Solution

Given expression $=(1-i)^{n} (1-\dfrac{1}i)^{n}$

$ \begin{aligned} & =(1-i)^{n}(i-1)^{n} \cdot i^{-n}=(1-i)^{n}(1-i)^{n}(-1)^{n} \cdot i^{-n} \\ \\ & =[(1-i)^{2}]^{n}(-1)^{n} \cdot i^{-n}=(1+i^{2}-2 i)^{n}(-1)^{n} i^{-n} \\ \\ & =(1-1-2 i)^{n}(-1)^{n} i^{-n} \quad[\because i^{2}=-1] \\ \\ &=(-2)^{n} \cdot i^{n}(-1)^{n} i^{-n} \\ \\ & =(-1)^{2 n} \cdot 2^{n}=2^{n} \end{aligned} $

Thinking Process

Use $i^{2}=-1, i^{4}=(-1)^{2}=1, i^{3}=-i \text {, and } i^{5}=i$ to solve it

Solution

Given that, $\displaystyle \sum_{n=1}^{13}(i^{n}+i^{n+1}), n \in N$

$=(i+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}+i^{13}) $

$\quad+(i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}+i^{13}+i^{14}) $

$=(i+2 i^{2}+2 i^{3}+2 i^{4}+2 i^{5}+2 i^{6}+2 i^{7}+2 i^{8}+2 i^{9}+2 i^{10}+2 i^{11}+2 i^{12}+2 i^{13}+i^{14}) $

$=i-2-2 i+2+2 i+2(i^{4}) i^{2}+2(i)^{4} i^{3}+2(i^{2})^{4}+2(i^{2})^{4} i+2(i^{2})^{5} $

$\quad+2(i^{2})^{5} \cdot i+2(i^{2})^{6}+2(i^{2})^{6} \cdot i+(i^{2})^{7} $

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

$=-1+i$

Alternate Method

$ \begin{aligned} & \sum_{n=1}^{13}(i^{n}+i^{n+1}), n \in N=\sum_{n=1}^{13} i^{n}(1+i) \\ \\ & =(1+i)[i+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+i^{10}+i^{11}+i^{12}+i^{13}] \\ \\ & =(1+i)[i^{13}] \quad[\because i^{n}+i^{n+1}+i^{n+2}+i^{n+3}=0, \text { where } n \in N \text { i.e., } \sum_{n=1}^{12} i^{n}=0] \\ \\ & =(1+i) i \quad [\because i^{13}=(i^4)^3 . i = i] \\ \\ & =(i^{2}+i)=i-1 \end{aligned} $

Thinking Process

If two complex numbers $z_1=x_1+i y_{1}$ and $z_2=x_2 +iy_{2}$ are equal

$ \text { i.e., } \quad z_1=z_2 \Rightarrow x_1+i y_1=x_2+i y_2 \text {, then } x_1=x_2 \text { and } y_1=y_2 \text {. } $

Solution

Given that, $\left(\dfrac{1+i}{1-i}\right)^{3}-\left(\dfrac{1-i}{1+i}\right)^{3}=x+i y \quad \ldots (i)$

$ \begin{aligned} \therefore \quad \left(\dfrac{1+i}{1-i}\right)^3 & =\dfrac{1+i^{3}+3 i(1+i)}{1-i^{3}-3 i(1-i)}=\dfrac{1-i+3 i+3 i^{2}}{1+i-3 i+3 i^{2}} \\ \\ & =\dfrac{2 i-2}{-2 i-2}=\dfrac{i-1}{-i-1}=\dfrac{1-i}{1+i} \\ \\ & =\dfrac{(1-i)}{(1+i)} \dfrac{(1-i)}{(1-i)}=\dfrac{1+i^{2}-2 i}{1+1}=\dfrac{1-1-2 i}{2} \end{aligned} $

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

Similarly, $\quad \left(\dfrac{1-i}{1+i}\right)^3=\dfrac{1+i}{1-i}=\dfrac{2i}{2}=i \quad \ldots (iii)$

Using Eqs. (ii) and (iii) in Eq. (i), we get

$ \begin{aligned} -i-i & =x+i y \\ \\ \Rightarrow -2 i & =x+i y \end{aligned} $

On comparing real and imaginary part of complex number, we get

$ x=0 \text { and } y=-2 $

So, $ \quad (x, y)=(0,-2)$

Solution

Given that, $ \dfrac{(1+i)^{2}}{2-i}=x+i y $

$ \begin{matrix} \Rightarrow & \dfrac{(1+i^{2}+2 i)}{2-i}=x+i y \\ \\ \Rightarrow & \dfrac{2 i}{2-i}=x+i y \\ \\ \Rightarrow & \dfrac{2 i(2+i)}{(2-i)(2+i)}=x+i y \\ \\ \Rightarrow & \dfrac{4 i+2 i^{2}}{4-i^{2}}=x+i y \end{matrix} $

$\Rightarrow \quad \dfrac{4 i-2}{4+1}=x+i y \\ \\ \Rightarrow \quad \dfrac{-2}{5}+\dfrac{4 i}{5}=x+i y$

On comparing real and imaginary part of complex number, we get

$ x=-\dfrac{2}{5} \Rightarrow y=\dfrac{4}{5} $

$ \Rightarrow \quad x+y=\dfrac{-2}{5}+\dfrac{4}{5}=\dfrac{2}{5} $

Solution

Given that, $\left(\dfrac{1-i}{1+i}\right)^{100}=a+i b$

$ \begin{aligned} & \Rightarrow \quad \left[\dfrac{(1-i)}{(1+i)} \cdot \dfrac{(1-i)}{(1-i)}\right]^{100}=a+i b \\ \\ & \Rightarrow \quad {\left(\dfrac{1+i^{2}-2 i}{1-i^{2}}\right)}^{100}=a+i b \\ \\ & \Rightarrow \quad \left(\dfrac{-2 i}{2}\right)^{100}=a+i b \quad[\because i^{2}=-1] \\ \\ & \Rightarrow \quad(i^{4})^{25}=a+i b \\ \\ & \Rightarrow \quad 1=a+i b \quad[\because i^{4}=1] \end{aligned} $

On comparing real and imaginary part of complex number, we get

$ \begin{aligned} & \quad a=1 \text { and } b=0 \\ \\ & \therefore \quad(a, b)=(1,0) \end{aligned} $

Thinking Process

To solve the above problem use the trigonometric formula

$\cos \theta=2 \cos ^{2} \theta /2$ $-1=1-2 \sin ^{2} \theta / 2$ and $\sin \theta=2 \sin \theta / 2 \cdot \cos \theta / 2$.

Solution

Given that, $a=\cos \theta+i \sin \theta$

$ \begin{aligned} \therefore \quad \dfrac{1+a}{1-a} & =\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta} \\ \\ & =\dfrac{1+2 \cos ^{2} \theta / 2-1+2 i \sin \theta / 2 \cdot \cos \theta / 2}{1-1+2 \sin ^{2} \theta / 2-2 i \sin \theta / 2 \cdot \cos \theta / 2}=\dfrac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 \sin \theta / 2(\sin \theta / 2-i \cos \theta / 2)} \\ \\ & =-\dfrac{2 \cos \theta / 2(\cos \theta / 2+i \sin \theta / 2)}{2 i \sin \theta / 2(\cos \theta / 2+i \sin \theta / 2)}=-\dfrac{1}{i} \cot \theta / 2 \\ \\ & =\dfrac{i^{2}}{i} \cot \theta / 2 \quad \left[\because \dfrac{-1}{i}=\dfrac{i^{2}}{i}\right] \\ \\ & =i \cot \theta / 2 \end{aligned} $

Solution

We have, $(1+i) z=(1-i) \bar{z} $

$\Rightarrow \dfrac{z}{\bar{z}}=\dfrac{(1-i)}{(1+i)}$

$\Rightarrow \dfrac{z}{\bar{z}}=\dfrac{(1-i)}{(1+i)} \dfrac{(1-i)}{(1-i)} $

$\Rightarrow \dfrac{z}{\bar{z}}=\dfrac{1+i^{2}-2 i}{1-i^{2}} \quad {[\because i^{2}=-1]} $

$\Rightarrow \dfrac{z}{\bar{z}}=\dfrac{1-1-2 i}{2}$

$ \Rightarrow \dfrac{z}{\bar{z}}=-i $

$\therefore \quad z=-i \bar{z}$

Hence proved.

Solution

Given that, $z=x+iy$

Then, $\bar{z} = x-iy$

Now, $ z \bar{z}+2(z+\bar{z})+b=0$

$\Rightarrow \quad(x+i y)(x-i y)+2(x+i y+x-i y)+b=0$

$\Rightarrow \quad x^{2}+y^{2}+4 x+b=0$, which is the equation of a circle.

Solution

Let $\quad z =x+i y $ and $\bar{z} = x-iy$

Now, $\quad \dfrac{\bar{z}+2}{\bar{z}-1} =\dfrac{x-i y+2}{x-i y-1} $

$=\dfrac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]} $

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

$ =\dfrac{(x-1)(x+2)+y^{2}+i[(x+2) y-(x-1) y]}{(x-1)^{2}+y^{2}} \quad[\because-i^{2}=1]$

According to the question, we have, real part $=4$

$\therefore \quad \dfrac{(x-1)(x+2)+y^{2}}{(x-1)^{2}+y^{2}}=4$

$\Rightarrow \quad x^{2}-x+2 x-2+y^{2}=4(x^{2}-2 x+1+y^{2})$

$\Rightarrow \quad 3 x^{2}+3 y^{2}-9 x+6=0$, which represents a circle.

