If $$z = \cos \dfrac { \pi } { 4 } + i \sin \dfrac { \pi } { 6 } ,$$ then

$$| z | = 1 , \arg ( z ) = \dfrac { \pi } { 4 }$$

$$| z | = 1 , \arg ( z ) = \dfrac { \pi } { 6 }$$

$$| z | = \dfrac { \sqrt { 3 } } { 2 } , \arg ( z ) = \dfrac { 5 \pi } { 24 }$$

$$| z | = \dfrac { \sqrt { 3 } } { 2 } , \arg ( z ) = \tan ^ { - 1 } \dfrac { 1 } { \sqrt { 2 } }$$

For any complex numbers $$z_{1}$$ and $$z_{2}$$ compare List I with with List II and choose the correct answer, using codes given below: List I List II $$arg (z_{1},z_{2})$$ $$\dfrac{\pi}{2}$$ $$arg \left(\dfrac{z_{1}}{z_{2}}\right)$$ $$arg (z_{1}-arg (z_{2})$$ $$arg (z)+arg (\bar{z})$$ $$arg (z_{1})+arg (z_{2})$$ $$arg (i)$$ $$2\pi$$

$$(i)-(r), (ii)-(q), (iii)-(s), (iv)-(p)$$

$$(i)-(q), (ii)-(r), (iii)-(s), (iv)-(p)$$

$$(i)-(r), (ii)-(q), (iii)-(p), (iv)-(s)$$

Compare List I with List II and choose the correct answer using codes given below: List I (Complex number) List II (Its modulus) $$(4-3i)$$ $$10$$ $$(8+6i)$$ $$\dfrac{1}{5}$$ $$\dfrac{1}{(3+4i)}$$ $$1$$ $$\dfrac{(3-4i)}{(3+4i)}$$ $$5$$

$$(i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)$$

$$(i)-(p), (ii)-(s), (iii)-(r), (iv)-(q)$$

$$(i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)$$

$$(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)$$

MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 9

$z_{1}=8 +4i,\ z_{2}=6+4i$ and

$arg \left(\dfrac {z-z_{1}}{z-z_{2}}\right)=\dfrac {\pi}{4}$ , then

$z$ satisfy

0%
$|z-7-4i|=1$
0%
$|z-7-5i|=\sqrt {2}$
0%
$|z-4i|=8$
0%
$|z-7i|=\sqrt {18}$

$a, b \notin R$ , then

$|e^{a + ib}|$ is equal to

0%
$e^a$
0%
$e^b$
0%
$1$
0%
None of these

Explanation

$|e^{a + ib}| = |e^a . e^{ib}|$

$= |e^a . (\cos b + i \sin b)|$

$= e^a |(\cos b + i \sin b)|$

$= e^a . \sqrt{(\cos b)^2 + (\sin b)^2}$

$= e^a . \sqrt{1}$

$= e^a$

$z_{1}$ and

$z_{2}$ two non-zero complex number such that

$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$ , then

$arg z_{1}-arg z_{2}$ is equal to

0%
$-p$
0%
$p/2$
0%
$-p/2$
0%
$0$

Explanation

$\Rightarrow z_1 \;\&\; z_2$ are collinear and in same direction.

$\Rightarrow arg \left(z_1\right)-arg\left(z_2\right)=0.$

Hence, the answer is

$0.$

$z$ is a complex number. If

$a = | x | + | y |$ and

$b = \sqrt { 2 } | x + i y |$ then which of the following is
true

0%
$a \leq b$
0%
$a > b$
0%
none of these
0%
$a - b + 2$

If arg

$\left( z \right) < 0,$ then arg

$\left( { - z} \right)-arg(z)$

0%
$\pi$
0%
$- \pi$
0%
$- \dfrac{\pi }{2}$
0%
$\dfrac{\pi }{2}$

Number of complex numbers

$z$ such that

$|z|=1$ and

$\left|\dfrac {z}{z}+\dfrac {\bar {z}}{z}\right|=1$ is

0%
$4$
0%
$1$
0%
$8$
0%
$more\ then\ 8$

Explanation

Since

$z$ is equidistant from

$1,-1,i$ , we can say that

$z$ is center of circle with

$1,-1,i$ lying on it's circumference and so there is one one circle possible passing through

$3$ points. So, the answer is

$1$ .

