MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 10

$\left ( x^{2} + 1 \right )^{2} - x^{2} =0$ has

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four real roots
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two real roots
0%
no real roots
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one real root

Explanation

Given equation is

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

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

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

Let

$x^{2} = y$

$\Rightarrow y^{2} +1y + 1 = 0$

Now,

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

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

$\Rightarrow D = -3$

$\Rightarrow D < 0$

So, the given equation

$y^{2} + y + 1 = 0$ has no real roots.

$\therefore$ the equation $$\left ( x^{2} + 1 \right )^{2} - x^{2} =0$ has no real roots.

Hence, the correct answer is option

$[C]$ .

The argument of the complex number

$\sin \frac{6\pi }{5}+i\left ( 1+\cos \frac{6\pi }{5} \right )$ is

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$\displaystyle \frac{6\pi }{5}$
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$\displaystyle\frac{5\pi }{6}$
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$\displaystyle\frac{9\pi }{10}$
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$\displaystyle\frac{2\pi }{5}$

Explanation

$\sin\dfrac{6\pi}{5}+i(1+\cos\dfrac{6\pi}{5})$
$=2\sin\dfrac{3\pi}{5} \cos\dfrac{3\pi}{5} +2i\cos^2\dfrac{3\pi}{5}$
$=-2\cos\dfrac{3\pi}{5}(-\sin\dfrac{3\pi}{5}-i\cos\dfrac{3\pi}{5})$
$=-2\cos\dfrac{3\pi}{5}(\cos\dfrac{9\pi}{10}+i\sin\dfrac{9\pi}{10})$
So argument is $\dfrac{9\pi}{10}$

The equation

$\sqrt {3x^2+x+5}=x-3$ , where

$x$ is real, has

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no solution.
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exactly one solution.
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exactly two solutions.
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exactly four solutions.

Explanation

Now,

$3x^2+x+5 \geqslant 0$

This is because

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

$=-59$ (roots are imaginary)

$3x^2+x+5 =(x-3)^2=x^2+9-6x$ and

$x-3 \geqslant 0$

$2x^2+7x-4=0$

$2x^2+8x-x-4=0$

$2x(x+4)-1(x+4)=0$

$(2x-1)(x+4)=0$

$x=\dfrac12 \ and -4$ but

$x\geq 3$

$\therefore$ No solution.

$\displaystyle z_{0}=\frac{1-i}{2}$ , then

$\displaystyle \left (1+z_{0} \right )\left (1+z_{0}^{{2}^{1}} \right )\left (1+z_{0}^{{2}^{2}} \right ).......... \left (1+z_{0}^{2^n} \right )$ must be

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$(1-i)(1+\dfrac{1}{2^{2^n}})$ for $n>1$
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$(1-i)(1-\dfrac{1}{2^{2^n}})$ for $n>1$
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$\dfrac{1+i}{2}$ for $n>1$
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$(1-i)(1-\dfrac{1}{2^{2^{n+1}}})$ for $n>1$

Explanation

Let

$P=(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=(1-z_0)(1+z_0)(1+{z_0}^{2^1})+(1+{z_0}^{2^2})....(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=(1-{z_0}^{2^2})(1+{z_0}^{2^2}).......(1+{z_0}^{2^n})$

$\Rightarrow (1-z_0)P=1-{z_0}^{2^{n+1}}$

$\Rightarrow P=\dfrac{1-{z_0}^{2^{n+1}}}{1-{z_0}}$

$\Rightarrow P=\dfrac{1-({z_0}^2)^{2n}}{1-z_0}$

$n>1$

Since,

$z_0=\dfrac{1-i}{2}$

${z_0}^2=\dfrac{1-1-2i}{4}$

$\Rightarrow {z_0}^2=\dfrac{-i}{2}$

$\therefore \ P=\dfrac{1-\left(\dfrac{-i}{2}\right)^{2^n}}{1-z_0}$

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

$\Rightarrow P=(1-i)\left(1-\dfrac{1}{2^{2^n}}\right)$

The number of solutions of

$log _{\frac{1}{5}}log_{\frac{1}{2}}(\left | z \right |^{2}+4\left | z \right |+3)< 0$ is/are?

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$0$
0%
$2$
0%
$4$
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infinite

Explanation

The glven inequality becomes

$log_{\frac{1}{2}}|\mathrm{z}|^{2}+4|\mathrm{z}|+3>1$ , since

$\displaystyle \frac{1}{5}<1$

$\displaystyle \Rightarrow|\mathrm{z}|^{2}+4|\mathrm{z}|+3<\frac{1}{2}$

$\Rightarrow(|\mathrm{z}|+2)^{2}$ + 1

$<\displaystyle \frac{1}{2}$

$(|\displaystyle \mathrm{z}|+2)^{2}<\frac{-1}{2}$
This is not possible. Hence, there is no solution.

Let

$\alpha,\ \beta$ be real and

$\mathrm{z}$ be a complex number. If

$\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$ has two distinct roots on the line

$Re(z) =1$ , then it is necessary that:

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$\beta\in(0,1)$
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$\beta\in(-1,0)$
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$|\beta|=1$
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$\beta\in(1, \infty)$

Let

$z$ be a complex number and

$c$ be a real number

$\geq$ 1 such that z +

$c\left | z+1 \right |+i=0 ,$ then

$c$ belongs to

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$[2, 3]$
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$(3, 4)$
0%
$[1,\sqrt{2}]$
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None of these

Explanation

Since

$c$

$\left | z+1 \right |$ is real

$z + i$ is real
let

$z-i = x$ , where

$x$ is real

$z+i=-c$

$\left | z+1 \right |$ (given)

$x^{2}=c^{2}\left \{ x+1 \right \}^{2}+1^{2})$

$(c^{2}-1)x^{2}+2c^{2}x+2c^{2}=0$
As

$x$ is real

$4c^{4}-8c^{2}(c^{2}-1)\geq 0$

$4c^{2}-(c^{2}-2)\leq 0$

$c^{2}\leq 2$

$c\leq \sqrt{2}$
But

$c\geq1$

$\Rightarrow c\, \, \epsilon [1,\sqrt{2}]$

$\displaystyle \log_{\tan 30^{\circ}}\left(\frac{2|Z|^{2}+2|Z|-3}{|z|+1}\right) <-2$ then