Hence, $z$ lies on the circle.

Thinking Process

First use, $\arg \left(\dfrac{z_1}{z_2}\right)=\arg (z_1)-\arg (z_2)$.

Also apply $\arg (z)=\theta=\tan ^{-1} (\dfrac{y}{x})$, where $z=x+i y$

and then use the property $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \left(\dfrac{x-y}{1+x y}\right)$

Solution

Let $\quad z =x+i y $

Given that, $\quad \arg \left(\dfrac{z-1}{z+1}\right) =\dfrac{\pi}{4}$

$\Rightarrow \arg (z-1)-\arg (z+1)=\dfrac{\pi}{4} $

$\Rightarrow \arg (x+i y-1)-\arg (x+i y+1)=\dfrac{\pi}{4} $

$\Rightarrow \arg (x-1+i y)-\arg (x+1+i y)=\dfrac{\pi}{4}$

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

$\Rightarrow \tan ^{-1} \left[\dfrac{\dfrac{y}{x-1}-\dfrac{y}{x+1}}{1+\left(\dfrac{y}{x-1} \right)\left(\dfrac{y}{x+1}\right)}\right]=\dfrac{\pi}{4} $

$\Rightarrow \dfrac{y \left[\dfrac{x+1-x+1}{x^{2}-1}\right]}{\dfrac{x^{2}-1+y^{2}}{x^{2}-1}}=\tan \left(\dfrac{\pi}{4}\right) $

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

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

$\Rightarrow x^{2}+y^{2}-2 y-1=0,$ which represents a circle.

Solution

The given equation is $|z|=z+1+2 i \quad \ldots (i)$

Let $\quad z=x+i y$

From Eq. (i), $\quad|x+i y|=x+i y+1+2 i$

$ \begin{matrix} \Rightarrow & \sqrt{x^{2}+y^{2}}=x+i y+1+2 i & [\because|z|+\sqrt{x^{2}=y^{2}}] \\ \\ \Rightarrow & \sqrt{x^{2}+y^{2}}=(x+1)+i(y+2) & \end{matrix} $

On squaring both sides, we get

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

$\Rightarrow x^{2}+y^{2}=x^{2}+2 x+1-y^{2}-4 y-4+2 i(x+1)(y+2)$

On comparing real and imaginary parts,

$x^{2}+y^{2} =x^{2}+2 x+1-y^{2}-4 y-4 $

i.e., $\quad 2y^2 = 2x-4y-3\quad \ldots (ii)$

and $\quad 2(x+1)(y+2) =0$

$(x+1)=0 \text { or }(y+2) =0$

$\Rightarrow x=-1, \text { or } y=-2$

For $\quad x= -1, 2y^2 = -2-4y-3$

$2y^2+ 4y + 5 =0 \qquad$ [using Eq. (ii)]

$\Rightarrow y =\dfrac{-4 \pm \sqrt{16-2 \times 4 \times 5}}{4} $

$y =\dfrac{-4 \pm \sqrt{-24}}{4} \notin R$

For $\quad y=-2, $

$2(-2)^{2} =2 x-4(-2)-3 \qquad$ [using Eq. (ii)]

$\Rightarrow 8 =2 x+8-3$

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

$\therefore \quad z= x+iy=\dfrac{3}{2}-2i$

Solution

Given that,

$|z+1| =z+2(1+i) \quad \ldots {(i)}$

Let, $z =x+i y$

Then,

$ |x+i y+1|=x+i y+2(1+i) $

$ \begin{aligned} & \Rightarrow \quad|x+1+i y|=(x+2)+i(y+2) \\ \\ & \Rightarrow \quad \sqrt{(x+1)^{2}+y^{2}}=(x+2)+i(y+2) \end{aligned} $

On comparing real and imaginary parts, we get

$\sqrt{(x+1)^2+y^2} = (x+2) \quad \ldots (ii)$

and $y+2 = 0$

$\Rightarrow y=-2$

From eq. (ii)

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

On squaring both sides, we get

$\Rightarrow \quad (x+1)^2+4 = x^2+4x+4$

$\Rightarrow \quad x^2+2x+1+4 = x^2+4x+4$

$\Rightarrow \quad -2x+1=0$

$\Rightarrow \quad x=\dfrac{1}{2}$

$\therefore \quad z=x+iy = \dfrac{1}{2}-2i$

Solution

Given that, $\arg (z-1)=\arg (z+3 i)$

$\text{and } \quad \text { let } z =x+i y$

$\text{Now } \quad \arg (z-1) =\arg (z+3 i) $

$\Rightarrow \arg (x+i y-1) =\arg (x+i y+3 i) $

$\Rightarrow \arg (x-1+i y) =\arg [x+i(y+3)] $

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

$\Rightarrow \dfrac{y}{x-1} =\dfrac{y+3}{x} $

$\Rightarrow x y=(x-1)(y+3) $

$\Rightarrow x y =x y-y+3 x-3 $

$\Rightarrow 3 x-3=y$

$\Rightarrow \dfrac{3(x-1)}{y}=1 $

$\Rightarrow \dfrac{x-1}{y}=\dfrac{1}{3} $

$\therefore (x-1): y=1: 3$

Thinking Process

We know that general equation of a circle is $x^{2}+y^{2}+2 g x+2 f y+c=0$, with centre $(-g,-f)$ and radius $=\sqrt{g^{2}+f^{2}-c}$.

Solution

Let $z=x+i y$

Given, equation is $\left|\dfrac{z-2}{z-3}\right|=2 $

$\Rightarrow \quad \left|\dfrac{x+i y-2}{x+i y-3}\right|=2$

$\Rightarrow \quad|x-2+i y|=2 | x-3+i y |$

$\Rightarrow \quad \sqrt{(x-2)^{2}+y^{2}}=2 \sqrt{(x-3)^{2}+y^{2}} \quad [\because|x+i y|=\sqrt{x^{2}+y^{2}}]$

On squaring both sides, we get

$ \begin{aligned} & & x^{2}-4 x+4+y^{2} & =4(x^{2}-6 x+9+y^{2}) \\ \\ \Rightarrow & & 3 x^{2}+3 y^{2}-20 x+32 & =0 \\ \\ \Rightarrow & & x^{2}+y^{2}-\dfrac{20}{3} x+\dfrac{32}{3} & =0 \end{aligned} $

On comparing the above equation with $x^{2}+y^{2}+2 g x+2 f y+c=0$, we get

$\Rightarrow \quad 2 g=\dfrac{-20}{3} \Rightarrow g=\dfrac{-10}{3} $

and $\quad 2 f=0 \Rightarrow f=0 \text { and } c=\dfrac{32}{3} $

$\therefore \quad \text { Centre }=(-g,-f)=\left(\dfrac{10}{3},0 \right)$

Also, $\quad \text { radius }(r)=\sqrt{\left(\dfrac{10}{3}\right)^{2}+0-\dfrac{32}3} \qquad[\because r=\sqrt{g^{2}+f^{2}-c}] $

$\qquad \qquad \qquad =\dfrac{1}{3} \sqrt{(100-96)}=\dfrac{2}{3}$

Thinking Process

If $z=x +iy$ is a purely imaginary number, then its real part must be equal to zero i.e., $x=0$,

Solution

Let $z =x+i y $

$\dfrac{z-1}{z+1} =\dfrac{x+i y-1}{x+i y+1}, z \neq-1 $

$=\dfrac{x-1+i y}{x+1+i y}=\dfrac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)}$

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

$\Rightarrow \quad \dfrac{z-1}{z+1} =\dfrac{(x^{2}-1)+y^{2}+i[y(x+1)-y(x-1)]}{(x+1)^{2}+y^{2}}$

Given that, $\dfrac{z-1}{z+1}$ is a purely imaginary numbers.

Then, $\quad \dfrac{(x^{2}-1)+y^{2}}{(x+1)^{2}+y^{2}}=0$

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

$\Rightarrow \sqrt{x^{2}+y^{2}}=\sqrt{1} $

$\Rightarrow|z|=1 \quad {[\because|z|=\sqrt{x^{2}+y^{2}}]}$

Solution

Let $z_1=r_1(\cos \theta_1+i \sin \theta_1)$ and $z_2=r_2(\cos \theta_2+i \sin \theta_2)$ are two complex numbers.

Given that, $\quad |z_1| =|z_2| $

and $\quad \arg (z_1)+\arg (z_2) =\pi $

If $\quad |z_1| =|z_2| $

$\Rightarrow r_1 =r_2 $

and if $\quad \arg (z_1)+\arg (z_2) =\pi $

$\Rightarrow \theta_1+\theta_2 =\pi$

$\Rightarrow \theta_1 =\pi-\theta_2$

Now $\quad z_1=r_1(\cos \theta_1+i \sin \theta_1)$

$\Rightarrow z_1=r_2(\cos (\pi-\theta_2)+i \sin (\pi-\theta_2)] \quad [\because r_1=r_2 \text { and } \theta_1=(\pi-\theta_2)] $

$\Rightarrow z_1=r_2(-\cos \theta_2+i \sin \theta_2) $

$\Rightarrow z_1=-r_2(\cos \theta_2-i \sin \theta_2) $

$\Rightarrow z_1=-[r_2(\cos \theta_2-i \sin \theta_2)] \quad {[\because \bar{z}_2=r_2(\cos \theta_2-i \sin \theta_2)]} $

$\Rightarrow z_1=-\bar{z}_2 $

Solution

Let $z_1 =x+i y $

$\Rightarrow |z_1| =\sqrt{x^{2}+y^{2}}=1 \qquad [\because |z_1| = 1, \text{given}] \quad \ldots (i)$

Now, $z_2 =\dfrac{z_1-1}{z_1+1}=\dfrac{x+i y-1}{x+i y+1} $

$=\dfrac{x-1+i y}{x+1+i y}=\dfrac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)} $

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

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

$=\dfrac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}=\dfrac{1-1+2 i y}{(x+1)^{2}+y^{2}} \quad[\because x^{2}+y^{2}=1] $

$=0+\dfrac{2 y i}{(x+1)^{2}+y^{2}}$

Hence, the real part of $z_2$ is zero.