The modulus of

$\dfrac { \left( 3+2i \right) ^{ 2 } }{ \left( 4-3i \right) }$ is:

0%
$\frac { 13 }{ 5 }$
0%
$\frac { 11 }{ 5 }$
0%
$\frac { 9 }{ 5 }$
0%
$\frac { 7 }{ 5 }$

Let P

$\left( x \right) ={ x }^{ 3 }-6{ x }^{ 2 }+Bx+C$ has 1+5i as a zero and B,C real number, then value of (B+C) is

0%
-70
0%
70
0%
24
0%
138

Explanation

Given

$1+5{i}$ is the root of

$P(x)=x^3-6{x^2}+B{x}+C$

$\implies 1-5{i}$ is also the root of

$P(x)$

Let the third root be

$h$

$\implies$ sum of roots is

$6$

$\implies 1+5{i}+1-5{i}+h=6\implies h=4$

$B=(1+5{i}+1-5{i})(4)+(1+5{i})(1-5 i)=8+26=34$

$C=-(1+5{i})(1-5{i})(4)=-104$

$B+C=34-104=-70$

Arg

$\left\{ {\sin \frac{{8\pi }}{5} + i\left( {1 + \cos \frac{{8\pi }}{5}} \right)} \right\}$ is equal to

0%
$\frac{{3\pi }}{{10}}$
0%
$\frac{{7\pi }}{{10}}$
0%
$\frac{{4\pi }}{5}$
0%
$\frac{{3\pi }}{5}$

A value of

$\theta$ for which

$\dfrac { 2+3isin\theta }{ 1-2isin\theta }$ is purely imaginary, is:

0%
${ sin }^{ -1 }\left( \dfrac { 1 }{ \sqrt { 3 } } \right)$
0%
$\dfrac { \pi }{ 3 }$
0%
$cos^{-1}\sqrt-1$
0%
$None of these$

Explanation

We have,

$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }$

On rationalize and we get,

$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta }$

$\Rightarrow \dfrac{2+3i\sin \theta +4i\sin \theta +6{{i}^{2}}{{\sin }^{2}}\theta }{{{1}^{2}}-4{{i}^{2}}{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{2-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+1-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{7\sin \theta }{1+4{{\sin }^{2}}\theta }$

Now,

Then purely imaginary is

Real part of equal to zero.

$\dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0$

$\Rightarrow 1+{{\cos }^{2}}\theta =0$

$\Rightarrow {{\cos }^{2}}\theta =-1$

$\Rightarrow \cos \theta =\sqrt{-1}$

$\Rightarrow \theta ={{\cos }^{-1}}\sqrt{-1}$

Hence, this is the answer.

Purely imaginary then find the sum of statement i

$a,b$

0%
$\dfrac {5\pi}{6}$
0%
$\pi$
0%
$\dfrac {3\pi}{4}$
0%
$\dfrac {2\pi}{3}$

$\alpha$ and

$\beta$ are the roots of

${ 4x }^{ 2 }-16x+c=0,$ c>0 such that

$1<\alpha <2<\beta <3$ , then the no.of integer values of c is

0%
$17$
0%
$14$
0%
$18$
0%
$15$

Explanation

Given,

$\alpha$ and

$\beta$ are roots of

$4x^2-16x+x=0$

where

$a=4$

$b=-16$

$x > 0$

now,

$\alpha\beta =\dfrac{a}{a}$

$\Rightarrow \alpha\beta =\dfrac{c}{4}$

$\Rightarrow \beta =\dfrac{c}{4\alpha}$ ………(i)

we have

$2 < \beta < 3$

$\Rightarrow 2 < \dfrac{c}{4\alpha} < 3$ [from

$(1)$ ]

$\Rightarrow 8 < \dfrac{c}{\alpha} < 12$

$\Rightarrow 8\alpha < c < 12\alpha$ ………(ii)

and

$1 < \alpha < 2$

$\Rightarrow \alpha > 1\Rightarrow 8\alpha > 8$ ……….(iii)

and

$\alpha < 2$

$\Rightarrow 12\alpha < 24$ ………..(iv)

Substituting (iii) and (iv) in (ii) we get

$8 < 8\alpha < c < 12\alpha < 24$

$[\alpha =9, 10, 11, 12, 13, 14, 15, 16, 13, 18, 19, 20, 21, 22, 23]$

$\Rightarrow 8 < c < 24$

Therefore number of integer values of c is

$15$ .

1196924_1318670_ans_ab0c4184d5f446658aeacf0a4a2d6ac4.JPG

The principle amplitude of

$(\sin 40^{o}+i \cos 40^{o})^{5}$ is

0%
$70^{o}$
0%
$-1100^{o}$
0%
$70^{110}$
0%
$70^{-70}$

Explanation

$40°=2\cfrac { \pi }{ 9 } ,50°=\cfrac { 5\pi }{ 18 }$

So,

$\left( \sin { 40° } +i\cos { 40° } \right) =\cos { 50° } +i\sin { 50° }$

Let

$T={ \left( \sin { 40° } +i\cos { 40° } \right) }^{ 5 }={ \left( \cos { 50° } +i\sin { 50° } \right) }^{ 5 }$

$T={ \left( { e }^{ i50° } \right) }^{ 5 }=\left( { e }^{ i\cfrac { 5\pi }{ 18 } \times 5 } \right) ={ e }^{ i\cfrac { 25\pi }{ 18 } }$

So,

$25\cfrac { \pi }{ 18 } =250°$

Thus, principal argument

$=-180°+250°=70°$

Let 'z' be a complex number satisfying

$|z-2-i|\le 5,$ Then |z-14-6i| lies in

0%
{8,18}
0%
{2,8}
0%
{0,2}
0%
{3,7}

Explanation

Here,

$|z-2-i|\leq 5$ ………..