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$|\displaystyle \mathrm{z}|<\frac{3}{2}$
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$|z|>\displaystyle \frac{3}{2}$
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$|z|>2$
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$|z|<2$

Explanation

we have,

$\Rightarrow log_{(\dfrac{1}{\sqrt{3}})}(\dfrac{2|z|^{2}+2|z|-3}{|z|+1})<-2$

$\Rightarrow \dfrac{2|z|^{2}+2|z|-3}{|z|+1}>(\dfrac{1}{\sqrt{3}})^{-2}=(\sqrt{3})^{2}=3$ [as

$\dfrac{1}{\sqrt{3}}<1$ so equality sign changes]

$\Rightarrow 2|z|^{2}+2|z|-3>3|z|+3$

$\Rightarrow 2|z|^{2}-|z|-6>0$

$\Rightarrow 2|z|^{2}-4|z|+3|z|-6>0$

$\Rightarrow 2|z| (|z|-2)+3 (|z|-2)>0$

$\Rightarrow (2|z|+3) (|z|-2)>0$

$\Rightarrow |z|>2$

$x = 2 + 5i($ where

$1 i = \sqrt{-1})$ and

$2(\displaystyle \frac{1}{1! 9!} + \frac{1}{3! 7!}) + \frac{1}{5! 5!} = \frac{2^{a}}{b!}$ then

$x^{3}-5x^{2}+33x-10 =$

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$a + b$
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$b - a$
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$a-b$
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$-a-b$
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$(a - b)(a + b)$

Explanation

$\displaystyle \frac{10! 2^{a}}{b!} = 2[^{10}C_{1} + ^{10}C_{3}] + ^{10} C_{5}$

$\displaystyle \frac{10! 2^{a}}{b!} = ^{10}C_{1} + ^{10}C_{3} + ^{10}C_{5} + ^{10}C_7 + ^{10}C_{9} = 2^{9}$

$\therefore a = 9, b = 10$

$x = 2 + 5i$

$(x - 2)^{2} = - 25$

$x^{2} - 4x + 29 = 0$

$\therefore x^{3} - 5x^{2} + 33x - 10 = (x - 1)(x^{2} - 4x + 29) + 19$

$\Rightarrow 19 = a + b$

$x=9^{\frac {1}{3}} 9^{\frac {1}{9}} 9^{\frac {1}{27}} .....\infty, y=4^{\frac {1}{3}} 4^{\frac {-1}{9}} 4^{\frac {1}{27}}....\infty,$ and

$z=\sum_{r=1}^{\infty} (1+i)^{-r}$ , then

$arg (x+yz)$ is equal to

0%
0
0%
$\tan^{-1}(\frac {\sqrt 2}{3})$
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$-\tan^{-1}(\frac {\sqrt 2}{3})$
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$- \tan^{-1}(\frac {2}{\sqrt 3})$

Explanation

$x=9^{\dfrac{1}{3}+\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$
$x=9^{\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}}$
$x=9^{\dfrac{1}{2}}$
$x=3$
$y=4^{\dfrac{1}{3}-\dfrac{1}{3^{2}}+\dfrac{1}{3^{3}}...\infty}$
$y=4^{\dfrac{\dfrac{1}{3}}{1+\dfrac{1}{3}}}$
$y=4^{\dfrac{1}{4}}$
$y=2^{\dfrac{1}{2}}$
$y=\sqrt{2}$
$z=\dfrac{1}{(1+i)^{1}}+\dfrac{1}{(1+i)^{2}}+\dfrac{1}{(1+i)^{3}}...\infty$
Since it is a G.P with a common ratio of $\dfrac{1}{(1+i)}$ we get the sum as
$=\dfrac{\dfrac{1}{1+i}}{1-\dfrac{1}{1+i}}$
$=\dfrac{1}{i}$
$=-i$ .
Hence $x+yz =3-\sqrt{2}i$
$\tan\theta=\dfrac{-\sqrt{2}}{3}$
$\theta=\tan^{-1}(\dfrac{-\sqrt{2}}{3})$
$=-\tan^{-1}(\dfrac{\sqrt{2}}{3})$

$\left | \log_{\sqrt{3}} \frac{\left | z \right |^{2}-\left | z \right |+1}{2+{}\left | z \right |}\right |< 2$ , then

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$|\displaystyle \mathrm{z}|<\frac{1}{3}$
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$|\mathrm{z}|=1$
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$|\mathrm{z}|=5$
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$1<|\mathrm{z}|<5$

Explanation

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

$\Rightarrow -2<\log_{{\sqrt{3}}}(\dfrac{|z|^{2}-|z|+1}{|z|+2})<2$

$\Rightarrow \dfrac{1}{3}<\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$

$\Rightarrow |z|+2<3|z|^{2}-3|z|+3$

$\Rightarrow 3|z|^{2}-4|z|+1>0$

$\Rightarrow (3|z|-1)(|z|-1)>0$

Hence

$|z|<\dfrac{1}{3}$ and

$|z|>1$ ...(i)

Similarly

$\dfrac{|z|^{2}-|z|+1}{|z|+2}<3$

$\Rightarrow |z|^{2}-|z|+1<3|z|+6$

$\Rightarrow |z|^{2}-4|z|-5<0$

$\Rightarrow (|z|+1)(|z|-5)<0$

$\Rightarrow 1<|z|<5$ ...(ii)

From (i) and (ii), we get

$1<|z|<5$

Given

$z$ is a complex number with modulus

$1$ . Then the equation in

$a$ ,

$\left(\dfrac{1+ia}{1-ia}\right )^4=z$ has

0%
all roots real and distinct.
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two real and two imaginary.
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three roots real and one imaginary.
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one root real and three imaginary.

Explanation

$\left (\dfrac{1+ia}{1-ia}\right )^4 = z$

$\left (\dfrac{-i(i-a)}{-i(i+a)}\right )^4 = z$

$\left (\dfrac{a-i}{a+i}\right )^4= z$

$\left |\dfrac{a-i}{a+i}\right |^4= |z| =1$

$|a-i| ^4 = |a+i|^4$

$|a-i| = |a+i|$

$\therefore a$ lies on the perpendicular bisector of

$i$ and

$-i$ .

$a$ lies on real axis. Hence the roots all are real and distinct.