Thinking Process

First let, $z=r(\cos \theta+i \sin \theta)$, then conjugate of $z$ i.e., $\bar{z}=r(\cos \theta-i \sin \theta)$.

Use the property $\arg (\dfrac{z_1}{z_2})=\arg (z_1)-\arg (z_2)$

Solution

Let $z_1=r_1(\cos \theta_1+i \sin \theta_1)$,

Then, $\quad z_2=\bar{z}_1=r_1(\cos \theta_1-i \sin \theta_1)=r_1[\cos (-\theta_1)+\sin (-\theta_1)]$

Also, $\quad$ let $\quad z_3=r_2(\cos \theta_2+i \sin \theta_2)$,

Then, $\quad z_4=\bar{z}_3=r_2(\cos \theta_2-i \sin \theta_2)=r_2 [\cos (- \theta _2)+i \sin (- \theta _2)]$

$ \begin{aligned} \arg \dfrac{z_1}{z_4}+\arg \dfrac{z_2}{z_3} & =\arg (z_1)-\arg (z_4)+\arg (z_2)-\arg (z_3) \\ \\ & =\theta_1-(-\theta_2)+(-\theta_1)-\theta_2 \quad[\because \arg (z)=\theta] \\ \\ & =\theta_1+\theta_2-\theta_1-\theta_2=0 \end{aligned} $

Solution

Given that,

$ |z_1| =|z_2|=\cdots=|z_{n}|=1 $

$\Rightarrow |z_1|^{2} =|z_2|^{2}=\cdots=|z_{n}|^{2}=1$

$\Rightarrow z_1 \bar z_1 =z_2 \bar z_2=z_3 \bar z_3=\cdots=z_{n} \bar z_n=1 $

$\Rightarrow z_1 =\dfrac{1}{\bar z_1}, z_2=\dfrac{1}{\bar z_2}=\cdots=z_{n}=\dfrac{1}{\overline{z_n}}$

Now, $|z_1+z_2+z_3+z_4+\cdots+z_{n}|$

$=|\overline{z_1+z_2+z_3+ \ldots + z_n}| \quad [\because |z|=|\bar{z}|]$

$=|\bar{z_1}+\bar{z_2}+\bar{z_3}+ \ldots + \bar{z_n}|$

$=\left|\dfrac{1}{z_1}+\dfrac{1}{z_2}+\dfrac{1}{z_3}+ \ldots +\dfrac{1}{z_n}\right|$

Hence, proved.

Solution

Let $ \ z_1=r_1(\cos \theta_1+i \sin \theta_1)$

and $\ z_2=r_2(\cos \theta_2+i \sin \theta_2)$

$\Rightarrow \arg (z_1)=\theta_1 \text { and } \arg (z_2)=\theta_2$

Given that, $\ \arg (z_1) - \arg(z_2) = 0$

$ \theta_1-\theta_2=0 \Rightarrow \theta_1=\theta_2 $

$ z_2=r_2(\cos \theta_1+i \sin \theta_1) $

$ z_1-z_2=(r_1 \cos \theta_1-r_2 \cos \theta_1)+i(r_1 \sin \theta_1-r_2 \sin \theta_1) $

$ \begin{aligned} |z_1-z_2| & =\sqrt{(r_1 \cos \theta_1-r_2 \cos \theta_1)^{2}+(r_1 \sin \theta_1-r_2 \sin \theta_1)^{2}} \\ \\ & =\sqrt{r_1^{2}+r_2^{2}-2 r_1 r_2 \cos ^{2} \theta_1-2 r_1 r_2 \sin ^{2} \theta_1} \\ \\ & =\sqrt{r_1^{2}+r_2^{2}-2 r_1 r_2(\sin ^{2} \theta_1+\cos ^{2} \theta_1)} \\ \\ & =\sqrt{r_1^{2}+r_2^{2}-2 r_1 r_2}=\sqrt{(r_1-r_2)^{2}} \end{aligned} $

$ \begin{aligned} \Rightarrow |z_1-z_2| & =r_1-r_2 \\ \\ & =|z_1|-|z_2| \end{aligned} $

Solution

Given that, $\ Re(z^{2})=0,|z|=2$

Let $z=x+iy$

$ \begin{matrix} \because & \sqrt{x^{2}+y^{2}}=2 \\ \\ \Rightarrow & x^{2}+y^{2}=4 \quad \ldots (i) \end{matrix} $

and $\ Re(z)=x$

Also, $\ z=x+i y$

$\Rightarrow z^{2}=x^{2}+2 i x y-y^{2}$

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

$\Rightarrow Re(z^{2})=x^{2}-y^{2}$

$\Rightarrow x^{2}-y^{2}=0$ $\qquad [\because Re(z^{2})=0]$ $\quad \ldots (ii)$

From Eqs. (i) and (ii),

$ x^{2}+x^{2} =4 $

$\Rightarrow 2 x^{2} =4 \Rightarrow x^{2}=2 $

$\Rightarrow x = \pm \sqrt{2} $

$\therefore y = \pm \sqrt{2} $

$\because z =x+i y $

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

Solution

Given equation is $\ z+\sqrt{2}|(z+1)|+i=0$

Let $\ z=x+iy$

$\Rightarrow \quad x+i y+\sqrt{2}|x+i y+1|+i=0$

$\Rightarrow \quad x+i(1+y)+\sqrt{2} \sqrt{(x+1)^{2}+y^{2}}=0$

$\Rightarrow \quad x+i(1+y)+\sqrt{2} \sqrt{(x^{2}+2 x+1+y^{2})}=0$

$\Rightarrow \quad x+\sqrt{2} \sqrt{(x^2+2x+1+y^2)}+i (1+y)=0 \quad \ldots (i)$

On Comparing real and imaginary parts, we get

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

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

$\Rightarrow \quad x^{2}+4 x+2 y^{2}+2=0 \quad \ldots (ii)$

And $ \ 1+y=0 $

$\Rightarrow \quad y=-1$

For $ \ y=-1, \quad x^{2}+4 x+2+2=0 \qquad$ [using Eq. (ii)]

$\Rightarrow \quad x^{2}+4 x+4=0 \Rightarrow(x+2)^{2}=0$

$\Rightarrow \quad x+2=0 \Rightarrow x=-2$

$\therefore \quad z=x+i y=-2-i$

Solution

Given that, $z=\dfrac{1-i}{\cos \left(\dfrac{\pi}{3}\right)+i \sin \left(\dfrac{\pi}{3}\right)}=\dfrac{-\sqrt{2} \left[\dfrac{-1}{\sqrt{2}}+i \dfrac{1}{\sqrt{2}}\right]}{\cos \left(\dfrac{\pi}{3}\right)+i \sin \left(\dfrac{\pi}{3}\right)}$

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

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

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

$=-\sqrt{2} \left[\cos \left(\dfrac{5 \pi}{12}\right)+i \sin \left(\dfrac{5 \pi}{12}\right)\right]$

Solution

Let $\ z=r_1(\cos \theta_1+i \sin \theta_1) \ $ and $ \ w=r_2(\cos \theta_2+i \sin \theta_2)$

Also, $\ |z w|=|z||w|=r_1 r_2=1$

$\therefore \ r_1 r_2=1$

Further, $\ \arg (z)=\theta_1 \ $ and $ \ \arg (w)=\theta_2$

But $\ \arg (z)-\arg (w) =\dfrac{\pi}{2}$

$\Rightarrow \quad \theta_1-\theta_2 =\dfrac{\pi}{2}$

$\Rightarrow \quad \arg (\dfrac{z}{w}) =\dfrac{\pi}{2} $

Now, to prove $\ \bar{z} w =-i $

$\text { LHS } =\bar{z} w $

$=r_1(\cos \theta_1-i \sin \theta_1) r_2(\cos \theta_2+i \sin \theta_2) $

$=r_1 r_2[\cos (\theta_2-\theta_1)+i \sin (\theta_2-\theta_1)] $

$=r_1 r_2[\cos (-\pi / 2)+i \sin (-\pi / 2)] $

$=1[0-i] $

$=-i=\text{RHS}$

Hence proved.

Solution

(i) $|a z_1-b z_2|^{2}+|b z_1+a z_2|^{2}$

$ \begin{aligned} & =|a z_1|^{2}+|b z_2|^{2}-2 Re(a z_1 \cdot b \bar{z}_2)+|b z_1|^{2}+|a z_2|^{2}+2 Re(a z_1 \cdot b \bar{z}_2) \\ \\ & =(a^{2}+b^{2})|z_1|^{2}+(a^{2}+b^{2})|z_2|^{2} \\ \\ & =(a^{2}+b^{2})(|z_1|^{2}+|z_2|^{2}) \end{aligned} $

(ii) $\sqrt{-25} \times \sqrt{-9}=i \sqrt{25} \times i \sqrt{9}=i^{2}(5 \times 3)=-15$

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

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

(iv) $i+i^{2}+i^{3}+\ldots$ upto 1000 terms $=i+i^{2}+i^{3}+i^{4}+\ldots i^{1000}=0$

$\displaystyle [\because i^{n}+i^{n+1}+i^{n+2}+i^{n+3}=0 \text {, where } n \in N \ i . e ., \sum_{n=1}^{1000} i^{n}=0] $

(v) Multiplicative inverse of $1+i=\dfrac{1}{1+i}=\dfrac{1-i}{1-i^{2}}=\dfrac{1}{2}(1-i)$

(vi) Let $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$

$z_1+z_2=(x_1+x_2)+i(y_1+y_2)$, which is real.