$(1)$

Now,

$|z-14-6i|=|z-2-i-12-5i|=|(z-2-i)-(12+5i)|$

$\leq |z-2-i|+|12+5i|$

$\leq 5+|12+5i|$

$=5+\sqrt{(12)^2+(5)^2}$

$=5+\sqrt{144+25}$

$=5+\sqrt{169}$

$=5+13$

$=18$ .

$\therefore |z-14-6i|\leq 18$

and

$|z-14-6i|=|z-2-i-12-5i|$

$=|-(12+5i)+(z-2-i)|$

$\geq ||-(12+5i)|-|z-2-i||$

$\geq |12+5i|-5$

$[\because |z-2-i|\leq 5$

$\Rightarrow -|z-2-i|\geq -5]$

$=13-5$

$=8$

$\therefore |z-14-6i|\geq 8$

$\therefore |z-14-6i|$ lies in

$\{8, 18\}$ .

1188470_1315915_ans_eed83e919159412d92c24e2d369fda96.JPG

$w = \dfrac { z } { z - \dfrac { 1 } { 3 } i }$ and

$| w | = 1$ then

$z$ lies on

0%
a circle
0%
an ellipse
0%
a parabola
0%
a straight line

Explanation

$\begin{array}{l} Given \\ \omega =\frac { { 3z } }{ { 3z-i } } \, \, \, \, \, \, \therefore \left| \omega \right| =\frac { { 3\left| z \right| } }{ { \left| { 3z-i } \right| } } \\ \Rightarrow \left| { 3z-i } \right| =3\left| z \right| \\ \Rightarrow \left| { 3\left( x \right) +i\left( { 3y-1 } \right) } \right| =\left| { 3\left( { x+iy } \right) } \right| \, \, \, \, \, \left( { z=x+iy } \right) \\ \Rightarrow { \left( { 3x } \right) ^{ 2 } }+{ \left( { 3y-1 } \right) ^{ 2 } }=9\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) \\ \Rightarrow 6y-1=0 \\ Which\, is\, straight\, line \\ Hence,\, option\, D\, is\, the\, correct\, answer. \end{array}$

If the roots of the equation

$mx^{2}+(2m-1)x+m-2=0$ are rational, then if

$m\in I$ it will be

0%
odd integer
0%
even integer
0%
zero only
0%
none of these

Explanation

$mx^{2}+(2m-1)x+(m-2)=0$

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

$D=(2m-1)^{2}-4m(m-2)$

$=4m^{2}+1-4-4m^{2}+8m$

$\Rightarrow 4m+1$

$4$ m will always be even, so

$(4m+1)$ will be odd integer.

1215268_1353354_ans_7e403c4a75bd466d90e7a150dbb497c6.jpg

The discriminant of

${ 2x }^{ 2 }-x-p=0$ is

$49$ , then

$p=$ ______

0%
$6$
0%
$24$
0%
$48$
0%
None of these

Explanation

Given a quadratic equation:

$2x^2-x-p=0$

Hence,

$D=b^2-4ac=(-1)^2+4(2)p=1+8p$

But

$D=49$ ......[Given]

$\implies 1+8p=49$

$\implies 8p=48$

$\implies p=\dfrac{48}{8}=6$

$|z-3i|<\sqrt{5}$ , then

$|i(z+1)+1|<2\sqrt{5}$ .

0%
True
0%
False

The value of the sum

$\sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) }$ , where

$i=\sqrt { -1 }$ , equals

0%
$i$
0%
$i-1$
0%
$-i$
0%
$0$

Explanation

As we know that

$i+i^2+i^3+i^4=0$

$\displaystyle\sum ^{13}_{n=1} (i^n+i^{n+1})=\sum ^{13}_{n=1}(1+i)(i^n)=(1+i)\sum ^{13}_{n=1} i^{n}=(1+i)(0+0+i^{13})=(1+i)i=i+i^2=i-1$

$z_1$ and

$z_2$ are two non-zero complex numbers such that

$z_1=2+4i\\z_2=5-6i$ , then

$z_2-z_1$ equals

0%
$3-10i$
0%
$3+10i$
0%
$7-2i$
0%
$10-24i$

Explanation

$z_1=2+4i\\\\z_2=5-6i\\\\z_2-z_1=5-6i-2-4i=3-10i$

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

$z_1z_2$

0%
$z_1+z_2$
0%
$z_1-z_2$
0%
$z_1/z_2$
0%
None.