Find the value of

$x$ such that

$\displaystyle \frac{(x + \alpha)^2 - (x + \beta)^2}{ \alpha + \beta} = \frac{sin 2 \theta}{sin^2 \theta}$ . when

$\alpha$ and

$\beta$ are the roots of

$t^2 - 2t + 2 = 0$

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$x = icot \, \, \, \theta - 1$
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$x = -(icot \, \, \, \theta + 1$ )
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$x = icot \, \, \, \theta$
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$x = itan \, \, \, \theta - 1$

Explanation

$t^{2}-2t+2=0$

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

$(t-1)^{2}=-1$

$t=1\pm i$
Hence

$\alpha+\beta=2$

$\alpha-\beta=2i$
Now let

$n=2$
We get

$\dfrac{2x(\alpha-\beta)+\alpha^{2}-\beta^{2}}{2}=\dfrac{sin2\theta}{sin^{2}\theta}$
Hence

$\dfrac{4ix+2i-(-2i)}{2}=\dfrac{2sin\theta.cos\theta}{sin^{2}\theta}$

$2ix+2i=2cot\theta$

$ix=cot\theta-i$

$-x=icot\theta+1$

$x=-(1+icot\theta)$

The complex numbers

$\sin x + i \cos 2x$ and

$\cos x - i \sin 2x$ are conjugate to each other, for

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$x = n\pi$
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$x \displaystyle = \left ( n + \frac{1}{2} \right ) \pi$
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$x = 0$
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No value of $x$

Explanation

We have,

$\sin x = \cos x$ - (i)

and

$\cos 2x = \sin 2x$ -(ii)

Hence,

$\cos^2x - \sin^2x = 2 \sin x \cos x$

$\Rightarrow 0 = 2 \sin^2 x$

$\Rightarrow \sin x =0$

However,

$\sin x \neq \cos x$

Hence, there is no value of

$x$ satisfying the given condition.

$z_1, z_2$ be two non zero complex numbers satisfying the equation

$\displaystyle \left | \frac{z_1 + z_2}{z_1 - z_2} \right | = 1$ then

$\displaystyle \frac{z_1}{z_2} + \left ( \frac{z_1}{z_2} \right )$ is

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zero
0%
1
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purely imaginary
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2

Explanation

let

$\frac{z_{1}}{z_{2}} =x$

given

$|\frac{x+1}{x-1}| = 1$

$\Rightarrow |x+1|=|x-1|$

Let

$x=a+ib$

We get

$|a+1+ib|=|a-1+ib|$

$\Rightarrow a=0$

Therefore

$x=ib$ and

$\overline { x } =-ib$

Therefore

$x+\overline { x } =ib-ib=0$

So the correct option is

$A$

Interpret the following equations geometrically on the Argand plane.

$1 < |z - 2 - 3 i| < 4$

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Annular
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Straight line
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A point
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Ring

Explanation

We have

$1 < |z - 2 - 3i| < 4$
Represents a circle with centre at

$(2, 3)$ and radius

$r \in (1, 4)$

$|z - 2 - 3 i| > 1$ represents the region in the plane outside the circle

$|z - 2 - 3i| = 1$
Similarly

$|z - 2 - 3i| < 4$ represents the region inside the circle

$|z - 2 - 3i| = 4$

$\therefore$

$1 < |z - 2 - 3i| < 4$ represents the annular space between the two circles.

If n is a natural number

$\geq 2$ , such that

$z^n=(z+1)^n$ , then

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Roots of equation lie on a straight line parallel to the $y-axis$
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Roots of equation lie on a straight line parallel to the $x-axis$
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Sum of the real parts of the roots is $-[(n-1)/2]$
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None of these

Explanation

Given,

$z^n=(z+1)^n\rightarrow |z^n|=|(z+1)^n|$

$\therefore |z|^n = |z+1|^n\Rightarrow |z|=|z+1|$

$\Rightarrow |z|^2=|z+1|^2$

$\Rightarrow x^2+y^2=(x+1)^2+y^2$ , where

$z=x-iy$

$\Rightarrow x=-\dfrac {1}{2}$
Hence,

$z$ lies on the line

$x=-1/2$ .
Hence, sum of real parts of the roots is

$-(n-1)/2$ (since equation has

$n-1$ roots).

Dividing f(z) by

$z-i$ , we obtain the remainder

$i$ and dividing it by

$z+i$ , we get the remainder

$1+i$ , then remainder upon the division of f(z) by

$z^2+1$ is

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$\displaystyle \frac {1}{2}(z+1)+i$
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$\displaystyle \frac {1}{2}(iz+1)+i$
0%
$\displaystyle \frac {1}{2}(iz-1)+i$
0%
$\displaystyle \frac {1}{2}(z+i)+1$

Explanation

$\textbf{Step -1: Using remainder theorem to get the following equations.}$

$\text{Given, Dividing }f(z) \text{ by } z - i \text{, we get remainder }i$

$\text{Therefore by remainder theorem, }f(i) = i$

$\dots \text{(i)}$

$\text{Given, Dividing }f(z) \text{ by } z + i \text{, we get remainder }1 + i$

$\text{Therefore by remainder theorem, }f(-i) = 1+i$

$\dots \text{(ii)}$

$\text{Let }R(x) \text{ be the remainder when }f(z)\text{is divided by }z^2 + 1$

$\text{degree of R(z) is 1, hence }R(z) = az+b$

$\text{we can simplify }z^2 + 1 = (z -i) (z+i)$

$\therefore f(z) = Q(z)(z-i)(z+i) + R(z)$

$\textbf{Step -2: Solving for a and b.}$

$\text{From (i) and (ii)}$

$f(i) = i = ai + b$

$\Rightarrow b = i - ai$

$f(-i) = 1+i = -ai +b$

$\Rightarrow -ai + i - ai = 1+i$

$\Rightarrow -2ai = 1$

$\Rightarrow a = -\dfrac{1}{2i}$

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

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

$\therefore a = \dfrac{i}{2}$

$b = i - ai$

$\Rightarrow b = i - \dfrac{i}{2} (i)$

$\Rightarrow b = i - \dfrac{(-1)}{2}$

$\Rightarrow b = i + \dfrac{1}{2}$

$\therefore R(z) = az+b$

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

$R(z) = \dfrac{1}{2}(iz + 1) + i$

$\textbf{Hence option B is correct}$

$b_1b_2=2(c_1+c_2)$ , then at least one of the equations

$x^2+b_1x+c_1=0$ and

$x^2+b_2x+c_2=0$ has

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imaginary roots
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real roots
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purely imaginary roots
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none of these