If $\quad z_1+z_2$ is real, then $y_1+y_2=0$

$\Rightarrow \quad y_1=-y_2$

$\because \quad z_2=x_2-i y_1$

$\Rightarrow \quad z_2=\bar{z}_1 \quad$ [when $x_1=x_2$ ]

(vii) $\arg (z)+\arg (\bar{z}),(\bar{z} \neq 0)$

$\Rightarrow \quad \theta+(-\theta)=0$

(viii) Given that, $|z+4| \leq 3$

For the greatest value of $|z+1|$.

$ \begin{aligned} \Rightarrow \quad|z+1| & =|z+4-3| \\ \\ & \leq|z+4|+|-3| \\ \\ & \leq 3+3 \\ \\ & \leq 6 \end{aligned} $

So, greatest value of $|z+1|=6$

For, least value of $|z+1|$, we know that the least value of the modulus of a complex number is zero. So, the least value of $|z+1|$ is zero.

(ix) Given that,

$ |\dfrac{z-2}{z+2}|=\dfrac{\pi}{6} $

$\Rightarrow \dfrac{|x+i y-2|}{|x+i y+2|}=\dfrac{\pi}{6} \Rightarrow \dfrac{|x-2+i y|}{|x+2+i y|}=\dfrac{\pi}{6} $

$\Rightarrow 6|x-2+i y|=\pi|x+2+i y| $

$\Rightarrow 6 \sqrt{(x-2)^{2}+y^{2}}=\pi \sqrt{(x+2)^{2}+y^{2}} $

$\Rightarrow 36[x^{2}+4-4 x+y^{2}]=\pi^{2}[x^{2}+4 x+4+y^{2}] $

$\Rightarrow(36-\pi^{2}) x^{2}+(36-\pi^{2}) y^{2}-(144+4 \pi^{2}) x+144+4 \pi^{2}=0, \text { which is a circle. }$

(x) Given that, $|z|=4$ and $\arg (z)=\dfrac{5 \pi}{6}$

$\text { Let } z =x+i y=r(\cos \theta+i \sin \theta) $

$\Rightarrow |z| =r=4 \text { and } \arg (z)=\theta =\dfrac{5 \pi}{6}$

$\Rightarrow z =4 \left[\cos \left(\dfrac{5 \pi}{6}\right)+i \sin \left(\dfrac{ 5 \pi}{6}\right)\right]$

$\qquad =4 \left[\dfrac{-\sqrt{3}}{2}+i \dfrac{1}{2}\right]=-2 \sqrt{3}+2 i$

Solution

(i) False

We can compare two complex numbers when they are purely real. Otherwise comparison of complex number is not possible.

(ii) False

$z=x+iy, \quad x > 0,y > 0$

$-iz = -i(x+iy)$

$\quad \quad =-xi +y$

(iii) True

Let $ \ z =x+i y $

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

If $x=0, y=0$, then the value of $|z|+|z-1|=1$.

(iv) True

Let $\ z =x+i y $

Given, $|z-1| =|z-i| $

Then, $ \ |x-1+i y| =|x-i(1-y)| $

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

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

$-2 x+1 =1-2 y $

$-2 x+2 y =0 $

$x-y =0 \quad \ldots (i)$

Equation of a line through the points $(1,0)$ and $(0,1)$,

$ \begin{aligned} & y-0=\dfrac{1-0}{0-1}(x-1) \\ \\ & \Rightarrow \quad y=-(x-1) \Rightarrow x+y=1 \qquad \ldots (ii) \end{aligned} $

which is perpendicular to the line $x-y=0$.

(v) False

Let $ \ z=x+i y, z \neq 0$ and $Re(z)=0$

i.e., $ \ x=0$

$\therefore \quad z=i y$

$Im(z^{2})=i^{2} y^{2}=-y^{2} \neq 0$

(vi) True

Given inequality, $ \ |z-4|<|z-2|$

Let $ \ z=x+i y$

$\therefore \quad |x-4+i y| <|x-2+i y| $

$\Rightarrow \sqrt{(x-4)^{2}+y^{2}} <\sqrt{(x-2)^{2}+y^{2}}$

$\Rightarrow (x-4)^{2}+y^{2}<(x-2)^{2}+y^{2}$

$\Rightarrow x^{2}-8 x+16+y^{2}<x^{2}-4 x+4+y^{2}$

$\Rightarrow -8 x+16<-4 x+4$

$\Rightarrow -8 x<-4 x-12$

$\Rightarrow -4 x<-12$

$\Rightarrow 4 x>12$

$\Rightarrow x>3$

(vii) False

Let $z_1=x_1+i y_1 \ \text { and } \ z_2=x_2+i y_2 $

Given that, $ \ |z_1+z_2|=|z_1|+|z_2|$

$|x_1+i y_1+x_2+i y_2| =|x_1+i y_1|+|x_2+i y_2| $

$\Rightarrow \quad \sqrt{(x_1+x_2)^{2}+(y_1+y_2)^{2}} =\sqrt{(x_1^{2}+y_1^{2})}+\sqrt{(x_2^{2}+y_2^{2})}$

On squaring both sides, we get

$(x_1+x_2)^{2}+(y_1+y_2)^{2}=(x_1^{2}+y_1^{2})+(x_2^{2}+y_2^{2})+2 \sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})} $

$\Rightarrow x_1^{2}+x_2^{2}+2 x_1 x_2+y_1^{2}+y_2^{2}+2 y_1 y_2=x_1^{2}+y_1^{2}+x_2^{2}+y_2^{2}+2 \sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})} $

$\Rightarrow 2 x_1 x_2+2 y_1 y_2=2 \sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})} $

$\Rightarrow x_1 x_2+y_1 y_2=\sqrt{(x_1^{2}+y_1^{2})(x_2^{2}+y_2^{2})}$

On squaring both sides, we get

$x_1^{2} x_2^{2}+y_1^{2} y_2^{2}+2 x_1 x_2 y_1 y_2 =x_1^{2} x_2^{2}+y_1^{2} x_2^{2}+x_1^{2} y_2^{2}+y_1^{2} y_2^{2} $

$\Rightarrow (x_1 y_2-x_2 y_1)^{2} =0 $

$\Rightarrow x_1 y_2 =x_2 y_1 $

$\Rightarrow \dfrac{y_1}{x_1} =\dfrac{y_2}{x_2} $

$\Rightarrow \left(\dfrac{y_1}{x_1}\right)-\left(\dfrac{y_2}{x_2}\right) =0 $

$\Rightarrow \arg (z_1)-\arg (z_2) =0$

(viii) False

Since, every real number is also a complex number.

Therefore, 2 is a complex number.

Solution

(a) Given, $\ z =i+\sqrt{3}=r(\cos \theta+\sin \theta) $

$\because \quad r \cos \theta =\sqrt{3}, r \sin \theta=1$

$\Rightarrow \quad r^{2} =1+3=4 \Rightarrow r=2 \quad [\because r>0]$

$\tan \theta = \left|\dfrac{r \sin \theta}{r \cos \theta}\right|=\dfrac{1}{\sqrt{3}}$

$\Rightarrow \quad \tan \theta =\dfrac{1}{\sqrt{3}} \Rightarrow \theta=\dfrac{\pi}{6}$

$\therefore \ \arg (z) =\theta=\dfrac{\pi}{6}$

So the polar form of $z$ is $2 \left[\cos \left(\dfrac{\pi}{6}\right)+i \sin \left(\dfrac{\pi}{6}\right)\right]$.

(b) Let $ \ z=x+iy$

Then amplitude $= \tan^{-1}\left(\left|\dfrac{y}{x}\right|\right)$

$\qquad = \pi- \tan^{-1}\left(\left|\dfrac{y}{x}\right|\right) \quad ( \text{when} \ x <0, y >0)$

Given $z= -1+\sqrt{-3}$

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

$\therefore \quad \text{amp}(z)= \pi- \tan^{-1} \left(\dfrac{\sqrt{3}}{1}\right)$

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

$\text{amp}(z)= \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$

(c) Given that, $ \ |z+2|=|z-2|$

Let $ \ z=x+iy$

$\therefore \quad |x+2+i y| =|x-2+i y| $

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

$\Rightarrow x^{2}+4 x+4 =x^{2}-4 x+4 \quad \Rightarrow \quad 8 x=0$

$\therefore x =0$

Which represent equation of $y-$ axis and it is perpendicular to the line joining the points $(-2,0)$ and $(0,2)$.

(d) Given that, $ \ |z+2 i|=|z-2 i|$

Let $ \ z=x+iy$

$\therefore \quad |x+i(y+2)| =|x+i(y-2)| $

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

$\Rightarrow 4 y =0 \Rightarrow y=0$

Which is the equation of $x-$ axis and it is perpendicular to the line segment joining $(0,-2)$ and $(0,2)$.