Explanation

$z_1=1+i$

$z_2=1-i$

$z_1z_2=1-(-1)=2$

$=1+1$

$=1-i+1+i$

$z_1z_2=z_1+z_2$

$z = \cos \dfrac { \pi } { 4 } + i \sin \dfrac { \pi } { 6 } ,$ then

0%
$| z | = 1 , \arg ( z ) = \dfrac { \pi } { 4 }$
0%
$| z | = 1 , \arg ( z ) = \dfrac { \pi } { 6 }$
0%
$| z | = \dfrac { \sqrt { 3 } } { 2 } , \arg ( z ) = \dfrac { 5 \pi } { 24 }$
0%
$| z | = \dfrac { \sqrt { 3 } } { 2 } , \arg ( z ) = \tan ^ { - 1 } \dfrac { 1 } { \sqrt { 2 } }$

Explanation

$z = \cos \dfrac{\pi}{4} + i \sin \dfrac{\pi}{6}$

$z = \dfrac{1}{\sqrt{2}} + i \dfrac{1}{2}$

$\theta = \tan^{-1} \dfrac{\dfrac{1}{2}}{\dfrac{1}{\sqrt{2}}} = \tan^{-1} \left(\dfrac{\sqrt{2}}{2} \right)$

$\theta = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$

$\therefore arg z = \tan^{-1} \dfrac{1}{\sqrt{2}}$

$|z| = \sqrt{\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{2}\right)^2} = \sqrt{\dfrac{1}{2} + \dfrac{1}{4}}$

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

$= \dfrac{\sqrt{3}}{2}$

$z = \cos \dfrac{\pi}{4} + i \sin \dfrac{pi}{4}$

$arg z = angle = \dfrac{\pi}{4}$

and

$|z| = \cos^2 \dfrac{\pi}{4} + \sin^2 \dfrac{\pi}{4} = 1$

The real value of

$'\theta '$ , for which the expression

$\frac { 1+i\cos { \theta } }{ 1-2i\cos { \theta } }$ is a real number is

0%
$2n\pi +\frac { 3\pi }{ 2 } ,n\in I$
0%
$2n\pi -\frac { 3\pi }{ 2 } ,n\in I$
0%
$2n\pi \pm \frac { \pi }{ 2 } ,n\in I$
0%
$2n\pi +\frac { \pi }{ 4 } ,n\in I$

Given

$z _ { 1 } + 3 z _ { 2 } - 4 z _ { 3 } = 0$ then

$z _ { 1 } , z _ { 2 } , z _ { 3 }$ are

0%
collinear
0%
can form sides of equilateral $\Delta$
0%
lie on circle
0%
none of these

The greatest and least value of

$\left | z \right |$ if

$z$ satisfies

$\left | z - 5 + 5i \right |$

$\leq 5$ are

0%
$10$ , $5\sqrt{2}$
0%
$5\sqrt{2}$ , $5$
0%
$10$ , $0$
0%
$5 + 5\sqrt{2}$ , $5\sqrt{2} - 5$

Explanation

$\left | z - 5 + 5i \right |$

$\leq 5$ represents equation of circle, where centre of circle is (5,-5)

Maximum value of

$z=c+\sqrt{a^2+b^2}$

Maximum value of

$z=c-\sqrt{a^2-b^2}$

Here,

$a=5, b=-5$ and

$c=5$

Maximum value of

$|z|=|5+\sqrt{25+25}|$

$=|5+\sqrt{50}|$

$=5+5\sqrt{2}$

Minimum value of

$|z|=|5-\sqrt{50}|$

$=|5-5\sqrt{2}|$

$=5\sqrt{2}-5$

If z be a complex number satisfying

$z^{4}+z^{3}+2z^{2}+z+1=0$ then

$\left|z\right|=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$
0%
$1$
0%
none of these

Explanation

$z^{4}+z^{3}+2z^{2}+z+1=0$

$\Rightarrow z^{4}+z^{2}+z^{3}+z+z^{2}+1=0$ by rearranging

$\Rightarrow z^{2}(z^{2}+1)+z(z^{2}+1)+1(z^{2}+1)=0$

$\Rightarrow z^{2}(z^{2}+1)+(z^{2}+1)(z+1)=0$

$\Rightarrow (z^{2}+1)(z^{2}+z+1)=0$

$\Rightarrow z^{2}+1=0, z^{2}+2+1=0$

$\Rightarrow z=\pm i, z=\frac{-1\pm i\sqrt{3}}{2}$

$\therefore z=i, -i, \frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$

$\therefore |z|=+1$

1240698_1507366_ans_b4f400c823594905ac18096a8656a951.PNG

The discriminant of the quadratic equation

$\left( { 2 }^{ \lambda } \right) { x }^{ 2 }+\left( { a }^{ 2 } \right) x-{ 8 }^{ \lambda }=0$ where