Explanation

Suppose the equations

$x^2+b_1x+c_1=0$ &

$x^2+b_2x+c_2=0$ have real roots.
Then

${ b_{ 1 } }^{ 2 }\ge 4c_{ 1 }$ ...(1)
and

${ b_{ 2 } }^{ 2 }\ge 4c_{ 2 }$ ...(2)
given that

$b_{ 1 }b_{ 2 }=2(c_{ 1 }+c_{ 2 })$
On squaring

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

$\Rightarrow { b_{ 1 } }^{ 2 }{ b_{ 2 } }^{ 2 }-16c_{ 1 }c_{ 2 }=4{ \left( c_{ 1 }-c_{ 2 } \right) }^{ 2 }\ge 0$
Multiplying (1) & (2), we get

${ b_{ 1 } }^{ 2 }{ b_{ 2 } }^{ 2 }\ge 16c_{ 1 }c_{ 2 }$
Therefore, at least one equation have real roots.

Ans: B

Find the regions of the z-plane for which

$\displaystyle \left | \frac{z - a}{z + \overline a} \right | < 1, = 1$ or

$> 1$ . when the real part of a is positive.

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The required regions are the right half of the z-pane, the imaginary axis and the left half of the z-plane respectively.
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The required regions are the left half of the z-pane, the imaginary axis and the right half of the z-plane respectively.
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The required regions are the right half of the z-pane the real axis and the left half of the z-plane respectively.
0%
The required regions are the left half of the z-pane the real axis and the right half of the z-plane respectively.

Explanation

Let

$z=x+iy$ and

$a$ be a fixed complex number

$(a+ib)$ .
Hence

$|\dfrac{z-a-ib}{z+a-ib}|=1$

$|(x-a)+i(y-b)|=|(x+a)+i(y-b)|$

$(x-a)^{2}=(x+a)^{2}$

$2x(a)=0$

$x=0$ ... imaginary axis.
If

$|\dfrac{z-a-ib}{z+a-ib}|<1$
Then

$(x-a)^{2}<(x+a)^{2}$
Or

$x>0$ ... positive real axis.
If

$|\dfrac{z-a-ib}{z+a-ib}|>1$

$(x-a)^{2}>(x+a)^{2}$

$x<0$ ... negative real axis.

$\alpha, \beta$ are the roots of equation

$(k + 1) x^{2} - (20k + 14)x + 91k + 40 = 0 ; (\alpha < \beta)$ ,

$k > 0$ , then

The nature of the roots of this equation is

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imaginary
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real and distinct
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equal real roots
0%
None of these

Explanation

Given equation is

$(k + 1) x^{2} - (20k + 14)x + 91k + 40 = 0$

$D=(20k+14)^{2}-4(k+1)(91k+40)=discriminant$

$\Rightarrow D=36(k^{2}+k+1)\ge0$ ..........

$(\because k^2 \ and \ k \ are \ always \ positive \ for \ all \ k>0)$

$\because D>0$

$\therefore \ the \ equation \ has \ real \ roots$

Hence, option

$B$ is correct.

If z be a complex number satisfying

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

$\displaystyle\ |z|$ is

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$\displaystyle\ \frac{1}{2}$
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$\displaystyle\ \frac{3}{4}$
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$\displaystyle\ 1$
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None of these

Explanation

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

$\left[ { z }^{ 4 }+2{ z }^{ 2 }+1 \right] +\left[ { z }^{ 3 }+z \right] =0$

$\left[ { z }^{ 2 }+1 \right] \left[ { z }^{ 2 }+1+z \right] =0$

${ z }^{ 2 }+1=0\Longrightarrow z=\pm i$

$\left| z \right| =1$
or

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

$z=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } =\cfrac { -1\pm \sqrt { 3i } }{ 2 }$

$\left| z \right| =\sqrt { \cfrac { 1 }{ 4 } +\cfrac { 3 }{ 4 } } =1$

$\therefore \boxed { \left| z \right| =1 }$

Find the range of real number

$\alpha$ for which the equation

$z + \alpha |z - 1| + 2i = 0; z= x + iy$ has a solution. Find the solution.

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$\displaystyle x = 5/2, y = - 2$
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$\displaystyle x = -2, y = 5/2$
0%
$\displaystyle x = -5/2, y = 2$
0%
$\displaystyle x = 2, y = -5/2$

Explanation

$x+iy+\alpha(\sqrt{(x-1)^{2}+y^{2}})+2i=0$

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

$x^{2}-(y+2)^{2}+i2x(y+2)=\alpha^{2}((x-1)^{2}+y^{2})$
Real part

$x^{2}-\alpha(x-1)^{2}-\alpha(y^{2})=0$ and
Imaginary part

$2x(y+2)=0$

$x(y+2)=0$
Either

$x=0$ or

$y=-2$ ...if the above equations have a solution.
Considering y=-2,
Hence

$\alpha\epsilon[-\dfrac{\sqrt{5}}{2},\dfrac{\sqrt{5}}{2}]$
Hence

$x=\dfrac{5}{2}$ if

$|\alpha|\neq1$ .

$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }=$

0%
$-2$
0%
$2$
0%
$-1$
0%
$1$

Explanation

We have

$\displaystyle \frac { \sqrt { 3 } +i }{ 2 } =\frac { i\sqrt { 3 } +{ i }^{ 2 } }{ 2i }$ (Here we multiplied the numerator and denominator by

$(-i)$ on the LHS)

We get:

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

Simliarly applying the same idea we get the steps below:

and

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

Hence

$\displaystyle { \left( \frac { \sqrt { 3 } +i }{ 2 } \right) }^{ 6 }+{ \left( \frac { i-\sqrt { 3 } }{ 2 } \right) }^{ 6 }={ \left( -i\omega \right) }^{ 6 }+{ \left( -i{ \omega }^{ 2 } \right) }^{ 6 }\\ \\ ={ i }^{ 6 }\left( { \omega }^{ 6 }+{ \omega }^{ 12 } \right) =-1\left( 1+1 \right) =-2$

Let

$\displaystyle\ z_{1}= a+ib, z_{2}= p+iq$ be two unimodular complex numbers such that