(e) Given that, $\quad |z+4 i| \geq 3 $

Let $\ z=x+iy$

$\therefore \quad |x+i y+4 i| \geq 3$

$\Rightarrow \quad |x+i(y+4)| \geq 3 $

$\Rightarrow \quad \sqrt{x^{2}+(y+4)^{2}} \geq 3 $

$\Rightarrow \quad x^{2}+(y+4)^{2} \geq 9 $

$\Rightarrow \quad x^{2}+y^{2}+8 y+16 \geq 9 $

$\Rightarrow \quad x^{2}+y^{2}+8 y+7 \geq 0$

Which represent a circle on or outside having centre $(0,-4)$ and radius 3 .

(f) Given that, $ \ |z+4| \leq 3$

Let $ \ z= x+iy$

$\therefore \quad |x+i y+4| \leq 3 $

$\Rightarrow |x+4+i y| \leq 3 $

$\Rightarrow \sqrt{(x+4)^{2}+y^{2}} \leq 3 $

$\Rightarrow (x+4)^{2}+y^{2} \leq 9 $

$\Rightarrow x^{2}+8 x+16+y^{2} \leq 9 $

$\Rightarrow x^{2}+8 x+y^{2}+7 \leq 0$

It represent the region which is on or inside the circle having the centre $(-4,0)$ and radius 3.

(g) Given that,

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

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

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

$ \therefore \quad \bar{z}=\dfrac{-1}{2}-\dfrac{3 i}{2} $

Hence, $\left(\dfrac{-1}{2}, \dfrac{-3}{2}\right)$ lies in third quadrant.

(h) Given that, $z=1-i$

$\therefore \quad \dfrac{1}{z}=\dfrac{1}{1-i}=\dfrac{1+i}{(1-i)(1+i)}$

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

Hence, $\quad \left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ lies in first quadrant.

Hence, the correct matches are

(a) $\rightarrow$(v),

(b) $\rightarrow$ (iii),

(c) $\rightarrow$ (i),

(d) $\rightarrow$(iv),

(e) $\rightarrow$(ii),

(f) $\rightarrow$ (vi),

(g) $\rightarrow$(viii),

(h) $\rightarrow$(vii)

Solution

$ \begin{aligned} &\text{Given that, } \ z=\dfrac{2-i}{(1-2 i)^{2}}=\dfrac{2-i}{1+4 i^{2}-4 i} \\ \\ & =\dfrac{2-i}{1-4-4 i}=\dfrac{2-i}{-3-4 i} \\ \\ & =\dfrac{(2-i)}{-(3+4 i)}=-\left[\dfrac{(2-i)(3-4 i)}{(3+4 i)(3-4 i)}\right] \\ \\ & =-\left(\dfrac{6-8 i-3 i+4 i^{2}}{9+16}\right)=-\dfrac{(-11 i+2)}{25} \\ \\ & =\dfrac{-1}{25}(2-11 i) \\ \\ & \Rightarrow z=\dfrac{1}{25}(-2+11 i) \\ \\ & \therefore \quad \bar{z}=\dfrac{1}{25}(-2-11 i)=\dfrac{-2}{25}-\dfrac{11}{25} i \end{aligned} $

Solution

Given that, $ \ |z_1| =|z_2| $

Let $ \ z_1 =x_1+i y_1 \text { and } z_2=x_2+i y_2 $

$\Rightarrow |x_1+i y_1| =|x_2+i y_2|$

$\Rightarrow x_1^{2}+y_1^{2} =x_2^{2}+y_2^{2}$

$\Rightarrow x_1^{2} =x_2^{2}, y_1^{2}=y_2^{2} $

$\Rightarrow x_1 = \pm x_2, y_1= \pm y_2 $

$\Rightarrow z_1 =x_1+i y_1 \text { or } z_1= \pm x_2 \pm i y_2$

$\therefore \quad z_1 \neq z_2$

Hence, it is not neccessary that $z_1=z_2$.

Solution

Given that, $\ \dfrac{(a^{2}+1)^{2}}{2 a-i}=x+i y $

On taking modulus both sides, we get

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

$\Rightarrow \quad \dfrac{(a^2+1)^2}{\sqrt{4a^2+1}} = \sqrt{x^2+y^2}$

On squaring both sides, we get

$\dfrac{(a^2+1)^4}{4a^2+1} = x^2+y^2$

Solution

Given $ \ |z|=4$ and $\text{arg}(z)= \dfrac{5 \pi}{6}$

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

$ \begin{matrix} \begin{aligned} \therefore \quad |z| & =r=4 \text { and } \theta=\dfrac{5 \pi}{6} \quad[\because \arg (z)=\theta] \\ \\ \text{Now,} & \quad z =4 \left[\cos \dfrac{5 \pi}{6}+i \sin \dfrac{5 \pi}{6}\right] \quad[\because z=r(\cos \theta+i \sin \theta)] \\ \\ & =4 \left[\cos \left(\pi-\dfrac{\pi}{6}\right)+i \sin \left(\pi-\dfrac{\pi}{6}\right)\right] \\ \\ & =4\left[-\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right] \\ \\ & =4\left[-\dfrac{\sqrt{3}}{2}+i \dfrac{1}{2}\right]=-2 \sqrt{3}+2 i \end{aligned} \end{matrix} $

Thinking Process

First, convert the given expression in the form $a+i b$, then use $|a+i b|=\sqrt{a^{2}+b^{2}}$.

Solution

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

$ \begin{aligned} & =\left|\dfrac{1+3 i}{3+i}\right|=\left|\dfrac{(1+3 i)(3-i)}{(3+i)(3-i)}\right| \\ \\ & =\left|\dfrac{3+9 i-i-3 i^{2}}{9-i^{2}}\right| \\ \\ &= \left|\dfrac{3+8 i+3}{9+1}\right|=\left|\dfrac{6+8 i}{10}\right| \\ \\ & =\sqrt{\dfrac{6^{2}}{100}+\dfrac{8^{2}}{100}} \\ \\ &=\sqrt{\dfrac{36+64}{100}}=\sqrt{\dfrac{100}{100}}=1 \end{aligned} $

Thinking Process

Let $z=a+i b$, then the polar form of $z$ is $r(\cos \theta+i \sin \theta)$, where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan \theta=\dfrac{b}{a} \cdot$ Here, $\theta$ is argument or amplitude of $z$ i.e., $\arg (z)=\theta$. The principal argument is a unique value of $\theta$ such that $-\pi \leq \theta \leq \pi$.

Solution

Given that,

$ \begin{aligned} & z=(1+i \sqrt{3})^{2} \\ \\ & z=1-3+2 i \sqrt{3} \\ \\ & z=-2+i 2 \sqrt{3} \end{aligned} $

$\Rightarrow \tan \alpha=\left|\dfrac{2 \sqrt{3}}{-2}\right|=\left|-\sqrt{3}\right|=\sqrt{3} \qquad \left[\because \tan \alpha = \left|\dfrac{Im(z)}{Re(z)}\right|\right] $

$\Rightarrow \tan \alpha=\tan \dfrac{\pi}{3} \Rightarrow \alpha=\dfrac{\pi}{3} $

$\because \quad Re(z)<0 \text { and } Im(z)>0 $

$\therefore \quad \arg (z)=\pi-\dfrac{\pi}{3} =\dfrac{2 \pi}{3}$

Thinking Process

$ \begin{aligned} & \text { If } z_1=x_1+iy_1 \ \text {and } \ z_2=x_2+i y_2 \text {, then }|z_1|=\sqrt{x_1^{2}+y_1^{2}} \text { and }|z_2|=\sqrt{x_2^{2}+y_2^{2}} . \\ \\ & \text { Also, use the modulus property i.e., } \left|\dfrac{z_1}{z_2}\right|=\dfrac{|z_1|}{|z_2|} \end{aligned} $

Solution

Let $z=x+iy$

Given that, $\quad \left|\dfrac{z-5 i}{z+5 i}\right|=\dfrac{|x+iy-i5|}{|x+iy+i5|}$

$\Rightarrow \quad \left|\dfrac{z-5 i}{z+5 i}\right|=\dfrac{|x+i(y-5)|}{|x+i(y+5)|}=1 \quad \Big[\because \left|\dfrac{z-5 i}{z+5 i} \right|=1 \Big]$

$\Rightarrow \quad \left|\dfrac{z-5 i}{z+5 i}\right|=\dfrac{\sqrt{x^{2}+(y-5)^{2}}}{\sqrt{x^{2}+(y+5)^{2}}} = 1$

On squaring both sides, we get

$\dfrac{x^2+(y-5)^2}{x^2+(y+5)^2} =1$

$ \Rightarrow x^{2}+(y-5)^{2}=x^{2}+(y+5)^{2} $

$\Rightarrow -10 y =+10 y $

$\Rightarrow 20 y =0 $

$\therefore \quad y =0$

So, $z$ lies on $ x-$ axis i.e. real axis.

Solution

Option (d) Let $z=\sin x+i \cos 2 x$

and $\bar{z}=\sin x-i \cos 2 x$

Given that, $\bar{z}=\cos x-i \sin 2 x$

$\therefore \sin x-i \cos 2 x=\cos x-i \sin 2 x$

$\Rightarrow \sin x=\cos x$ and $\cos 2 x=\sin 2 x$

$\Rightarrow \tan x=1$ and $\tan 2 x=1$

$\Rightarrow \tan x=\tan \dfrac{\pi}{4}$ and $\tan 2 x=\tan \dfrac{\pi}{4}$

$\Rightarrow x=n \pi+\dfrac{\pi}{4}$ and $2 x=n \pi+\dfrac{\pi}{4}$

$\Rightarrow 2 x-x=0 \Rightarrow x=0$

• Option (a) $ x = n\pi $: If $ x = n\pi $, then $ \sin x = 0 $ and $ \cos x = (-1)^n $.
  This would make $ z = 0 + i \cos 2x $ and $ \bar{z} = 0 - i \cos 2x $.
  However, for $ \bar{z} $ to be equal to $ \cos x - i \sin 2x $, $ \cos x $ and $ \sin 2x $ must satisfy the given conditions, which they do not for $ x = n\pi $.

