Solution

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

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

Solution

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

$ \begin{gathered} =(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 \end{gathered} $

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 \\ & [\because (i^4)^3 . i = 1] \\ & =(i^{2}+i)=i-1 \end{aligned} $

Solution

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

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

$\Rightarrow \quad \left(\frac{1+i}{1-i}\right)^{3}=-i$

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

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,

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

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

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

On comparing both sides, we get

$ x=-2 / 5 \Rightarrow y=4 / 5 $

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

Solution

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

$ \begin{aligned} & \Rightarrow \quad \left[\frac{(1-i)}{(1+i)} \cdot \frac{(1-i)}{(1-i)}\right]^{100}=a+i b \Rightarrow{\left(\frac{1+i^{2}-2 i}{1-i^{2}}\right)}^{100}=a+i b \\ & \Rightarrow \quad \left(\frac{-2 i}{2}\right)^{100}=a+i b \quad[\because i^{2}=-1] \\ & \Rightarrow \quad(i^{4})^{25}=a+i b \Rightarrow 1=a+i b \\ & \text { Then, } \quad a=1 \text { and } b=0 \quad[\because i^{4}=1] \\ & \therefore \quad(a, b)=(1,0) \end{aligned} $

Solution

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

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

Solution

We have,

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

$ \begin{matrix} \Rightarrow & \frac{z}{\bar{z}}=\frac{(1-i)}{(1+i)} \frac{(1-i)}{(1-i)} \Rightarrow \frac{z}{\bar{z}}=\frac{1+i^{2}-2 i}{1-i^{2}} & {[\because i^{2}=-1]} \\ \Rightarrow & \frac{z}{\bar{z}}=\frac{1-1-2 i}{2} \Rightarrow \frac{z}{\bar{z}}=-i & \end{matrix} $

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

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

$ \begin{aligned} \text{Let } \qquad z & =x+i y \\ \text{Now, }\qquad \frac{\bar{z}+2}{\bar{z}-1} & =\frac{x-i y+2}{x-i y-1} \\ & =\frac{[(x+2)-i y][(x-1)+i y]}{[(x-1)-i y][(x-1)+i y]} \\ & =\frac{(x-1)(x+2)-i y(x-1)+i y(x+2)+y^{2}}{(x-1)^{2}+y^{2}} \\ & =\frac{(x-1)(x+2)+y^{2}+i[(x+2) y-(x-1) y]}{(x-1)^{2}+y^{2}} \quad[\because-i^{2}=1] \end{aligned} $

Taking real part, $\quad \frac{(x-1)(x+2)+y^{2}}{(x-1)^{2}+y^{2}}=4$

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

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

Hence, $z$ lies on the circle.

Solution

Let $\qquad z =x+i y $

Given that, $\qquad \arg \left(\frac{z-1}{z+1}\right) =\pi / 4$

$ \begin{matrix} \Rightarrow & \arg (z-1)-\arg (z+1)=\pi / 4 \\ \Rightarrow & \arg (x+i y-1)-\arg (x+i y+1)=\pi / 4 \\ \Rightarrow & \arg (x-1+i y)-\arg (x+1+i y)=\pi / 4 \end{matrix} $

$ \begin{matrix} \Rightarrow & \tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y}{x+1}=\pi / 4 \\ \\ \Rightarrow & \tan ^{-1} \left[\frac{\frac{y}{x-1}-\frac{y}{x+1}}{1+\left(\frac{y}{x-1} \right)\left(\frac{y}{x+1}\right)}\right]=\pi / 4 \\ \\ \Rightarrow & \frac{y \left[\frac{x+1-x+1}{x^{2}-1}\right]}{\frac{x^{2}-1+y^{2}}{x^{2}-1}}=\tan \pi / 4 \\ \\ \Rightarrow & \frac{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, \text { which represents a circle. } \end{matrix} $

Solution

The given equation is $|z|=z+1+2 i \qquad$ …(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., $\qquad 2y^2 = 2x-4y-3\qquad$ …(ii)

