrt { 14 } \right) =\dfrac{1}{ \left( 15+4\sqrt { 14 } \right) }$
Using this in eqn (1), we get

${ \left( 15+4\sqrt { 14 } \right) }^{ t }+\left( {\dfrac{1}{ 15+4\sqrt { 14 } }}\right)^{ t }=30$
Put

${ \left( 15+4\sqrt { 14 } \right) }^{ t }=u$

$\Rightarrow u^2-30u+1=0$

$\Rightarrow u=15\pm 4\sqrt{14}$

$\Rightarrow { \left( 15+4\sqrt { 14 } \right) }^{ t }=15\pm 4\sqrt{14}$

$\Rightarrow t=1, -1$

$\Rightarrow x^2-2|x|=1 ; x^2-2|x|=-1$
Now, if

$x\ge 0$

$\Rightarrow x^2-2x-1=0$ ;

$x^2-2x+1=0$

$\Rightarrow x=1\pm \sqrt{2}$ ;

$x=1$

$\therefore x=1+\sqrt2;\,x=1$
Now, if

$x<0$

$x^2+2x-1=0 ; x^2+2x+1=0$

$\Rightarrow x=-1\pm \sqrt{2}$ ;

$x=-1$

$\therefore x=-1-\sqrt2;\,x=-1$
So, the number of roots of the eqn are 4

show that: The modulus and argument of the complex number $\displaystyle z_{1}=z^{2}-z,\space$ if $z=\cos \phi +i\sin \phi .$ is $2|\sin \phi/2|, \left (\displaystyle \dfrac{3\pi+3\phi }{2} \right )$ .

$z=e^{i\phi}$
Hence

$z^{2}-z$

$=e^{2i\phi}-e^{i\phi}$

$=cos2\phi+isin2\phi-cos\phi-isin\phi$

$=-2sin\dfrac{3\phi}{2}.sin\dfrac{\phi}{2}+i(2sin\dfrac{\phi}{2}.cos\dfrac{3\phi}{2})$

$=2sin\dfrac{\phi}{2}[-sin\dfrac{3\phi}{2}+icos\dfrac{3\phi}{2}]$

$=2sin\dfrac{\phi}{2}[cos(\dfrac{\pi}{2}+\dfrac{3\phi}{2})+isin(\dfrac{\pi}{2}+\dfrac{3\phi}{2})]$ or

$2sin\dfrac{\phi}{2}[cos(\dfrac{3\pi}{2}+\dfrac{3\phi}{2})+isin(\dfrac{3\pi}{2}+\dfrac{3\phi}{2})]$ depending on the value of

$\phi$ .
Hence

$|z|=2|sin\dfrac{\phi}{2}|$
And

$arg(z)=\dfrac{3\pi}{2}+\dfrac{3\phi}{2}$ or

$\dfrac{\pi}{2}+\dfrac{\phi}{2}$ depending on the value of

$\phi$ .

Express the complex number $\dfrac {(1 + \sqrt {3}i)^{2}}{\sqrt {3} - i}$ in the form of $a + ib$ . Hence, find the modulus and argument of the complex number

$z = \dfrac {(1 + \sqrt {3}i)^{2}}{(\sqrt {3} - i)}$

$= \dfrac {1 + 3i^{2} + 2\sqrt {3} i}{(\sqrt {3} - i)}$

$= \dfrac {1 - 3 + 2\sqrt {3} i}{\sqrt {3} - i}$

$= \dfrac {-2 + 2\sqrt {3}i}{\sqrt {3} - i} \times \dfrac {\sqrt {3} + i}{\sqrt {3} + i}$

$= \dfrac {-2\sqrt {3} + 6i - 2i + 2\sqrt {3} i^{2}}{3 - i^{2}}$

$= \dfrac {-4\sqrt {3} + 4i}{3 + 1}$

$= \dfrac {4(-\sqrt {3} + i)}{4}$

$\therefore z = -\sqrt {3} + i$
Here,

$a = -\sqrt {3}, b = 1$
Modulus,

$|z| = \sqrt {a^{2} + b^{2}}$

$= \sqrt {(-\sqrt {3})^{2} + (1)^{2}}$

$= \sqrt {3 + 1} = \sqrt {4}$

$= 2$

$Argument = -\pi + \tan^{-1}\left |\dfrac {b}{a}\right |$

$= -\pi + \tan^{-1} \left |\dfrac {1}{-\sqrt {3}}\right |$

$= -\pi + \tan^{-1} \dfrac {1}{\sqrt {3}}$

$= -\pi + \dfrac {\pi}{6} = \dfrac {-5\pi}{6}$

Find the modulus, argument and the principal argument of the complex numbers.
$\dfrac{2+i}{4i+(1+i)^2}$