• Option (b) $ x = (n + \dfrac{1}{2}) \dfrac{\pi}{2} $: This expression is not correctly formatted. At $ x=\left(n+\dfrac{1}{2}\right) \dfrac{\pi}{2}, \sin x $ and $ \cos x $ are not equal for integer $ n $, as these points are odd multiples of $ \dfrac{\pi}{4} $. The imaginary parts $ \cos 2 x $ and $ -\sin 2 x $ will also not satisfy the conjugate condition at these points.

• Option (c) $ x = 0 $: If $ x = 0 $, then $ \sin x = 0 $ and $ \cos x = 1 $.
  This would make $ z = 0 + i \cos 0 = i $ and $ \bar{z} = 0 - i \cos 0 = -i $.
  However, for $ \bar{z} $ to be equal to $ \cos x - i \sin 2x $, $ \cos x $ and $ \sin 2x $ must satisfy the given conditions, which they do not for $ x = 0 $.

Thinking Process

First, convert the given expansion into $a +ib$ form and then check whether the complex number $a +ib$ is purely real.

Solution

Option (c) Given expression, $z=\dfrac{1-i \sin \alpha}{1+2 i \sin \alpha}$

$ \begin{aligned} & =\dfrac{(1-i \sin \alpha)(1-2 i \sin \alpha)}{(1+2 i \sin \alpha)(1-2 i \sin \alpha)} \\ \\ & =\dfrac{1-i \sin \alpha-2 i \sin \alpha+2 i^{2} \sin ^{2} \alpha}{1-4 i^{2} \sin ^{2} \alpha} \\ \\ & =\dfrac{1-3 i \sin \alpha-2 \sin ^{2} \alpha}{1+4 \sin ^{2} \alpha} \\ \\ & =\dfrac{1-2 \sin ^{2} \alpha}{1+4 \sin ^{2} \alpha}-\dfrac{3 i \sin \alpha}{1+4 \sin ^{2} \alpha} \end{aligned} $

It is given that $z$ is a purely real.

$\therefore \ \dfrac{-3 \sin \alpha}{1+4 \sin ^{2} \alpha} =0 $

$\Rightarrow \ -3 \sin \alpha =0 \Rightarrow \sin \alpha=0 $

$\quad \alpha =n \pi$

• Option (a) $(n+1) \dfrac{\pi}{2}$: This option suggests that $\alpha$ is of the form $(n+1) \dfrac{\pi}{2}$. For these values, $\sin \alpha$ would be either 1 or -1, which would not satisfy the condition $\sin \alpha = 0$. Therefore, the expression $\dfrac{1-i \sin \alpha}{1+2 i \sin \alpha}$ would not be purely real.

• Option (b) $(2 n+1) \dfrac{\pi}{2}$: This option suggests that $\alpha$ is of the form $(2 n+1) \dfrac{\pi}{2}$. For these values, $\sin \alpha$ would be either 1 or -1, which would not satisfy the condition $\sin \alpha = 0$. Therefore, the expression $\dfrac{1-i \sin \alpha}{1+2 i \sin \alpha}$ would not be purely real.

• Option (d) None of these: This option is incorrect because there is indeed a correct value for $\alpha$ that makes the expression purely real, which is $\alpha = n \pi$ as shown in the solution.

Solution

Option (b) Given that, $z=x+i y$ lies in third quadrant.

$ x<0 \ \text { and } \ y<0 \text {. } $

Now, $\quad \dfrac{\bar{z}}{z}=\dfrac{x-i y}{x+i y}=\dfrac{(x-i y)(x-i y)}{(x+i y)(x-i y)}=\dfrac{x^{2}-y^{2}-2 i x y}{x^{2}+y^{2}}$

$ \dfrac{\bar{z}}{z}=\dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}-\dfrac{2 i x y}{x^{2}+y^{2}} $

Since, $\dfrac{\bar{z}}{z}$ also lies in third quadrant.

$\therefore \quad \dfrac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \ \text { and } \ \dfrac{-2 x y}{x^{2}+y^{2}}<0 $

$x^{2}-y^{2}<0 \ \text { and } \ -2 x y<0 $

$\Rightarrow x^{2}<y^{2} \ \text { and } \ x y>0 $

$\text { So, } x<y<0$

• Option (a) $x>y>0$: This option is incorrect because if $x$ and $y$ are both positive, then $z = x + iy$ would lie in the first quadrant, not the third quadrant. For $z$ to lie in the third quadrant, both $x$ and $y$ must be negative.

• Option (c) $y<x<0$: This option is incorrect because while it correctly states that both $x$ and $y$ are negative (placing $z$ in the third quadrant), it does not satisfy the condition $x^2 < y^2$. If $y < x < 0$, then $|y| > |x|$, which means $y^2 > x^2$, contradicting the requirement $x^2 < y^2$.

• Option (d) $y>x>0$: This option is incorrect because if $x$ and $y$ are both positive, then $z = x + iy$ would lie in the first quadrant, not the third quadrant. For $z$ to lie in the third quadrant, both $x$ and $y$ must be negative.

Solution

Option (a) Given that, $(z+3)(\bar{z}+3)$

Let $\ z=x+i y$ and $\bar{z} = x-iy$

$\therefore \quad(z+3)(\bar{z}+3)=(x+i y+3)(x-iy+3)$

$=(x+3)^{2}-(i y)^{2}$

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

$=|x+3+i y|^{2}=|z+3|^{2}$

• Option (b) $|z-3|$: This option is incorrect because the expression $(z+3)(\bar{z}+3)$ involves the sum of $z$ and 3, not the difference. The magnitude $|z-3|$ represents the distance between $z$ and 3 in the complex plane, which is not related to the given expression.

• Option (c) $z^{2}+3$: This option is incorrect because the expression $(z+3)(\bar{z}+3)$ involves the product of $z+3$ and its conjugate, which results in a real number. The expression $z^2 + 3$ does not account for the conjugate and does not simplify to the same form as $(z+3)(\bar{z}+3)$.

• Option (d) None of these: This option is incorrect because the correct answer is provided in option (a), which is $|z+3|^2$. Therefore, “None of these” is not applicable.

Solution

Option (b) Given that, $ \ \left(\dfrac{1+i}{1-i}\right)^{x}=1$

$\Rightarrow \left[\dfrac{(1+i)(1+i)}{(1-i)(1+i)}\right]^x =1 $

$\Rightarrow \left[{\dfrac{1+2 i+i^{2}}{1-i^{2}}}\right]^{x}=1 $

$\Rightarrow \left[\dfrac{2 i}{1+1}\right]^{x} =1 $

$\Rightarrow \left[\dfrac{2 i}2\right]^{x}=1$

$\Rightarrow i^{x} =1 \Rightarrow i^{x}=(i^{4 n}) $

$\Rightarrow x =4 n \ $ where, $n \in N$

• Option (a) $ x = 2n + 1 $: This option is incorrect because $ i^{2n+1} $ does not equal 1 for all natural numbers $ n $. The powers of $ i $ cycle through $ i, -1, -i, 1 $ every four terms. For $ i^{2n+1} $, the result will be either $ i $ or $ -i $, not 1.

• Option (c) $ x = 2n $: This option is incorrect because $ i^{2n} $ does not equal 1 for all natural numbers $ n $. The powers of $ i $ cycle through $ i, -1, -i, 1 $ every four terms. For $ i^{2n} $, the result will be either $ 1 $ or $ -1 $, not consistently 1.

• Option (d) $ x = 4n + 1 $: This option is incorrect because $ i^{4n+1} $ does not equal 1 for all natural numbers $ n $. The powers of $ i $ cycle through $ i, -1, -i, 1 $ every four terms. For $ i^{4n+1} $, the result will be $ i $, not 1.

Solution

Option (a) Given equation, $\left(\dfrac{3-4 i x}{3+4 i x}\right)=\alpha-i \beta(\alpha, \beta \in R)$

$\text { Now, } (\alpha-i \beta)=\dfrac{(3-4 i x)(3-4 i x)}{(3+4 i x)(3-4 i x)}=\dfrac{9+16 i^{2} x^{2}-24 i x}{9-16 i^{2} x^{2}} $

$\Rightarrow \alpha-i \beta=\dfrac{9-16 x^{2}-24 i x}{9+16 x^{2}} $

$\Rightarrow \alpha-i \beta=\dfrac{9-16 x^{2}}{9+16 x^{2}}-\dfrac{i 24 x}{9+16 x^{2}} \quad \ldots (i)$

$\therefore \ \alpha+i \beta=\dfrac{9-16 x^{2}}{9+16 x^{2}}+\dfrac{i 24 x}{9+16 x^{2}} \quad \ldots (ii)$

So, $ \ (\alpha -i \beta)(\alpha +i \beta) = \left(\dfrac{9-16x^2}{9+16x^2}- i \dfrac{24x}{9+16x^2}\right) \left( \dfrac{9-16x^2}{9+16x^2}+i \dfrac{24x}{9+16x^2}\right)$

$\alpha ^2+ \beta ^2 = \left(\dfrac{9-16x^2}{9+16x^2}\right)^2+ \left(\dfrac{24x}{9+16x^2}\right)^2$

$= \dfrac{(9-16x^2)^2+(24x)^2}{(9+16x^2)^2}$

$= \dfrac{81+256x^4+288x^2}{(9+16x^2)^2}$

$\alpha ^2+ \beta ^2 = \dfrac{(9+16x^2)^2}{(9+16x^2)^2} =1$

• Option (b) $-1$: The expression $\alpha^2 + \beta^2$ represents the sum of the squares of the real and imaginary parts of a complex number. This sum is always non-negative because it is the magnitude squared of the complex number. Therefore, $\alpha^2 + \beta^2$ cannot be negative, making $-1$ an impossible value.