$a,\lambda ,\in N$ is surely

0%
A perfect square
0%
A prime number
0%
A composite number
0%
An even number

Express the following complex numbers in the standard form

$a+ib$ :

$\left ( \dfrac{1}{1-4i}-\dfrac{2}{1+i} \right )\left ( \dfrac{3-4i}{5+i} \right )$

0%
$\dfrac{307}{442}+i \dfrac{599}{442}i$
0%
$\dfrac{307}{442}-i \dfrac{599}{442}i$
0%
$\dfrac{-307}{442}+i \dfrac{599}{442}i$
0%
None of the above

Let z be a complex number such that the principal value of argument, arg

$z < 0$ . Then

$arg(-z) - arg (z)$ is

0%
$\dfrac{\pi}{2}$
0%
$\pm \pi$
0%
$\pi$
0%
$-\pi$

Explanation

Consider

$z=|z|e^{iθ}$

Now,

$−z=|z|e^{i(θ±π)}=|z|e^{i\bar{\theta}}$

where

$\bar{\theta}=arg(−z)$

Then

$arg(−z)−arg(z)=\bar{\theta}−θ=±π$

This has been verified numerically for random values of

$z$ .

Express the following complex numbers in the standard form

$a+ib$ :

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

0%
$\dfrac{37}{13}-\dfrac{16}{13}i$
0%
$\dfrac{-37}{13}+\dfrac{16}{13}i$
0%
$\dfrac{37}{13}+\dfrac{16}{13}i$
0%
None of the above

The imaginary roots of the equation

${ ({ x }^{ 2 }+2) }^{ 2 }+8{ x }^{ 2 }=6x({ x }^{ 2 }+2)$ are ____________.

0%
$1+i$
0%
$2\pm i$
0%
$-1\pm i$
0%
$none of these$

Explanation

${\left({x}^{2}+2\right)}^{2}+8{x}^{2}=6x\left({x}^{2}+2\right)$

${\left({x}^{2}+2\right)}^{2}-6x\left({x}^{2}+2\right)+8{x}^{2}=0$

Let

$t={x}^{2}+2$

$\Rightarrow\,{t}^{2}-6xt+8{x}^{2}=0$

$\Rightarrow\,{t}^{2}-2t\times 3x+9{x}^{2}-{x}^{2}=0$

$\Rightarrow\,{\left(t-3x\right)}^{2}-{x}^{2}=0$

$\Rightarrow\,{\left({x}^{2}+2-3x\right)}^{2}-{x}^{2}=0$ where

$t={x}^{2}+2$

$\Rightarrow\,{\left({x}^{2}-3x+2\right)}^{2}-{x}^{2}=0$

$\Rightarrow\,\left({x}^{2}-3x+2+x\right)\left({x}^{2}-3x+2-x\right)=0$

$\Rightarrow\,\left({x}^{2}-2x+2\right)\left({x}^{2}-4x+2\right)=0$

$\Rightarrow\,{x}^{2}-2x+2=0,\,or\,{x}^{2}-4x+2=0$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-4\times\,1\times 2}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-4\times 1\times 2}}{2}$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-8}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-8}}{2}$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{-4}}{2},\,or\,x=\dfrac{-4\pm\sqrt{8}}{2}$

$\Rightarrow\,x=\dfrac{2\pm 2i}{2},\,or\,x=\dfrac{-4\pm\,2\sqrt{2}}{2}$

$\Rightarrow\,x=1\pm i,\,or\,x=-2\pm\,\sqrt{2}$

Hence the roots are

$\left\{1+i,\,1-i,\,-2+\sqrt{2},\,-2-\sqrt{2}\right\}$

Imaginary roots are

$\left\{1+i,\,1-i\right\}$

Express the following complex numbers in the standard form

$a+ib$ :

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

0%
$\dfrac{1}{4}+\dfrac{3}{4}i$
0%
$\dfrac{1}{4}-\dfrac{3}{4}i$
0%
$\dfrac{-1}{4}-\dfrac{3}{4}i$
0%
None of the above

Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:

$1-i$

0%
$\sqrt{2}(cos\,\pi /4+i\, sin\, \pi /4)$
0%
$\sqrt{2}(cos\,\pi /3-i\, sin\, \pi /3)$
0%
$\sqrt{2}(cos\,\pi /4-i\, sin\, \pi /4)$
0%
$\sqrt{2}(cos\,\pi /3+i\, sin\, \pi /3)$

Explanation

Let

$z=1-i$ . Then,

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

Let

$\alpha$ be the acute angle given by

$tan\, \alpha =\dfrac{\left | Im(z) \right |}{\left | Re(z) \right |}$ .