$=\dfrac{2+i}{4i+(1+i)2}$

$\Rightarrow \dfrac{2+i}{4i+1-1+2i}$

$\dfrac{2+i}{6i}$

$\dfrac{1}{3i}+\frac{1}{6}$

$\dfrac{1}{6}-\frac{i}{3}$

$z=\dfrac{1}{6}-\frac{i}{3}$

$|z|=\sqrt{\dfrac{1}{36}+\dfrac{1}{9}}=\dfrac{\sqrt{5}}{6}$

principal Arg(2)=

$\pi tan^{-1}\left ( \dfrac{-V_{3}}{V6} \right )$

$\pi tan^{-1}(-2)$

$=\pi -tan^{-1}(2)$

$=\pi -63.44$

and arg(2)

$= 2K\pi +\pi -63.44$

$=2(K+1)\pi -63.44$

Convert the complex number $\dfrac {-16}{1+i\sqrt {3}}$ into polar form.

Let $z$ be the given number,

$z = \dfrac{{ - 16}}{{1 + i\sqrt 3 }}$

$= \dfrac{{ - 16}}{{1 + i\sqrt 3 }} \times \dfrac{{1 - i\sqrt 3 }}{{1 - i\sqrt 3 }}$

$= \dfrac{{ - 16\left( {1 - i\sqrt 3 } \right)}}{{{{\left( 1 \right)}^2} - {{\left( {i\sqrt 3 } \right)}^2}}}$

$= \dfrac{{ - 16\left( {1 - i\sqrt 3 } \right)}}{{1 + 3}}$

$= - 4\left( {1 - i\sqrt 3 } \right)$

$= - 4 + 4i\sqrt 3$

$r = \sqrt {{a^2} + {b^2}}$

$= \sqrt {{{\left( { - 4} \right)}^2} + {{\left( {4\sqrt 3 } \right)}^2}}$

$= \sqrt {64}$

$= 8$

$\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$

$= {\tan ^{ - 1}}\left( {\dfrac{{4\sqrt 3 }}{{ - 4}}} \right)$

$= {\tan ^{ - 1}}\left( { - \sqrt 3 } \right)$

$= - \dfrac{\pi }{3}$

The polar form of the complex number is,

$z = r\left( {\cos \theta + i\sin \theta } \right)$

$z = 8\left( {\cos \left( { - \dfrac{\pi }{3}} \right) + i\sin \left( {\dfrac{{ - \pi }}{3}} \right)} \right)$

$z = 8\left( {\cos \dfrac{\pi }{3} - i\sin \dfrac{\pi }{3}} \right)$

Therefore, the complex form of the given complex number is $z = 8\left( {\cos \dfrac{\pi }{3} - i\sin \dfrac{\pi }{3}} \right)$ .

Find the modulus and argument of the complex number:
$\dfrac{1+i}{1-i}$

Let

$z=\dfrac { 1+i }{ 1-i }$

Rationalizing the same,

$z=\dfrac { 1+i }{ 1-i } *\dfrac { 1+i }{ 1+i }$

$z=\dfrac { (1+i)\quad (1+i) }{ (1-i)\quad (1+i) }$

$Using (a-b)(a+b)= { a }^{ 2 }-{ b }^{ 2 }$

$=\dfrac { { (1+i) }^{ 2 } }{ { 1 }^{ 2 }-{ i }^{ 2 } }$

$Using { (a+b) }^{ 2= }{ a }^{ 2 }+{ b }^{ 2 }+2ab$

$=\dfrac { { 1 }^{ 2 }+{ i }^{ 2 }+2i }{ { 1 }^{ 2 }-{ i }^{ 2 } }$

$Putting\quad { i }^{ 2 }=-1$

$=\dfrac { { 1 }^{ 2 }+(-1)+2i }{ 1-(-1) }$

$=\dfrac { 2i }{ 2 }$

$=i$

$=0+i$

$Hence z=(0+i)$

To calculate modulus of z,

$z=(0+i)$

$Complex number z is of the form x+iy$

$Hence x=0 and y=1$

Modulus of

$z=\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

$z=\sqrt { { 0 }^{ 2 }+{ 1 }^{ 2 } }$

$z=\sqrt { 0+1 }$

$z=\sqrt { 1 }$

To find the argument,

$0+i=rcos\theta +irsin\theta$

Comparing real part,

$0=rcos\theta$

Put r=1

$0=1*cos\theta$

$0=cos\theta$

$cos\theta =0$

Comparing imaginary part,

$1=rsin\theta$

Put r=1

$1=1*sin\theta$

$1=sin\theta$

$sin\theta =1$

$Hence cos\theta =0\quad and\quad sin\theta =1$

Express the complex number given in the form $a+ib$ .
$i^{-39}$ .