$\displaystyle\ Im(z_{1}z_{2})=1$ . If

$\displaystyle\ \omega_{1}= a+ip, \omega_{2}=b+iq$ then

0%
$\displaystyle\ Re(\omega_{1}\omega_{2})=1$
0%
$\displaystyle\ Im(\omega_{1}\omega_{2})=1$
0%
$\displaystyle\ Rm(\omega_{1}\omega_{2})=0$
0%
$\displaystyle\ Im(\omega_{1}\bar{\omega_{2}})=0$

Explanation

$z_1 = a + ib$ and

$z_2 = p + iq$
Given that

$Im(z_1z_2) = 1$ , so

$aq + bp = 1$
Now,

$\omega_1 = a + ip, \omega_2 = b + iq$

$\Rightarrow \omega_1\omega_2 = ab - pq + i(aq + bp)$

$\Rightarrow Im(\omega_1\omega_2) = 1$

Find all complex numbers satisfying the equation

$2|z|^2 + z^2 - 5 + i \sqrt{3} = 0$

0%
$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} + \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} + \frac{3}{2} i \right )$
0%
$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{2}{\sqrt{6}} - \frac{3}{2} i \right )$
0%
$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{3}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{2} i \right )$
0%
$\displaystyle \pm \left ( \frac{\sqrt{6}}{2} - \frac{1}{\sqrt{2}} i \right ); \pm \left ( \frac{1}{\sqrt{6}} - \frac{3}{\sqrt{2}} i \right )$

Explanation

$Let\quad z=a+ib\\ Thus:\quad 2({ a }^{ 2 }+{ b }^{ 2 })+({ a }^{ 2 }-{ b }^{ 2 }+2abi)-5+i\sqrt { 3 } =0\\ =>3{ a }^{ 2 }+{ b }^{ 2 }-5=0\quad and\quad 2ab+\sqrt { 3 } =0=>a=-\frac { \sqrt { 3 } }{ b } \\ Substituting\quad in\quad the\quad first:\quad \frac { 9 }{ 4{ b }^{ 2 } } +{ b }^{ 2 }=5=>{ b }^{ 2 }=\frac { 9 }{ 2 } \quad or\quad \frac { 1 }{ 2 } \\ =>b=\pm \frac { 3 }{ \sqrt { 2 } } \quad or\quad \pm \frac { 1 }{ \sqrt { 2 } } =>a=\mp \frac { 1 }{ \sqrt { 6 } } \quad or\quad \mp \frac { \sqrt { 3 } }{ \sqrt { 2 } } =\mp \frac { \sqrt { 6 } }{ 2 } \\ =>z=\pm \left( \frac { \sqrt { 6 } }{ 2 } -\frac { i }{ \sqrt { 2 } } \right) \quad or\quad \pm \left( \frac { 1 }{ \sqrt { 6 } } -\frac { 3i }{ \sqrt { 2 } } \right) \\$
Hence, (d) is correct.

Let

$\displaystyle z=1+i\:b=(a,b)$ be any complex number,

$\displaystyle a,b,\epsilon R$ and

$\displaystyle \sqrt{-1}=i.$ Let

$\displaystyle z\neq 0+0i,arg z=\tan^{-1}\left (\frac{Im\:z}{Re\:z}\right)$ where

$\displaystyle -\pi<arg z\leq \pi$

$\displaystyle arg(\bar{z})+arg(-z)=\left\{\begin{matrix}\pi, \; if\: arg (z)<0 & \\ -\pi, \; if\: arg (z)>0 & \end{matrix}\right.$

Let

$z$ &

$w$ be non-zero complex numbers such that they have equal modulus values and

$\displaystyle arg z- arg \bar{w} =\pi,$ then z equals

0%
$-w$
0%
$w$
0%
$\displaystyle -\bar{w}$
0%
$\displaystyle \bar{w}$

Explanation

Let

$\overline{w}=cos\theta+isin\theta=e^{i\theta}$

Therefore

$arg(z)=\pi+\theta$

Hence

$z=e^{i(\pi+\theta)}$

$=cos(\pi+\theta)+isin(\pi+\theta)$

$=-cos\theta-isin\theta$

$=-\overline{w}$

The root of

$(x + a) (x + b) - 8k = (k - 2)^2$ are real and equal, when

$a,b,c$

$\epsilon$ R, then

0%
$a + b = 0$
0%
$a =b$
0%
$k = -3$
0%
$k = 0$

Explanation

The equation:

$(x + a) (x + b) - 8k = (k - 2)^2$

$x^2 + x (a + b) - 8k + ab = k^2 + 4 - 4k$

$x^2 + x (a + b) - k^2 - 4k - 4 + ab = 0$

The roots of the equation are real and equal.
Thus, discriminant =

$0$

$D = 0$

$(a + b)^2- 4(1)(ab- k^2- 4k - 4)= 0$

$a^2 +b^2 + 2ab - 4ab + 4k^2 + 16 k + 16 = 0$

$(a - b)^2 + 4(k +2)^2 = 0$
Thus, both

$(a -b) =0$ and

$(k + 2) = 0$
Hence,

$a = b$ and

$k = -2$

$z = x+iy$ and

$w = \dfrac{(1-iz)}{(z-i)}$ , then

$|w| = 1$ implies that, in the complex plane

0%
$z$ lies on the imaginary axis
0%
$z$ lies on the real axis
0%
$z$ lies on the unit circle
0%
None of these

Explanation

Given

$|w|$

$=1$

$1=|\dfrac{1-iz}{z-i}|$

$\Rightarrow |z-i|=|1-iz|$ .....(i)

Now let

$z=x+iy$

$\Rightarrow iz=-y+ix$

Put in (i)

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

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

$\Rightarrow (2y)(2)=0$

$\Rightarrow y=0$ .

This is the equation of the x-axis.