Then,

$tan\, \alpha =\dfrac{|-1|}{|1|}=1\Rightarrow \alpha =\dfrac{\pi }{4}$

Clearly, z lies in the fourth quadrant. Therefore,

$arg(z)= -\alpha =-\dfrac{\pi }{4}$ .

$a$ and

$b$ are the nonzero distinct roots of

$x^2 + ax + b =0$ , then the minimum vlue of

$x^2 + ax+b$ is

0%
$\dfrac{2}{3}$
0%
$\dfrac{9}{4}$
0%
$\dfrac{-9}{4}$
0%
$\dfrac{-2}{3}$

Explanation

Given,

$x^2 + ax + b = 0$

For distinct now zero roots

$D > 0$

$\Rightarrow a^2 - 4b > 0$

Now,

$x^2 + ax + b$

$= \left(x + \dfrac{a}{2}\right)^2 + \left(b-\dfrac{a^2}{4}\right)$

$= \left(x + \dfrac{a}{2}\right)^2 - \left(a^2-\dfrac{4b}{4}\right)$

We know, sum of roots

$a + b = -a$

$\Rightarrow 2a + b = 0$

Product of roots

$a\times b = b$

$\Rightarrow b(a-1) = 0$

$\Rightarrow a = 1, b \neq 0$

From Equation (i),

$2a+b = 0$ $

$2(1) + b = 0$

$b = -2$

Now,

$\left( x + \dfrac{A}{2}\right)^2 - \left(\dfrac{1^2 + 4 \times 2}{4}\right)$

$= \left(x+\dfrac{a}{2}\right)^2 - \dfrac{9}{4}$

$\therefore$ Maximum value

$= -\dfrac{9}{4}$

Express the following complex numbers in the standard from

$a+ib$ :

$\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$

0%
$1-2\sqrt{2}i$
0%
$1+\sqrt{2}i$
0%
$1+2\sqrt{2}i$
0%
$1-\sqrt{2}i$

Explanation

$\dfrac{5+\sqrt{2}i}{1-\sqrt{2}i}$

Rationalizing the denominator, we get

$= \dfrac{5+\sqrt{2}i}{1-\sqrt{2}i} \times \dfrac{1+\sqrt{2}i}{1+\sqrt{2}i}$

$= \dfrac{5+2i^2+(5+1)\sqrt{2}i}{1+2}$

$= \dfrac{3+6\sqrt{2}i}{1+2}$ ......................

$(\because i^2=-1)$

$= 1+2\sqrt{2}i$

Let z be a complex number such that

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

$|z|=\dfrac{5}{2}$ . Then the value of

$|z+3i|$ is?

0%
$\dfrac{7}{2}$
0%
$\dfrac{15}{4}$
0%
$2\sqrt{3}$
0%
$\sqrt{10}$

Explanation

Given

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

$\Rightarrow |z - i| = |z + 2i|$

Let

$z = x + iy$

So,

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

$-2y + 1 = 4y + 4$

$6y = -3 \Rightarrow y = \dfrac{-1}{2}$

$x^2 + y^2 = \dfrac{25}{4} \Rightarrow x^2 = \dfrac{24}{4} = 6$

$z = \pm \sqrt{6} - \dfrac{i}{2}$

$|z + 3i| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$

$\therefore$

$\boxed{|2 + 3i| = \dfrac{7}{2}}.....Answer$

Hence option

$'A'$ is the answer.

The real part of

$(i - \sqrt{3})^{13}$ is

0%
$2^{-10}\sqrt3$
0%
$-2^{12}\sqrt3$
0%
$2^{-12}\sqrt3$
0%
$-2^{-12}\sqrt3$
0%
$-2^{10}\sqrt3$

Explanation

$w^2=\dfrac{-1-i\sqrt3}{2}$

$\Rightarrow iw^2=\dfrac{-i+\sqrt3}{2}$

$\Rightarrow -2iw^2=i-\sqrt3$

$\Rightarrow (-2iw^2)^{13}=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}i^{13}w^{26}=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}iw^2=(i-\sqrt3)^{13}$

$\Rightarrow -2^{13}\left({\dfrac{-i+\sqrt3}{2}}\right)=(i-\sqrt3)^{13}$

$\Rightarrow -2^{12}(-i+\sqrt3)=(i-\sqrt3)^{13}$

$\therefore (i-\sqrt3)^{13}=-2^{12}\sqrt3+i2^{12}$

$(1-\sqrt{-1})(1+\sqrt{-1})(5-\sqrt{-7})(5+\sqrt{-7})=?$

0%
$(25+7i)$
0%
$(32+5i)$
0%
$(29-3i)$
0%
$none\ of\ these$

Explanation

Given expression

$=(1-i)(1+i)(5-\sqrt 7 i)(5+\sqrt 7i)$

$=(1-i^2)(25-7i^2)\\=(1+1)(25+7)\\=(2\times 32)=64$ .