• Option (c) $2$: The calculation in the solution shows that $\alpha^2 + \beta^2$ simplifies to $1$. There is no algebraic manipulation or value of $x$ that would change this result to $2$. Hence, $2$ is not a correct value for $\alpha^2 + \beta^2$.

• Option (d) $-2$: Similar to option (b), $\alpha^2 + \beta^2$ is always non-negative because it represents the magnitude squared of a complex number. Therefore, it cannot be negative, making $-2$ an impossible value.

Solution

Option (a) Let $z_1=r_1(\cos \theta_1+i \sin \theta_1)$

$\Rightarrow \quad|z_1|=r_1 \quad \ldots (i)$

and $z_2=r_2(\cos \theta_2+i \sin \theta_2)$

$\Rightarrow \quad|z_2|=r_2 \quad \ldots (ii)$

Now, $\quad z_1 z_2=r_1 r_2[\cos \theta_1 \cos \theta_2+i \sin \theta_1 \cos \theta_2+i \cos \theta_1 \sin \theta_2+i^{2} \sin \theta_1 \sin \theta_2]$ $=r_1 r_2[\cos (\theta_1+\theta_2)+i \sin (\theta_1+\theta_2)]$

$\Rightarrow \quad|z_1 z_2|=r_1 r_2$

$\therefore \quad|z_1 z_2|=|z_1||z_2| \qquad$ [using Eqs. (i) and (ii)]

• Option (b) $\arg (z_1 z_2)=\arg (z_1) \cdot \arg (z_2)$: The argument of the product of two complex numbers is the sum of their arguments, not the product. Therefore, $\arg (z_1 z_2) = \arg (z_1) + \arg (z_2)$, not $\arg (z_1) \cdot \arg (z_2)$.

• Option (c) $|z_1+z_2|=|z_1|+|z_2|$: The magnitude of the sum of two complex numbers is not necessarily equal to the sum of their magnitudes. In fact, $|z_1+z_2| \leq |z_1| + |z_2|$ by the triangle inequality.

• Option (d) $|z_1+z_2| \geq|z_1|-|z_2|$: The correct inequality involving the magnitudes of the sum of two complex numbers is $|z_1+z_2| \leq |z_1| + |z_2|$. The given inequality $|z_1+z_2| \geq |z_1| - |z_2|$ is not always true.

Thinking Process

Here, $z<\alpha$ is a complex number, where modulus is $r$ and argument $(\theta+\alpha)$. If $P(z)$ rotates in clockwise sense through an angle $\boldsymbol{\alpha}$, then its new position will be $z(\boldsymbol{\theta}-i \boldsymbol{\alpha})$.

Solution

Option (b) Given that, $z=2-i$

It is rotated about origin through an angle $\dfrac{\pi}{2}$ in the clockwise direction

$\therefore \quad$ New position $=z e^{-i \pi / 2}=(2-i) e^{-i \pi / 2}$

$ \begin{aligned} & =(2-i) \left[\cos \left(\dfrac{-\pi}{2}\right)+i \sin \left(\dfrac{-\pi}{2}\right)\right]=(2-i)[0-i] \\ \\ & =-2 i-1=-1-2 i \end{aligned} $

• Option (a) $1+2i$: This option is incorrect because rotating the complex number $(2-i)$ by $\dfrac{\pi}{2}$ in the clockwise direction should result in a complex number with a negative real part and a negative imaginary part, not a positive real part and a positive imaginary part.

• Option (c) $2+i$: This option is incorrect because rotating the complex number $(2-i)$ by $\dfrac{\pi}{2}$ in the clockwise direction should result in a complex number with both real and imaginary parts being negative, not a positive real part and a positive imaginary part.

• Option (d) $-1+2i$: This option is incorrect because rotating the complex number $(2-i)$ by $\dfrac{\pi}{2}$ in the clockwise direction should result in a complex number with both real and imaginary parts being negative, not a negative real part and a positive imaginary part.

Solution

Option (d) Given that, $x, y \in R$

Then, $x+i y$ is non-real complex number if and only if $y \neq 0$.

• Option (a) x=0 : If $x=0,$ the complex number becomes $iy$. The number is non-real if $y \neq 0$. This option does not necessarily guarantee $x+iy$ is non-real unless $y \neq 0$.

• Option (b) $y=0$ : If $y=0,$ the complex number becomes $x$. This means it is a real number. This option results in a real number, not a non-real complex number.

• Option (c) $x \neq 0$: If $x \neq 0,$ the complex number can still be real if $y=0$. This option alone does not ensure the number is non-real.

Thinking Process

If two complex numbers $z_1=x_1+i y_1$ and $z_2=x_2+i y_2$ are equal, then

$ |z_1|=|z_2| \Rightarrow \sqrt{x_1^{2}+y_1^{2}}=\sqrt{x_2^{2}+y_2^{2}} $

Solution

Option (d) Given that,

$a+i b=c+i d $

$\Rightarrow |a+i b|=|c+i d| $

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

On squaring both sides, we get

$ a^{2}+b^{2}=c^{2}+d^{2} $

• Option (a) $a^{2}+c^{2}=0$ is incorrect because it implies that both $a$ and $c$ must be zero, which is not necessarily true given the equation $a + i b = c + i d$.

• Option (b) $b^{2}+c^{2}=0$ is incorrect because it implies that both $b$ and $c$ must be zero, which is not necessarily true given the equation $a + i b = c + i d$.

• Option (c) $b^{2}+d^{2}=0$ is incorrect because it implies that both $b$ and $d$ must be zero, which is not necessarily true given the equation $a + i b = c + i d$.

Solution

Option (b) Given that, $ \ \left|\dfrac{i+z}{i-z}\right|=1$

Let $z = x+iy$

$ \begin{aligned} & \therefore \quad \left|\dfrac{x+i(y+1)}{-x-i(y-1)}\right|=1 \\ \\ & \Rightarrow \quad \dfrac{x^{2}+(y+1)^{2}}{x^{2}+(y-1)^{2}}=1 \\ \\ & \Rightarrow \quad x^{2}+(y+1)^{2}=x^{2}+(y-1)^{2} \\ \\ & \Rightarrow \quad 4 y=0 \Rightarrow y=0 \end{aligned} $

So, $z$ lies on $X$-axis (real axis).

• Option (a) circle $x^{2}+y^{2}=1$: This option is incorrect because the condition $\left|\dfrac{i+z}{i-z}\right|=1$ simplifies to $y=0$, which means $z$ lies on the real axis. Points on the circle $x^{2}+y^{2}=1$ generally have non-zero $y$-coordinates, except for the points $(1,0)$ and $(-1,0)$. However, the given condition does not restrict $x$ to $\pm 1$, so not all points on the circle satisfy the condition.

• Option (c) the $Y$-axis: This option is incorrect because the condition $\left|\dfrac{i+z}{i-z}\right|=1$ simplifies to $y=0$, which means $z$ lies on the real axis. Points on the $Y$-axis have $x=0$ and non-zero $y$-coordinates, which contradicts the condition that $y=0$.

• Option (d) the line $x+y=1$: This option is incorrect because the condition $\left|\dfrac{i+z}{i-z}\right|=1$ simplifies to $y=0$, which means $z$ lies on the real axis. Points on the line $x+y=1$ generally have non-zero $y$-coordinates, except for the point $(1,0)$. However, the given condition does not restrict $x$ to 1, so not all points on the line satisfy the condition.

Solution

Option (b) If $z$ is a complex number, then $z=x+i y$

$ \begin{aligned} & |z|=|x+i y| \text { and }|z|^{2}=|x+i y|^{2} \\ \\ & \Rightarrow \quad|z|^{2}=x^{2}+y^{2} \quad \ldots (i)\\ \\ & \text { and } \quad z^{2}=(x+i y)^{2}=x^{2}+i^{2} y^{2}+i 2 x y \\ \\ & z^{2}=x^{2}-y^{2}+i 2 x y \\ \\ & \Rightarrow \quad|z^{2}|=\sqrt{(x^{2}-y^{2})^{2}+(2 x y)^{2}} \\ \\ & \Rightarrow \quad|z^{2}|=\sqrt{x^{4}+y^{4}-2 x^{2} y^{2}+4 x^{2} y^{2}} \\ \\ & \Rightarrow \quad|z^{2}|=\sqrt{x^{4}+y^{4}+2 x^{2} y^{2}}=\sqrt{(x^{2}+y^{2})^{2}} \\ \\ & \Rightarrow \quad|z^{2}|=x^{2}+y^{2} \quad \ldots (ii) \end{aligned} $

From Eqs. (i) and (ii),

$ |z|^{2}=|z^{2}| $

• Option (a) $|z^{2}|>|z|$: This option is incorrect because the magnitude of $ z^2 $ is equal to the square of the magnitude of $ z $. Specifically, $ |z^2| = |z|^2 $. Since $ |z|^2 $ is not necessarily greater than $ |z| $ (it could be equal or less depending on the value of $ |z| $), this statement is false.