$a, b, c$ are real distinct numbers satisfying the condition

$a + b + c = 0$ , then the roots of the quadratic equation

$3ax^2 + 5bx + 7c = 0$ are

0%
positive
0%
negative
0%
real and distinct
0%
imaginary

Explanation

Given equation is :

$3a{ x }^{ 2 }+5bx+7c=0$
Also given is that

$a+b+c=0\dots(1)$

Discriminant of the equation

$\Rightarrow D =b^2-4ac$

$= { \left( 5b \right) }^{ 2 }-4\left( 3a \right) \left( 7c \right)$

$= 25{ b }^{ 2 }-84ac\dots(2)$

Substituting the value of b from equation (1) in equation (2),we get

$=25{ \left( -a-c \right) }^{ 2 }-84ac$

$=25{ \left( a+c \right) }^{ 2 }-84ac$

$= 25{ a }^{ 2 }+25{ c }^{ 2 }-34ac$

$= 4{ a }^{ 2 }+4{ c }^{ 2 }+8ac$ +

$21{ a }^{ 2 }+{ 21c }^{ 2 }-42ac$

$= 4{ \left( a+c \right) }^{ 2 }+21{ \left( a-c \right) }^{ 2 }> 0$

Since D>0,both the roots will be real and distinct.

Also,in the absence of information on the coefficients a,b and c in the equation we can't conclude whether both the roots will be positive or negative.

Hence,option (C) is the correct alternative.

Find the modulus, argument and the principal argument of the complex numbers.

$z=1+cos\frac {10\pi}{9}+i sin \left (\frac {10\pi}{9}\right )$

0%
Principal Arg $z=-\frac {4\pi}{9}; |z|=2 cos \frac {4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$
0%
Principal Arg $z=-\frac {10\pi}{9}; |z|=2 cos \frac {10\pi}{9}; Arg z=2 k\pi -\frac {10\pi}{9} k\epsilon l$
0%
Principal Arg $z=-\frac {-10\pi}{9}; |z|=2 cos \frac {-10\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$
0%
Principal Arg $z=-\frac {-4\pi}{9}; |z|=2 cos \frac {-4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$

Explanation

Simplifying, we get

$z=2\cos^{2}\left(\dfrac{5\pi}{9}\right)+2i\sin\left(\dfrac{5\pi}{9}\right)\cos \left(\dfrac{5\pi}{9}\right)$

$=2\cos \left(\dfrac{5\pi}{9}\right)[\cos \left(\dfrac{5\pi}{9}\right)+i\sin\left(\dfrac{5\pi}{9}\right)]$

$=-2\cos \left(\dfrac{4\pi}{9}\right)[-\cos \left(\dfrac{4\pi}{9}\right)+i\sin\left(\dfrac{4\pi}{9}\right)]$

$=2\cos \left(\dfrac{4\pi}{9}\right)[\cos \left(\dfrac{4\pi}{9}\right)-i\sin\left(\dfrac{4\pi}{9}\right)]$
Hence

$|z|=2\cos \left(\dfrac{4\pi}{9}\right)$
And
Principal Arg

$\left(z\right)=-\dfrac{4\pi}{9}$

Find the modulus, argument and the principal argument of the complex numbers.

$(tan 1-i)^2$

0%
$Modulus =\sec^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2 -\pi)$
0%
$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$
0%
$Modulus =\sec^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$
0%
$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2- \pi)$

Explanation

Let

$z=\tan1-i$
Then

$|z|=\sqrt{\tan^{2}1+1}=\sec1$

$Arg(z)=\tan^{-1}(\dfrac{-1}{\tan1})$

$=-\tan^{-1}(cot(1))$

$=-(\dfrac{\pi}{2}-\cot^{-1}(cot(1)))$

$=1-\dfrac{\pi}{2}$

Hence

$Z=(\tan1-i)^{2}=z^{2}=|z|^{2}.e^{2i.(1-\frac{\pi}{2})}$

$=\sec^{2}(1)e^{i.(2-\pi)}$
Hence

$|Z|=\sec^{2}(1)$ and principal Arg(z)=

$(2-\pi)$ .

The set of all real values of

$p$ for which the equation

$x + 1 = \displaystyle \sqrt{px}$ has exactly one root is

0%
{0}
0%
{4}
0%
{0, 4}
0%
{0,2}

Explanation

The equation can be written as

$x^{ 2 }+(2-p)x+1$

The equation will have exactly 1 root if the equation is a perfect square,

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

$\Longrightarrow (2-p)^{ 2 }-4=0$

$\Longrightarrow |2-p|=2$

$\Longrightarrow 2-p=2$ or

$2-p=-2$ ,

$p=0,4$

$z_1, z_2, ..., z_n$ lie on

$|z|=r$ and

$Re\left(\displaystyle\sum_{j=1}^n\displaystyle\sum_{k=1}^n{\displaystyle\frac{z_j}{z_k}}\right) = 0$ , then

0%
$\displaystyle\sum_{j=1}^n{z_j}=0$
0%
$\left|\displaystyle\sum_{j=1}^n{z_j}\right|=0$
0%
$\displaystyle\sum_{j=1}^n{\displaystyle\frac{1}{z_j}}=0$
0%
None of these

Explanation

Let

${ z }_{ i }=r(cos{ \theta }_{ i }+i.sin{ \theta }_{ i })$

$Re\left(\displaystyle\sum_{j=1}^n\displaystyle\sum_{k=1}^n{\displaystyle\frac{z_j}{z_k}}\right) = 0$
Thus,

$\displaystyle ({ z }_{ 1 }+{ z }_{ 2 } + ... + { z }_{ n }) \times \left (\frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } ...\frac { 1 }{ { z }_{ n } } \right )=0$
Thus,

$[(cos{ \theta }_{ 1 }+...cos{ \theta }_{ n })+i.(sin{ \theta }_{ 1 }+...s{ in\theta }_{ n })] \times [(cos{ \theta }_{ 1 }+...cos{ \theta }_{ n })-i.(sin{ \theta }_{ 1 }+...s{ in\theta }_{ n })]=0$
So,

${ (cos{ \theta }_{ 1 }+...cos{ \theta }_{ n }) }^{ 2 }+{ (sin{ \theta }_{ 1 }+...s{ in\theta }_{ n }) }^{ 2 }=0$
So,

${ (cos{ \theta }_{ 1 }+...cos{ \theta }_{ n }) }={ (sin{ \theta }_{ 1 }+...s{ in\theta }_{ n }) }=0$
Hence (a), (b), (c) are correct.

Given that

$i = \sqrt {-1}$ , find the multiplicative inverse of

$5 - i$ .