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

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{2\pi}{3}$
0%
$\pi$
0%
$none\ of\ these$

Explanation

$(-1+i\sqrt{3})\\ =\dfrac{-1}{2}+i\dfrac{\sqrt 3}{2} \\=2\left(\cos \dfrac{2\pi}{3}+i\sin \dfrac{2\pi}{3}\right)\\ \Rightarrow arg (-1+i\sqrt{3})=\dfrac{2\pi}{3}$

$(2-3i)(-3+4i)=?$

0%
$(6+17i)$
0%
$(6-17i)$
0%
$(-6+17i)$
0%
$none\ of\ these$

Explanation

$(2-3i)(-3+4i)\\=(-6+8i+9i-12i^2)\\=(-6+17i+12)\\=(6+17i)$ .

$arg (1+i)=?$

0%
$\pi$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$

Explanation

$(1+i)\\=\sqrt 2 \left(\dfrac {1}{\sqrt2}+\dfrac{i}{\sqrt2}\right)\\=\sqrt{2}\left(\cos \dfrac{\pi}{4}+i\sin \dfrac{\pi}{4}\right)\\ \Rightarrow arg (1+i)=\dfrac{\pi}{4}$

$arg \left(\dfrac{2+6\sqrt{3}i}{5+\sqrt{3}i}\right)=?$

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{2\pi}{3}$
0%
$\pi$

Explanation

$\dfrac{(2+6\sqrt{3}i)}{(5+\sqrt{3}i)}\\=\dfrac{(2+6\sqrt{3}i)}{(5+\sqrt{3}i)}\times \dfrac{(5-\sqrt{3}i)}{(5-\sqrt{3}i)}\\=\dfrac{(2+6\sqrt{3}i)(5-\sqrt{3}i)}{(5-\sqrt{3}i)(5-\sqrt{3}i)}$

$=\dfrac{(28+28\sqrt{3}i)}{28}\\=\dfrac{28(1+\sqrt{3}i)}{28}=\dfrac{28(1+\sqrt{3}i)}{28}=(1+\sqrt{3}i)=2\left(\cos\dfrac{\pi}{3}+i\sin\dfrac{\pi}{3}\right)$

$\therefore arg\left(\dfrac{2+6\sqrt{3}i}{5+\sqrt{3}i}\right)=\dfrac{\pi}{3}$

Let

$z, w$ be complex numbers such that

$\bar z + i\bar w =0$ and arg

$zw = \pi$ . then

$arg \ z$ equals

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{3\pi}{4}$
0%
$\dfrac{5\pi}{4}$

Explanation

$\overline{z} + i\overline{w} = 0$

$\Rightarrow z - iw = 0$

$\Rightarrow z = iw$

$arg \space zw = \pi$

$\Rightarrow arg\space z + arg \space w = \pi$

$\Rightarrow arg \space z + arg \space \dfrac{z}{i} = \pi$ [Using (1)]

$\Rightarrow arg\space z + arg\space z - arg \space i = \pi$

$\Rightarrow 2 arg\space z - \dfrac{\pi}{2} = \pi$

$\Rightarrow 2 arg \space z = \dfrac{3\pi}{2}$

$\Rightarrow arg \space z = \dfrac{3\pi}{4}$

For any complex numbers

$z_{1}$ and

$z_{2}$ compare List I with with List II and choose the correct answer, using codes given below:

List I	List II
$arg (z_{1},z_{2})$	$\dfrac{\pi}{2}$
$arg \left(\dfrac{z_{1}}{z_{2}}\right)$	$arg (z_{1}-arg (z_{2})$
$arg (z)+arg (\bar{z})$	$arg (z_{1})+arg (z_{2})$
$arg (i)$	$2\pi$

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$(i)-(q), (ii)-(r), (iii)-(s), (iv)-(p)$
0%
$(i)-(r), (ii)-(q), (iii)-(p), (iv)-(s)$
0%
$(i)-(r), (ii)-(q), (iii)-(s), (iv)-(p)$
0%
$none\ of\ these$

The principal argument of the complex number

$[(1 + i)^5 (1 + \sqrt{3}i)^2] / [-2i(-\sqrt{3} +i)]$ is

0%
$\frac{19\pi}{12}$
0%
$-\frac{17\pi}{12}$
0%
$-\frac{5\pi}{12}$
0%
$\frac{5\pi}{12}$