${ i }^{ -39 }\\ =i({ i }^{ -40 })\\ =i(({ i }^{ 2 })^{ -20 })\\ =i((-1)^{ -20 })\quad \quad \quad \quad [\because { i }^{ 2 }-1]\\ =i\left( \cfrac { 1 }{ (-1)^{ 20 } } \right) \\ =i\left( \cfrac { 1 }{ 1 } \right) =i=0+i(1)$

Comparing with

$a+ib$ ,

$a=0$ ,

$b=1$

Ans=

$0+i(1)$

Express the complex number given in the form $a+ib$ .
$i^9+i^{19}$ .

${ i }^{ 9 }+{ i }^{ 19 }\\ =i({ i }^{ 8 }+{ i }^{ 18 })\\ =i(({ i }^{ 2 })^{ 4 }+({ i }^{ 2 })^{ 9 }\\ =i((-1)^{ 4 }+(-1)^{ 9 })\quad \quad [\because { i }^{ 2 }=-1]\\ =i(1-1)\\ =i(0)\\ =0\\ =0+i(0)$

Comparing with

$a+ib$ ,

$a=0$ ,

$b=0$

Ans

$0+i(0)$

Find the modulus and argument of $z=\dfrac{3+2i}{-2+i}$ .

$z=\dfrac{3+2i}{-2+i}$

$z=\dfrac{3+2i}{-2+c}\times \dfrac{(i+2)}{(i+2)}=\dfrac{3i+4i-2+6}{-1-4}$

$z=\dfrac{4+7i}{-5}$

$z=-\dfrac{4}{5}+(\dfrac{-7i}{5})$

$121=\sqrt{\dfrac{16}{25}+\dfrac{49}{25}}=\sqrt{\dfrac{65}{25}}=\dfrac{\sqrt{65}}{5}$

$Arg(2)=-\pi +\tan^{-1}\left(\dfrac{7/5}{4/5}\right)$

$=-\pi +\tan^{-1}\left(\dfrac{7}{4}\right)$

$arg(2)=2k\pi -\pi +\tan^{-1}(7/4)$

$arg(2)=2(k-1)\pi +\tan^{-1}(7/4)$

Prove that the equation ${x}^{2}+px-1=0$ has real and distinct roots for all real values of $p$ .

To prove: The equation

$x^2+px-1=0$ has real and distinct roots for all real values of

$p$ .

Consider

$x^2+px-1=0$

Discriminant

$D=p^2-4(1)(-1)=p^2+4$

We know

$p^2\geq 0$ for all values of

$p$

$\Rightarrow p^2+4\geq 0$ (since

$4>0$ )

Therefore

$D\geq 0$

Hence the equation

$x^2+px-1=0$ has real and distinct roots for all real values of

$p$ .

If $(2+i)(2+2i)(2+3i).....(2+ni)=x+iy$ then the value of $5.8.13.....(4+n^2)$ .

Given,

$(2+i)(2+2i)(2+3i).....(2+ni)=x+iy$

Now taking modulus both sides we get,

$|(2+i)(2+2i)(2+3i).....(2+ni)|=|x+iy|$

or,

$|(2+i)|.|(2+2i)|.|(2+3i)|.....|(2+ni)|=|x+iy|$ [ Since

$|ab|=|a|.|b|$ ]

Now squaring both sides we get,

or,

$|(2+i)|^2.|(2+2i)|^2.|(2+3i)|^2.....|(2+ni)|^2=|x+iy|^2$

or,

$5.8.13......(4+n^2)=x^2+y^2$ .

Real numbers $x$ and $y$ satisfy the equation
$\dfrac{x}{1+i}+\dfrac{y}{1-i}=\dfrac{148}{12+2i}.$
What is the value of $xy$ ?