0%
$5 + i$
0%
$\dfrac {5 + i}{26}$
0%
$\dfrac {1}{5 + i}$
0%
$\dfrac {5 + i}{24}$
0%
$\dfrac {5 - i}{24}$

Explanation

Let the multiplicative inverse of

$5-i$ be

$a+bi$

Therefore we have

$(5-i)(a+bi) = 1$

$\Rightarrow 5a+b +i(5b-a) = 1$

$\Rightarrow 5a+b = 1$ and

$5b-a = 0$

$\Rightarrow a=5b$

Substitute this in

$5a+b =1$ , we get

$b = \dfrac {1}{26}$ and

$a = \dfrac {5}{16}$

Therefore, multiplicative inverse is

$\dfrac {(5+i)}{26}$ .

Write the complete number

$- 2 - 2i$ in polar form.

0%
$2(cos\dfrac{\pi}{4}+i\,sin\dfrac{\pi}{4})$
0%
$-2(cos\dfrac{\pi}{4}-i\,sin\dfrac{\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{3\pi}{4}+i\,sin\dfrac{3\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{7\pi}{4}+i\,sin\dfrac{7\pi}{4})$
0%
$2\sqrt{2}(cos\dfrac{5\pi}{4}+i\,sin\dfrac{5\pi}{4})$

Explanation

given complex number is $-2-2i$
the magnitude of given number is $2\sqrt2$
So the number can be written as $2\sqrt2(-1/\sqrt2 -i/\sqrt2)$
which gives $2\sqrt2(cos(225)+isin(225)) = 2\sqrt2 (cos\frac{5\pi}{4}+isin\frac{5\pi}{4})$

$a> 0$ and discriminant of

$a{ x }^{ 2 }+2bx+c$ is -ve then

$\begin{vmatrix} a & b & ax+b \\ b & c & bx+c \\ ax+b & bx+c & 0 \end{vmatrix}$ is equal to

0%
+ve
0%
$\left( ac-{ b }^{ 2 } \right) \left( a{ x }^{ 2 }+2bx+c \right)$
0%
-ve
0%
$0$

$(cos\theta+i\, sin \theta)\,\,( cos\,2\theta+i\,sin\,\theta)$ ....

$(cos\, n\theta +i\,sin\,n\theta) = 1$ , then the value of

$\theta$ is

0%
$\dfrac{2m\pi}{n\,(n+1)}$
0%
$4\,m\,\pi$
0%
$\dfrac{4\,m\pi}{n\,(n+1)}$
0%
$\dfrac{m\pi}{n\,(n+1)}$

Explanation

We have,

$(cos \theta+ i\, sin \theta) (cos \,2\theta + i \,sin 2\theta) ...$

$(cos \,n\theta + i \,sin\, n\theta) = 1$

$\therefore cos (\theta + 2\theta + 3\theta + .... + n\theta)+ i\, sin (\theta + 2\theta + 3\theta + ......... + n\theta) = 1$

$\Rightarrow cos\dfrac{n\,n+1}{2}\theta+i\,sin\dfrac{n\,n+1}{2}\theta=1$
On comparing the coefficients of real and imaginary parts on both sides; we get

$cos\dfrac{n\,(n+1)}{2}\theta=1$
And

$sin\,\dfrac{n(n+1)}{2}\theta=0$

$\therefore \dfrac{n\,n+1}{2}\therefore = 2\,m\pi$

$\Rightarrow \theta = \dfrac{4\,m\pi}{n\,(n+1)}$

Add and express in the form of a complex number

$a+bi$

$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$

0%
$-4+9i$
0%
$-5+9i$
0%
$2+9i$
0%
$-5+8i$

Explanation

Adding the given expression as follows:

$(2+3i)+(-4+5i)-\dfrac {(9-3i)}{3}$

$=(2+3i)+(-4+5i)-(3-i)$

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

$=-5+9i$

$z$ is a complex number such that

$|z|\geq 2$ then the minimum value of

$\left |z + \dfrac {1}{2}\right |$ is

0%
Is strictly greater than $\dfrac {5}{2}$
0%
Is strictly greater than $\dfrac 32$ but less than $\dfrac {5}{2}$
0%
Is equal to $\dfrac {5}{2}$
0%
Lies in the interval $(1, 2)$

Explanation

$\left| z \right| \ge 2$ is the region on or out side the circle whose center is

$\left( 0,0 \right)$ and radius is

$2$ .

Minimum

$\left| z+\dfrac { 1 }{ 2 } \right|$ is distance of

$z$ which lie on circle

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

$\left( -\dfrac { 1 }{ 2 } ,0 \right)$

Thus minimum

$\left| z+\dfrac { 1 }{ 2 } \right| =$ Distance between

$\left( -\dfrac { 1 }{ 2 } ,0 \right)$ to

$\left( 0,0 \right)$

$= \sqrt { { \left( -\dfrac { 1 }{ 2 } +2 \right) }^{ 2 }{ +\left( 0-0 \right) }^{ 2 } }$

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

Apply radical rule

$\sqrt [ n ]{ a^{ n } } =a,$ assuming

$a\ge \: 0$

$=-\dfrac { 1 }{ 2 } +2$

$=\dfrac { 2 }{ 1 } -\dfrac { 1 }{ 2 }$

$=\dfrac { 2\cdot \: 2 }{ 2 } -\dfrac { 1 }{ 2 }$

$=\dfrac { 2\cdot \: 2-1 }{ 2 }$

$=\dfrac { 3 }{ 2 }$

Hence, option B is correct.

$\alpha$ and

$\beta$ are two different complex numbers with

$|\beta|=1$ , then

$\left | \dfrac{\beta -\alpha}{1-\bar{\alpha }\beta } \right |$ is equal to.

0%
$0$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$-1$

Explanation

Let

$t = \left|\cfrac { \beta -\alpha }{ 1-\overline { \alpha } \beta } \right|$

Multiply the numerator and denominator with

$\overline { \beta }$

$\Rightarrow t = \left|\cfrac { 1-\alpha \overline { \beta } }{ \overline { \beta } -\overline { \alpha } } \right| = \left|\cfrac { \overline { 1- \alpha \overline { \beta } } }{ \overline { \overline{\beta} -\overline{\alpha} } } \right| = \left|\cfrac { 1-\overline { \alpha } \beta }{ \beta -\alpha } \right|=\cfrac{1}{t}$

$\Rightarrow {t}^{2} = 1$

$\Rightarrow t = 1$

If the equation

$\displaystyle 4x^{2}+x\left ( p+1 \right )+1=0$ has exactly two equal roots , then one of the value of

$p$ is

0%
$5$
0%
$-3$
0%
$0$
0%
$3$

Explanation

Since the equation

$\displaystyle 4x^{2}+x(p+1)+1=0$ has two equal roots therefore
Discriminant

$=0$
or

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

$\displaystyle (p+1)^{2}-4\times 4\times 1=0$
or

$\displaystyle p^{2}+2p+1-16=0$
or

$\displaystyle p^{2}+2p-15=0$
or

$\displaystyle \left ( p+5 \right )\left ( p-3 \right )=0$

$\displaystyle \therefore p=-5$ or

$p=3$
Hence, one of the values of

$p = 3$ .