The argument and the principle value of the complex number

$\dfrac {2+i}{4i+(1+i)^2}$ are

0%
$\tan^{-1}(-2)$
0%
$-\tan^{-1} 2$
0%
$\tan^{-1}\left(\dfrac {1}{2}\right)$
0%
$-\tan^{-1}\left(\dfrac {1}{2}\right)$

Compare List I with List II and choose the correct answer using codes given below:

List I (Complex number)	List II (Its modulus)
$(4-3i)$	$10$
$(8+6i)$	$\dfrac{1}{5}$
$\dfrac{1}{(3+4i)}$	$1$
$\dfrac{(3-4i)}{(3+4i)}$	$5$

0%
$(i)-(p), (ii)-(s), (iii)-(r), (iv)-(q)$
0%
$(i)-(s), (ii)-(p), (iii)-(q), (iv)-(r)$
0%
$(i)-(s), (ii)-(p), (iii)-(r), (iv)-(q)$
0%
$(i)-(r), (ii)-(p), (iii)-(s), (iv)-(q)$

$b_{1} b_{2}=2\left(c_{1}+c_{2}\right),$ then at least one of the equations

$x^{2}+b_{1} x$

$+c_{1}=0$ and

$x^{2}+b_{2} x+c_{2}=0$ has

0%
imaginary roots
0%
real roots
0%
purely imaginary roots
0%
none of these

Explanation

Let

$D_{1}$ and

$D_{2}$ be discriminants of

$x^{2}+b_{1} x+c_{1}=0$ and

$x^{2}+b_{2} x$

$+c_{2}=0,$ respectively.

Then,

$D_{1}+D_{2} =b_{1}^{2}-4 c_{1}+b_{2}^{2}-4 c_{2}$

$=\left(b_{1}^{2}+b_{2}^{2}\right)-4\left(c_{1}+c_{2}\right)$

$=b_{1}^{2}+b_{2}^{2}-2 b_{1} b_{2} \quad\left[\because b_{1} b_{2}=2\left(c_{1}+c_{2}\right)\right]$

$=\left(b_{1}-b_{2}\right)^{2} \geq 0$

$\Rightarrow \quad D_{1} \geq 0$ or

$D_{2} \geq 0$ or

$D,$ and

$D_{2}$ both are positive

Hence, at least one of the equations has real roots.

Which of the following equations has no real roots.

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

Explanation

${\textbf{Step -1: Considering option A.}}$

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

$\therefore D = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( {3\sqrt 2 } \right)$

$\therefore D = 16 - 12\sqrt 2 < 0$

${\text{Therefore, it has no real roots}}{\text{.}}$

${\textbf{Step -2: Considering option B.}}$

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

$\therefore D = {\left( 4 \right)^2} - 4\left( 1 \right)\left( { - 3\sqrt 2 } \right)$

$\therefore D = 16 + 12\sqrt 2 > 0$

${\text{Therefore, it has real roots}}{\text{.}}$

${\textbf{Step -3: Considering option C.}}$

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

$\therefore D = {\left( { - 4} \right)^2} - 4\left( 1 \right)\left( { - 3\sqrt 2 } \right)$

$\therefore D = 16 + 12\sqrt 2 > 0$

${\text{Therefore, it has real roots}}{\text{.}}$

${\textbf{Step -4: Considering option D.}}$

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

$\therefore D = {\left( {4\sqrt 3 } \right)^2} - 4\left( 3 \right)\left( 4 \right)$

$\therefore D = 48 - 48$

$\therefore D = 0$

${\text{Therefore, it has real and equal roots}}{\text{.}}$

${\textbf{Hence, option A. is correct answer.}}$

The modulus and amplitude of the complex number

$\left[e^{{3}-i\dfrac{\pi}{4}}\right]^{3}$ are respectively.

0%
$e^{6},\dfrac{-3\pi}{4}$
0%
$e^{9},\dfrac{\pi}{2}$
0%
$e^{9},\dfrac{-3\pi}{4}$
0%
$e^{9},\dfrac{-\pi}{2}$

Explanation

Let

$z=\left[e^{3-i\dfrac{\pi}{4}}\right]^3$

$\implies z=\left[e^3\times e^{-i\dfrac{\pi}{4}}\right]^3$

$\implies z=[e^3]^3\times \left[e^{-i\dfrac{\pi}{4}}\right]^3$

$\implies z=e^9\times e^{-i\dfrac{3\pi}{4}}$

This is in the form of

$z=|z|e^{i \text{arg} (z)}$ where

$|z|$ is the modulus and

$arg(z)$ is the argument of

$z$

$\implies |z|=e^9,\text{arg}(z)=-\dfrac{3\pi}{4}$

$\implies$ modulus is

$e^9$ and argument is

$-\dfrac{3\pi}{4}$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 9 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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