$\cfrac { x }{ 1+i } +\cfrac { y }{ 1-i } =\cfrac { 148 }{ 12+2i } =\cfrac { 148 }{ 2\left( 6+i \right) }$

$\Rightarrow \cfrac { x }{ 1+i } +\cfrac { y }{ 1-i } =\cfrac { 74 }{ 6+i }$

$\Rightarrow \cfrac { x }{ 1+i } \times \cfrac { 1-i }{ 1-i } +\cfrac { y }{ 1-i } \times \cfrac { 1+i }{ 1+i } =\cfrac { 74 }{ 6+i } \times \cfrac { 6-i }{ 6-i }$

$\Rightarrow \cfrac { x\left( 1-i \right) }{ 1-{ i }^{ 2 } } +\cfrac { y\left( 1+i \right) }{ 1-{ i }^{ 2 } } =\cfrac { 74\left( 6-i \right) }{ 36-{ i }^{ 2 } }$

${ i }^{ 2 }=-1$

$\Rightarrow \cfrac { x\left( 1-i \right) }{ 2 } +\cfrac { y\left( 1+i \right) }{ 2 } =\cfrac { 74\left( 6-i \right) }{ 37 }$

$\Rightarrow \cfrac { x\left( 1-i \right) +y\left( 1+i \right) }{ 2 } =2\left( 6-i \right)$

$\Rightarrow x-xi+y+yi=4(6-i)$

$\Rightarrow (x+y)+(y-x)i=24-4i$

Comparing the real and imaginary parts

$\Rightarrow x+y=24;$ and

$y-x=-4$

adding both

$\Rightarrow 2y = 20$

$\Rightarrow y = 10$

$\Rightarrow x = 24 - y = 24 - 10$

$\Rightarrow x = 14$ and

$y = 10$

$\Rightarrow xy = 14(10) = 140$

Evaluate : $[i^{19}+(\frac {1}{i}^{32}]^2$

1338015_1274798_ans_18e1eeafdb144b0d8fab50b7c58e3d30.PNG

Find the real value of $\theta$ such that $\dfrac{3+2i\sin{\theta}}{1-2i\sin{\theta}}$ , where $i=\sqrt{-1}$ is
$(i)$ purely real
$(ii)$ Purely imaginary.

1350068_1241063_ans_b6932e2fc391478eb85a8b1359f9779b.PNG

Find the value of $k$ for which the equation $x ^ { 2 } + k ( 2 x + k - 1 ) + 2 = 0$ has real and equal roots.

The given equation is

${x}^{2}+k\left(2x+k-1\right)+2=0$

$\Rightarrow {x}^{2}+2kx+\left({k}^{2}-k+2\right)=0$

So,

$a=1,b=2k,c={k}^{2}-k+2$

We know

$D={b}^{2}-4ac$

$={\left(2k\right)}^{2}-4\times 1\times\left({k}^{2}-k+2\right)$

$=4{k}^{2}-4\left({k}^{2}-k+2\right)$

$=4{k}^{2}-4{k}^{2}+4k-8$

$=4k-8=4\left(k-2\right)$

For equal roots,

$D=0$

Thus,

$4\left(k-2\right)=0$

So,

$k=2$

What is the nature of the roots of the quadratic equation $4 x^{2}-12 x-9=0?$

2040870_1453033_ans_0e8d40b2239049bb8836e8fb5dcbace8.JPG

Complex Numbers And Quadratic Equations - Class 11 Engineering Maths - Extra Questions

Number of roots of the quadratic equation 8sec2x−6secx+1=0\displaystyle 8\sec ^{2}x-6\sec x+1= 0 is/are:

|z|≤5\left | z \right |\leq 5 & Arg(z−1−i)>π/6Arg\left ( z-1-i \right )> \pi /6

Arg(z−1−i)≥π/3Arg \left ( z-1-i \right )\geq \pi /3

Multiply 2√−3+3√−22\sqrt { -3 } +3\sqrt { -2 } by 4√−3−5√−24\sqrt { -3 } -5\sqrt { -2 }

If the value of the discriminant of the quadratic equation ax2+bx+c=0a{ x }^{ 2 }+bx+c=0 is less than 00, then the nature of the roots is ____

If Z1=−1{Z}_{1}=-1 and Z2=i{Z}_{2}=i, then find Arg(Z1Z2)Arg\left( \cfrac { { Z }_{ 1 } }{ { Z }_{ 2 } } \right)

Write an example for a quadratic polynomial that has no real zeroes.

Without solving the following quadratic equation, find the value of mm for which the given equation has real and equal roots.x2+2(m−1)x+(m+5)=0x^2+2(m-1)x+(m+5)=0

Calculate √i\displaystyle \sqrt{i}

What is cis 0cis\ 0?

Find the value of kk so that the following equation has equal roots.2x2−(k−2)x+1=02x^{2} - (k - 2) x+1 = 0

Calculate 3√−i\displaystyle \sqrt[3]{- \, i}

Represent (−1−√3i)(-1-\sqrt 3i) in the polar form.