• Option (c) $|z^{2}|<|z|^{2}$: This option is incorrect because, as derived, $ |z^2| = |z|^2 $. Therefore, $ |z^2| $ is exactly equal to $ |z|^2 $, not less than $ |z|^2 $.

• Option (d) $|z^{2}| \geq|z|^{2}$: This option is incorrect because it implies that $ |z^2| $ could be greater than $ |z|^2 $. However, as shown, $ |z^2| = |z|^2 $, so the correct relationship is equality, not inequality.

Solution

Option (c) Given that, $ \ |z_1+z_2|=|z_1|+|z_2| \quad \ldots (i)$

Let, $z_1= r_1(\cos \theta _1 + i \sin \theta _1)$

and $z_2 = r_2(\cos \theta _2 + i \sin \theta _2)$

Now, $|z_1|=r_1$ and $|z_2|=r_2 \quad \ldots (ii)$

$\text{arg}|z_1|= \theta _1$ and $\text{arg}|z_2|= \theta _2 \quad \ldots (iii)$

$\therefore$ From eq. (i) $\ |r_1(\cos \theta_1+i \sin \theta_1)+r_2(\cos \theta_2+i \sin \theta_2)|=|r_1(\cos \theta_1+i \sin \theta_1)| +|r_2(\cos \theta_2+i \sin \theta_2)|$

$\Rightarrow \ |(r_1 \cos \theta _1 + r_2 \cos \theta _2)+ i (r_1 \sin \theta _1+r_2 \sin \theta _2)| = r_1+r_2 \quad $ [Using eq. (ii)]

$\Rightarrow \ \sqrt{r_1^2 \cos ^2 \theta _1+ r_2^2 \cos^2 \theta _2 +2r_1 r_2 \cos \theta _1 \cos \theta _2 + r_1^2 \sin ^2 \theta _1+ r_2^2 \sin ^2 \theta _2 + 2r_1r_2 \sin \theta _1 \sin \theta _2} = r_1+r_2$

$\Rightarrow \ \sqrt{r_1^2+r_2^2+2r_1r_2 [\cos \theta _1 \cos \theta _2 + \sin \theta _1 \sin \theta _2]} = r_1+r_2$

$\Rightarrow \ \sqrt{r_1^2+r_2^2+2r_1r_2 \cos (\theta _1 - \theta _2)} = r_1+r_2$

On squaring both sides, we get

$ \begin{aligned} & r_1^{2}+r_2^{2}+2 r_1 r_2 \cos (\theta_1-\theta_2)=r_1^{2}+r_2^{2}+2 r_1 r_2 \\ \\ \Rightarrow & \ 2 r_1 r_2[1-\cos (\theta_1-\theta_2)] =0 \\ \\ \Rightarrow & \ 1-\cos (\theta_1-\theta_2) =0 \\ \\ \Rightarrow & \ \cos (\theta_1-\theta_2) =1 \\ \\ \Rightarrow & \ \cos (\theta_1-\theta_2) =\cos 0^{\circ} \\ \\ \Rightarrow & \ \theta_1-\theta_2 =0^{\circ} \\ \\ \Rightarrow & \ \theta_1 =\theta_2 \\ \\ \therefore & \ \arg (z_1) =\arg (z_2) \quad [\text{Using eq. (iii)}] \end{aligned} $

• Option (a) $z_2=\bar{z}_1$: This option is incorrect because if $ z_2 = \bar{z}_1 $, then $ z_1 + z_2 $ would be a real number (since $ z_1 + \bar{z}_1 $ is real). However, the magnitudes $ |z_1| $ and $ |z_2| $ would not necessarily add up to the magnitude of the sum $ |z_1 + z_2| $, unless $ z_1 $ is purely imaginary, which is not a general case.

• Option (b) $z_2=\dfrac{1}{z_1}$: This option is incorrect because if $ z_2 = \dfrac{1}{z_1} $, then $ z_1 z_2 = 1 $. The magnitudes $ |z_1| $ and $ |z_2| $ would satisfy $ |z_1| \cdot |z_2| = 1 $, but this does not imply that $ |z_1 + z_2| = |z_1| + |z_2| $. The equality $ |z_1 + z_2| = |z_1| + |z_2| $ holds only when $ z_1 $ and $ z_2 $ are in the same direction (i.e., have the same argument), which is not guaranteed by $ z_2 = \dfrac{1}{z_1} $.

• Option (d) $|z_1|=|z_2|$: This option is incorrect because having equal magnitudes $ |z_1| = |z_2| $ does not necessarily mean that $ |z_1 + z_2| = |z_1| + |z_2| $. For the equality $ |z_1 + z_2| = |z_1| + |z_2| $ to hold, $ z_1 $ and $ z_2 $ must be in the same direction (i.e., have the same argument), which is not implied by just having equal magnitudes.

Solution

Option (c) Given expression $=\dfrac{1+i \cos \theta}{1-2 i \cos \theta}=\dfrac{(1+i \cos \theta)(1+2 i \cos \theta)}{(1-2 i \cos \theta)(1+2 i \cos \theta)}$

$ \begin{aligned} & =\dfrac{1+i \cos \theta+2 i \cos \theta+2 i^{2} \cos ^{2} \theta}{1-4 i^{2} \cos ^{2} \theta} \\ \\ & =\dfrac{1+3 i \cos \theta-2 \cos ^{2} \theta}{1+4 \cos ^{2} \theta} \end{aligned} $

For real value of $\theta, \dfrac{3 \cos \theta}{1+4 \cos ^{2} \theta}=0$

$\Rightarrow 3 \cos \theta =0 $

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

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

• Option (a): $n \pi+\dfrac{\pi}{4}$ is incorrect because it implies that $\cos \theta = \cos \left(n \pi + \dfrac{\pi}{4}\right)$. However, $\cos \left(n \pi + \dfrac{\pi}{4}\right)$ is not zero for any integer $n$, which contradicts the requirement that $\cos \theta = 0$ for the given expression to be real.

• Option (b): $n \pi+(-1)^{n} \dfrac{\pi}{4}$ is incorrect because it implies that $\cos \theta = \cos \left(n \pi + (-1)^n \dfrac{\pi}{4}\right)$. Similar to option (a), $\cos \left(n \pi + (-1)^n \dfrac{\pi}{4}\right)$ is not zero for any integer $n$, which does not satisfy the condition $\cos \theta = 0$.

• Option (d): None of these is incorrect because there is a correct option provided, which is option (c). The correct value of $\theta$ that makes the given expression real is indeed $2 n \pi \pm \dfrac{\pi}{2}$.

Solution

Option (c) Let $ \ z =x+i(0) \text { and } x<0 $

$\text{arg}(z) = \tan^{-1} \left(\dfrac{y}{x}\right)$

$\qquad = \tan^{-1} \left(\dfrac{0}{x}\right)$

$\qquad = \tan^{-1} (0)$

If $x <0,$ the $\text{arg}(z)$ is $ \pi $ because the number lies on the negative real axis.

• Option (a) 0: This is incorrect because the argument of a negative real number is not 0. The argument 0 corresponds to positive real numbers on the positive side of the real axis.

• Option (b) $\dfrac{\pi}{2}$: This is incorrect because the argument $\dfrac{\pi}{2}$ corresponds to purely imaginary numbers on the positive imaginary axis, not negative real numbers.

• Option (d) None of these: This is incorrect because there is a correct option provided, which is (c) $\pi$. The argument of a negative real number is indeed $\pi$.

Solution

Option (a)

$ \begin{aligned} \text{Let } \quad z & =1+2 i \\ \\ \Rightarrow \quad |z| & =\sqrt{1+4}=\sqrt{5} \\ \\ \text{Now, } \quad f(z) & =\dfrac{7-z}{1-z^{2}}=\dfrac{7-1-2 i}{1-(1+2 i)^{2}} \\ \\ & =\dfrac{6-2 i}{1-1-4 i^{2}-4 i}=\dfrac{6-2 i}{4-4 i} \\ \\ & =\dfrac{(3-i)(2+2 i)}{(2-2 i)(2+2 i)} \\ \\ & =\dfrac{6-2 i+6 i-2 i^{2}}{4-4 i^{2}}=\dfrac{6+4 i+2}{4+4} \\ \\ & =\dfrac{8+4 i}{8}=1+\dfrac{1}{2} i \\ \\ f(z) & =1+\dfrac{1}{2} i \end{aligned} $

$ \begin{aligned} & \therefore \quad|f(z)|=\sqrt{1+\dfrac{1}{4}}=\sqrt{\dfrac{4+1}{4}}=\dfrac{\sqrt{5}}{2}=\dfrac{|z|}{2} \end{aligned} $

• Option (b) $|z|$: This option is incorrect because the magnitude of $ f(z) $ is calculated to be $ \dfrac{|z|}{2} $, not $ |z| $. The calculation shows that $ |f(z)| = \dfrac{\sqrt{5}}{2} $, which is half of $ |z| $.

• Option (c) $2 |z|$: This option is incorrect because the magnitude of $ f(z) $ is not twice the magnitude of $ z $. The correct magnitude of $ f(z) $ is $ \dfrac{|z|}{2} $, not $ 2|z| $.

• Option (d) None of these: This option is incorrect because there is an option (a) that correctly matches the calculated magnitude of $ f(z) $, which is $ \dfrac{|z|}{2} $. Therefore, “None of these” is not the correct answer.