Evaluate in standard form:

$\dfrac {(2-3i)}{(2-2i)}$ , where

${i}^{2}=-1$ .

0%
$\dfrac {5}{4}-\dfrac {i}{4}$
0%
$\dfrac {5}{4}+\dfrac {i}{4}$
0%
$-\dfrac {5}{4}-\dfrac {i}{4}$
0%
$-\dfrac {5}{4}+\dfrac {i}{4}$

Explanation

Consider

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

Rationalise and solve considering that

$i^2=-1$ as shown below:

$\Rightarrow \dfrac { 2-3i }{ 2-2i } \times \dfrac { 2+2i }{ 2+2i } =\dfrac { 4+4i-6i-{ 6i }^{ 2 } }{ 4-{ 4i }^{ 2 } }$

$=\dfrac { 4-2i+6 }{ 4+4 }$

$=\dfrac { -2i+10 }{ 8 }$

$=\dfrac { -i }{ 4 } +\dfrac { 5 }{ 4 }$

${z}_{1},{z}_{2},..{z}_{n}$ lie on the circle

$|z|=2$ then the value of

$|{z}_{1},{z}_{2},..{z}_{n}|-4|\dfrac {1}{{z}_{1}}+\dfrac {1}{{z}_{2}}++\dfrac {1}{{z}_{n}}|=$

0%
$0$
0%
$n$
0%
$-n$
0%
$1$

Let

$z_1$ = 18 + 83i,

$z_2$ = 18 + 39i, ana

$z_3$ = 78 + 99i. where i =

$\sqrt-1$ . Let z be a unique comlpex number with the properties that

$\dfrac{z_3 - z_1}{z_2 - z_1}$

$\cdot$

$\dfrac{z - z_2}{z - z_3}$ is a real number and the imaginary part of the size z is the greatest possible.

0%
$Re (z) = 56$
0%
$Re (z) = 61$
0%
$Re (z) = 54$
0%
$Re (z) = 59$

Explanation

Using the property that

$\dfrac{z_3-z_1}{z_2-z_1}.\dfrac{z-z_2}{z-z_3}$ is a real number then the four points are concyclic i.e. are present on a circle

since it is told the imaginary part is maximum so,it the top point of the circle

so we take the perpendicular bisector of any two line and take its intersection.

As shown in the figure,

Equation of perpendicular bisector of

$z_1z_2$

$y=61$

Equation of perpendicular bisector of

$z_3z_2$

$y+x=117$

Solving the equation we get,

$x=56$

So,

$Re(z)=56$

$z=\sqrt{20i-21}+\sqrt{21+20i}$ , then the principal value of arg 'z' can be

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

What is

${ i }^{ 1000 }+{ i }^{ 1001 }+{ i }^{ 1002 }+{ i }^{ 1003 }$ equal to (where

$i=\sqrt { -1 }$ )?

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

Explanation

$i^{1000} + i^{100} + i^{1002} + i^{1003}$

$i^{2} = -1, i^{3} = -1, i^{4} = 1$

$i^{1000} = (i^{4})^{250} = 1\ i^{1002} = i^{2} - i^{1000} = -1$

$i^{1001} = i\cdot i^{1000} = i\ i^{1003} = i^{3}\cdot i^{1000} = -i$

$\therefore i^{1000} + i^{1001} + i^{1002} + i^{1003} = 1 - 1 + i - i = 0$

$Z_{1},Z_{2}$ are two complex numbers satisfying

$|\dfrac{Z_{1}-3Z_{2}}{3-Z_{1}Z_{2}}|=1|z_{1}|\neq 3$ then

$|z_{2}|=$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

A complex number z is said to be unimodular if

$|z| =$ . Suppose

$z_1$ and

$z_2$ are complex numbers such that

$\frac{z_1-2z_2}{2-z_1\overline {z}_2}$ is unimolecolar and

$z_2$ is not unimodular. Then the point

$z_1$ lies on a:

0%
straight line parallel to y-axis
0%
circle of radius 2.
0%
Circle of radius $\sqrt2$ .
0%
Staright line parallel to x-axis.

$z_1+ z_2 + z_3 = 0$ and

$|z_1|=|z_2|=|z_3|= 1$ , then area of triangle whose vertices are

$z_1, z_2$ and

$z_3$ is:

0%
$\dfrac{3 \sqrt{3}}{4}$
0%
$\dfrac{\sqrt{3}}{4}$
0%
$1$
0%
$2$

Explanation

${|z^{}_{1}+z^{}_{2}|}^2+{|z^{}_{1}-z^{}_{2}|}^2=2({|z^{}_{1}|}^2+{|z^{}_{2}|}^2)$ .......

$(1)$

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

$z^{}_{3}=-z^{}_{2}-z^{}_{1}$

By taking modulus both sides

$|-z^{}_{3}|=|z^{}_{2}+z^{}_{1}|$ .........

$(2)$

Using

$(2)$ in

$(1),$ we have

${|-z_{3}|}^2+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$

$1+{|z^{}_{1}-z^{}_{2}|^2}=2{(1+1)}$

${|z^{}_{1}-z^{}_{2}|^2}=3$

$|z^{}_{1}-z^{}_{2}|=\sqrt{3}$

Similarily

$|z^{}_{2}-z^{}_{3}|$ =

$|z^{}_{3}-z^{}_{1}|=\sqrt{3}$

Hence sides of triangle formed are

$\sqrt{3},$

$\sqrt{3}$ and are equal.

So triangle formed is an equilateral triangle

Area of an equilateral triangle

$=\dfrac{\sqrt{3}}{4}s^2$

where

$s$ is side of equilateral triangle

So area of triangle formed by

$z^{}_{1} ,z^{}_{2},z^{}_{3}=3\dfrac{\sqrt{3}}{4}$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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