For the real numbers x and y, if (x−iy)(3+5i)(x-iy)(3+5i) is the conjugate of −6−24i-6-24i then find the value of xx and yy.

Indicate the point of the complex plane zz which satisfy the following equation.\displaystyle z^2 \, + \, | \, \overline{z} \, | \, = \, 0\displaystyle z^2 \, + \, | \, \overline{z} \, | \, = \, 0

Find the values of p for which the quadratic equation 4x^2+px+3=04x^2+px+3=0 has equal roots.

Find the modulus of the complex number 2-5!2-5!.

Determine the nature of the roots of the following quadratic equation2x^2+ 6x + 3 = 02x^2+ 6x + 3 = 0

Find the length of the segments connecting the points represented by the following pairs of numbers :3, -4i

locate the point representing the complex numbers zz on the Argand diagram for which\left| z \right| \ge 3,\left| z \right| \ge 3,

Simplify the following :\dfrac{3 \, - \, i}{2 \, + \, i} \, + \, \dfrac{3 \, + \, i}{2 \, - \, i}\dfrac{3 \, - \, i}{2 \, + \, i} \, + \, \dfrac{3 \, + \, i}{2 \, - \, i}

Find the length of the segments connecting the points represented by the following pairs of numbers :-1 - i, 2 + 3i

x \, +\, i \sqrt x^\, 4 \, +, x^2 \, +\, 1x \, +\, i \sqrt x^\, 4 \, +, x^2 \, +\, 1

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :{x^2} = 9{x^2} = 9

What is value of (-i)^{12}(-i)^{12}

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.{ x }^{ 2 }+5x+5=0{ x }^{ 2 }+5x+5=0

Determine the nature of the roots of the given quadratic equation2{ x }^{ 2 }+x-1=02{ x }^{ 2 }+x-1=0

Determine the nature of the roots of the given quadratic equation4{ x }^{ 2 }-4x+1=0\quad 4{ x }^{ 2 }-4x+1=0\quad

Solvez = \left( {2 + i} \right)+\left( {1 - 3i} \right)z = \left( {2 + i} \right)+\left( {1 - 3i} \right)

Find the modulus of: (i) {{21} \over 5} - {{12} \over 5}i{{21} \over 5} - {{12} \over 5}i (ii) 3+4i

Find the value of (1+i)(1+{i}^{2})(1+{i}^{3})(1+{i}^{4})(1+i)(1+{i}^{2})(1+{i}^{3})(1+{i}^{4}).

Find the amplitude and modulus of following(i) \displaystyle {{21} \over 5} - {{12} \over 5}i\displaystyle {{21} \over 5} - {{12} \over 5}i (ii) 3 + 4{\rm{ i}}3 + 4{\rm{ i}}

Express the given complex number 3(7+i7)+i(7+i7)3(7+i7)+i(7+i7) in the form a+iba+ib,

Find the amplitude of -4-4.

Find modulus and argument of ii

Write the nature of roots of the quadratic equation 5x^2-2x+3=05x^2-2x+3=0.

Write the argument of (1+\sqrt {3})(1+i)(\cos \theta+i\sin \theta)(1+\sqrt {3})(1+i)(\cos \theta+i\sin \theta).

The values of xx and yy satisfying the equation \dfrac{(1+i)x - 2i}{3+ i} + \dfrac{(2 -3i)y + i}{3 - i} = i\dfrac{(1+i)x - 2i}{3+ i} + \dfrac{(2 -3i)y + i}{3 - i} = i, are

Solve for xx : 4\sqrt 3 x^{2} + 5x-2\sqrt 3 = 04\sqrt 3 x^{2} + 5x-2\sqrt 3 = 0 Write about the nature of roots.

Express (1-i)-(1+i6)(1-i)-(1+i6) as a+iba+ib

If \dfrac {a-ib}{a+ib}=\dfrac {1+i}{1-i}\dfrac {a-ib}{a+ib}=\dfrac {1+i}{1-i}, then prove that a+b=0a+b=0

Solve the equation \left| z \right| + z = 2 + i\left| z \right| + z = 2 + i

Convert the given complex number into the polar form:z = -3z = -3

If {z_1} = 1 + i = \sqrt 3 + i{z_1} = 1 + i = \sqrt 3 + i, then the principle arg \left( {\frac{{{z_1}}}{{{z_2}}}} \right)\left( {\frac{{{z_1}}}{{{z_2}}}} \right)

If a+ib=\cfrac { { (x+i) }^{ 2 } }{ 2{ x }^{ 2 }+1 } a+ib=\cfrac { { (x+i) }^{ 2 } }{ 2{ x }^{ 2 }+1 } , prove that { a }^{ 2 }+{ b }^{ 2 }=\cfrac { { (x+i) }^{ 2 } }{ { (2{ x }^{ 2 }+1) }^{ 2 } } { a }^{ 2 }+{ b }^{ 2 }=\cfrac { { (x+i) }^{ 2 } }{ { (2{ x }^{ 2 }+1) }^{ 2 } } .

Solve the problem:-\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right)\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right)

For what value of kk, the equation kx^2 - 6x - 2 = 0kx^2 - 6x - 2 = 0 has equal roots.

Evaluate: \left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3.

If z=x+iyz=x+iy and z_1=1+2iz_1=1+2i, determine the region in the complex plane represented by 1 < |z-z_1 |< 31 < |z-z_1 |< 3. Represent it on the Argand plane.

Determine the nature of roots of the following quadratic equation :2x^2+5x+5=02x^2+5x+5=0

If z_1=2-i, z_2=1+iz_1=2-i, z_2=1+i, find \left|\dfrac{z_1+z_2+1}{z_1-z_2+i}\right|\left|\dfrac{z_1+z_2+1}{z_1-z_2+i}\right|.

Find the values of the following:{ \left( 1+i\sqrt { 3 } \right) }^{ 3 }{ \left( 1+i\sqrt { 3 } \right) }^{ 3 }

Solve (3-7i)+(-2+4i)(3-7i)+(-2+4i).

If \alpha\alpha and \beta\beta are the zeroes of the equation 6{x^2} + x - 2 = 06{x^2} + x - 2 = 0, find \dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }

\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}.

Write principal argument of \dfrac {-\sqrt {11}i}{17}\dfrac {-\sqrt {11}i}{17}.

Find the modulus:-4-4i-4-4i

If a=\frac { -1+\sqrt { 3i } }{ 2 } ,b=\frac { -1-\sqrt { 3i } }{ 2 }a=\frac { -1+\sqrt { 3i } }{ 2 } ,b=\frac { -1-\sqrt { 3i } }{ 2 } then show that {a}^{2}=b{a}^{2}=b and {b}^{2}=a{b}^{2}=a.

Represent following complex numbers z_1=1+2iz_1=1+2i and z_2=5-7iz_2=5-7i by points in Argand's diagram and determine their amplitudes approximately.

If g\left( x \right) ={ x }^{ 4 }{ -x }^{ 3 }+{ x }^{ 2 }+3x-5g\left( x \right) ={ x }^{ 4 }{ -x }^{ 3 }+{ x }^{ 2 }+3x-5, find g(2+3i)g(2+3i)

What will be the nature of roots of the quadratic equation 2{x^2} + 4x + 7 = 02{x^2} + 4x + 7 = 0.

Find the modulus of \dfrac {3 -4i}{5 + 7i}\dfrac {3 -4i}{5 + 7i}.

Write the nature of roots of quadratic equation 4{ x }^{ 2 }+6x+3=04{ x }^{ 2 }+6x+3=0

Find the amplitude of 1 + i \sqrt{3} 1 + i \sqrt{3}.

Find the modulus of the complex number z=2+3iz=2+3i.

Find discriminant of the equation 5x - 6 = \dfrac{1}{x}5x - 6 = \dfrac{1}{x}.

If a+ib=\dfrac{c+i}{c-i}a+ib=\dfrac{c+i}{c-i}, where cc is a real number, then prove that : a^{2}+b^{2}=1a^{2}+b^{2}=1 and \dfrac{b}{a}=\dfrac{2c}{c^{2}-1}\dfrac{b}{a}=\dfrac{2c}{c^{2}-1}.

Write the nature of roots from the given values of the discriminants and complete the activity.

Find the modulus and amplitude of 4+3i4+3i.

Write (i) -1 -i-1 -i (ii) 1 - i1 - i in polar form.

Simplify and hence find the modulus and argument \dfrac {(3 + i)(3 - i)}{2 + i} \dfrac {(3 + i)(3 - i)}{2 + i} .

In -3+3i-3+3i, find amplitude and modulus

If the discriminant is 13, how many solutions and of what type?

Express \dfrac{-1+i}{\sqrt{2}}\dfrac{-1+i}{\sqrt{2}} in the polar form

If z=(\cos \theta, \sin \theta))z=(\cos \theta, \sin \theta)), find \left(z-\dfrac{1}{z}\right)\left(z-\dfrac{1}{z}\right)

Find the nature of roots of the quadratic equation { y }^{ 2 }-7y+2=0{ y }^{ 2 }-7y+2=0.

Simplify: \left(\dfrac{1}{3}+3i\right)^{3}\left(\dfrac{1}{3}+3i\right)^{3}

If \omega =\frac { Z }{ \bar { Z } } \omega =\frac { Z }{ \bar { Z } } , then |\omega |=|\omega |=

If z_1=5+3i\\z_2=2-3iz_1=5+3i\\z_2=2-3i Find z_1+z_2z_1+z_2

Number of roots of the quadratic equation $\displaystyle 8\sec ^{2}x-6\sec x+1= 0$ is/are:

$\left | z \right |\leq 5$ & $Arg\left ( z-1-i \right )> \pi /6$

$Arg \left ( z-1-i \right )\geq \pi /3$

Multiply $2\sqrt { -3 } +3\sqrt { -2 }$ by $4\sqrt { -3 } -5\sqrt { -2 }$

If the value of the discriminant of the quadratic equation $a{ x }^{ 2 }+bx+c=0$ is less than $0$ , then the nature of the roots is ____

If ${Z}_{1}=-1$ and ${Z}_{2}=i$ , then find $Arg\left( \cfrac { { Z }_{ 1 } }{ { Z }_{ 2 } } \right)$

Without solving the following quadratic equation, find the value of $m$ for which the given equation has real and equal roots.
$x^2+2(m-1)x+(m+5)=0$

Calculate $\displaystyle \sqrt{i}$

What is $cis\ 0$ ?

Find the value of $k$ so that the following equation has equal roots.
$2x^{2} - (k - 2) x+1 = 0$

Calculate $\displaystyle \sqrt[3]{- \, i}$

Represent $(-1-\sqrt 3i)$ in the polar form.

For the real numbers x and y, if $(x-iy)(3+5i)$ is the conjugate of $-6-24i$ then find the value of $x$ and $y$ .

Indicate the point of the complex plane $z$ which satisfy the following equation.
$\displaystyle z^2 \, + \, | \, \overline{z} \, | \, = \, 0$

Find the values of p for which the quadratic equation $4x^2+px+3=0$ has equal roots.

Find the modulus of the complex number $2-5!$ .

Determine the nature of the roots of the following quadratic equation
$2x^2+ 6x + 3 = 0$

Find the length of the segments connecting the points represented by the following pairs of numbers :
3, -4i

locate the point representing the complex numbers $z$ on the Argand diagram for which
$\left| z \right| \ge 3,$

Simplify the following :
$\dfrac{3 \, - \, i}{2 \, + \, i} \, + \, \dfrac{3 \, + \, i}{2 \, - \, i}$

Find the length of the segments connecting the points represented by the following pairs of numbers :
-1 - i, 2 + 3i

$x \, +\, i \sqrt x^\, 4 \, +, x^2 \, +\, 1$

Find the discriminant of the following quadratic equations and hence determine the nature of the roots of the equation :
${x^2} = 9$

What is value of $(-i)^{12}$

In the given, determine whether the given quadratic equation has real roots and if so, find the roots.
${ x }^{ 2 }+5x+5=0$

Determine the nature of the roots of the given quadratic equation
$2{ x }^{ 2 }+x-1=0$

Determine the nature of the roots of the given quadratic equation
$4{ x }^{ 2 }-4x+1=0\quad$

Solve
$z = \left( {2 + i} \right)+\left( {1 - 3i} \right)$

Find the modulus of: (i) ${{21} \over 5} - {{12} \over 5}i$ (ii) 3+4i

Find the value of $(1+i)(1+{i}^{2})(1+{i}^{3})(1+{i}^{4})$ .

Find the amplitude and modulus of following
(i) $\displaystyle {{21} \over 5} - {{12} \over 5}i$ (ii) $3 + 4{\rm{ i}}$

Express the given complex number $3(7+i7)+i(7+i7)$ in the form $a+ib$ ,

Find the amplitude of $-4$ .

Find modulus and argument of $i$

Write the nature of roots of the quadratic equation $5x^2-2x+3=0$ .

Write the argument of $(1+\sqrt {3})(1+i)(\cos \theta+i\sin \theta)$ .

The values of $x$ and $y$ satisfying the equation $\dfrac{(1+i)x - 2i}{3+ i} + \dfrac{(2 -3i)y + i}{3 - i} = i$ , are

Solve for $x$ : $4\sqrt 3 x^{2} + 5x-2\sqrt 3 = 0$ Write about the nature of roots.

Express $(1-i)-(1+i6)$ as $a+ib$

If $\dfrac {a-ib}{a+ib}=\dfrac {1+i}{1-i}$ , then prove that $a+b=0$

Solve the equation $\left| z \right| + z = 2 + i$

Convert the given complex number into the polar form: $z = -3$

If ${z_1} = 1 + i = \sqrt 3 + i$ , then the principle arg $\left( {\frac{{{z_1}}}{{{z_2}}}} \right)$

If $a+ib=\cfrac { { (x+i) }^{ 2 } }{ 2{ x }^{ 2 }+1 }$ , prove that ${ a }^{ 2 }+{ b }^{ 2 }=\cfrac { { (x+i) }^{ 2 } }{ { (2{ x }^{ 2 }+1) }^{ 2 } }$ .

Solve the problem:-
$\left( {\frac{1}{5} + i\frac{2}{5}} \right) - \left( {4 + i\frac{5}{2}} \right)$

For what value of $k$ , the equation $kx^2 - 6x - 2 = 0$ has equal roots.

Evaluate: $\left[i^{18}+\left(\dfrac{1}{i}\right)^{25}\right]^3$ .

If $z=x+iy$ and $z_1=1+2i$ , determine the region in the complex plane represented by $1 < |z-z_1 |< 3$ . Represent it on the Argand plane.

Determine the nature of roots of the following quadratic equation : $2x^2+5x+5=0$

If $z_1=2-i, z_2=1+i$ , find $\left|\dfrac{z_1+z_2+1}{z_1-z_2+i}\right|$ .

Find the values of the following:
${ \left( 1+i\sqrt { 3 } \right) }^{ 3 }$

Solve $(3-7i)+(-2+4i)$ .

If $\alpha$ and $\beta$ are the zeroes of the equation $6{x^2} + x - 2 = 0$ , find $\dfrac{\alpha }{\beta } + \dfrac{\beta }{\alpha }$

$\dfrac{2+5i}{3-2i}+\dfrac{2-5i}{3+2i}$ .

Write principal argument of $\dfrac {-\sqrt {11}i}{17}$ .

Find the modulus:
$-4-4i$

If $a=\frac { -1+\sqrt { 3i } }{ 2 } ,b=\frac { -1-\sqrt { 3i } }{ 2 }$ then show that ${a}^{2}=b$ and ${b}^{2}=a$ .

Represent following complex numbers $z_1=1+2i$ and $z_2=5-7i$ by points in Argand's diagram and determine their amplitudes approximately.

If $g\left( x \right) ={ x }^{ 4 }{ -x }^{ 3 }+{ x }^{ 2 }+3x-5$ , find $g(2+3i)$

What will be the nature of roots of the quadratic equation $2{x^2} + 4x + 7 = 0$ .

Find the modulus of $\dfrac {3 -4i}{5 + 7i}$ .

Write the nature of roots of quadratic equation $4{ x }^{ 2 }+6x+3=0$

Find the amplitude of $1 + i \sqrt{3}$ .

Find the modulus of the complex number $z=2+3i$ .

Find discriminant of the equation $5x - 6 = \dfrac{1}{x}$ .

If $a+ib=\dfrac{c+i}{c-i}$ , where $c$ is a real number, then prove that : $a^{2}+b^{2}=1$ and $\dfrac{b}{a}=\dfrac{2c}{c^{2}-1}$ .

Find the modulus and amplitude of $4+3i$ .

Write (i) $-1 -i$ (ii) $1 - i$ in polar form.

Simplify and hence find the modulus and argument $\dfrac {(3 + i)(3 - i)}{2 + i}$ .

In $-3+3i$ , find amplitude and modulus

Express $\dfrac{-1+i}{\sqrt{2}}$ in the polar form

If $z=(\cos \theta, \sin \theta))$ , find $\left(z-\dfrac{1}{z}\right)$

Find the nature of roots of the quadratic equation ${ y }^{ 2 }-7y+2=0$ .

Simplify: $\left(\dfrac{1}{3}+3i\right)^{3}$

If $\omega =\frac { Z }{ \bar { Z } }$ , then $|\omega |=$

If $z_1=5+3i\\z_2=2-3i$ Find $z_1+z_2$

$3 \sqrt { 2 } x ^ { 2 } - 2 \sqrt { 6 } x + 2$ find the nature of roots.

Check whether the following equation will have two distinct real or imaginary or two real equal roots.
$x^2-3x+5=0$ .

Prove that the roots of the following quadratic equation has two real distinct roots.
$2x^2-6x+3=0$ .

Find the condition for which the roots of the equation $ax^{2}+bx+c=0$ be real and distinct.