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

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

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

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

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

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

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

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

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

$\Rightarrow 2 x =3 \Rightarrow x=3 / 2$

$\therefore z= x+iy=3/2-2i$

Solution

Given that,

$ \begin{aligned} |z+1| & =z+2(1+i) \qquad \text{…(i)}\\ z & =x+i y \end{aligned} $

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 squaring both sides, we get

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

On comparing real and imaginary parts, we get

$x^2+y^2+2x+1=x^2-y^2+4x-4y \qquad$ …(ii)

$ \Rightarrow \quad 2 y^{2}-2 x+4 y+1=0 \qquad \text{…(ii)} $

$ \begin{matrix} \text { and } & 2(x+2)(y+2) =0 \\ \Rightarrow & x+2=0 \text { or } y+2 & =0 \\ x & =-2 \text { or } y=-2 \text{…(iii)} \end{matrix} $

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

$\Rightarrow 2 y^{2}+4 y+5=0$

$\Rightarrow \quad 16-4 \times 2 \times 5<0$

$\therefore$ Discriminant, $\quad D=b^{2}-4 a c<0$

$\Rightarrow \quad 2 y^{2}+4 y+5$ has no real roots.

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

$\Rightarrow \quad 8-2 x-8+1=0$

$\Rightarrow x=1 / 2$

$\therefore \quad z=x+i y=\frac{1}{2}-2 i$

Solution

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

$ \begin{aligned} \text{and } \qquad \text { let } z & =x+i y \\ \text{Now } \qquad \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} \frac{y}{x-1} & =\tan ^{-1} \frac{y+3}{x} \end{aligned} $

$ \begin{aligned} \Rightarrow \frac{y}{x-1} & =\frac{y+3}{x} \Rightarrow x y=(x-1)(y+3) \\ \Rightarrow x y & =x y-y+3 x-3 \Rightarrow 3 x-3=y \end{aligned} $

$ \begin{matrix} \Rightarrow & \frac{3(x-1)}{y}=1 \Rightarrow \frac{x-1}{y}=\frac{1}{3} \\ \therefore & (x-1): y=1: 3 \end{matrix} $

Solution

Let $z=x+i y$

Given, equation is $|\frac{z-2}{z-3}|=2 \Rightarrow|\frac{z-2}{z-3}|=2$

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

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

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

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}-\frac{20}{3} x+\frac{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

$ \begin{aligned} & \Rightarrow \quad 2 g=\frac{-20}{3} \Rightarrow g=\frac{-10}{3} \\ & \text { and } \quad 2 f=0 \Rightarrow f=0 \text { and } c=\frac{32}{3} \\ & \therefore \quad \text { Centre }=(-g,-f)=(10 / 3,0) \\ & \text { Also, } \quad \text { radius }(r)=\sqrt{(10 / 3)^{2}+0-32 / 3} \quad[\because r=\sqrt{g^{2}+f^{2}-c}] \\ & =\frac{1}{3} \sqrt{(100-96)}=2 / 3 \end{aligned} $

Solution

Let

$ \begin{aligned} z & =x+i y \\ \frac{z-1}{z+1} & =\frac{x+i y-1}{x+i y+1}, z \neq-1 \\ & =\frac{x-1+i y}{x+1+i y}=\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)} \end{aligned} $

$ \begin{aligned} & =\frac{(x^{2}-1)+i y(x+1)-i y(x-1)-i^{2} y^{2}}{(x+1)^{2}-(i y)^{2}} \\ \Rightarrow \quad \frac{z-1}{z+1} & =\frac{(x^{2}-1)+y^{2}+i[y(x+1)-y(x-1)]}{(x+1)^{2}+y^{2}} \end{aligned} $

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

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

$ \begin{matrix} \Rightarrow & x^{2}-1+y^{2}=0 \Rightarrow x^{2}+y^{2}=1 & \\ \Rightarrow & \sqrt{x^{2}+y^{2}}=\sqrt{1} \Rightarrow|z|=1 & {[\because|z|=\sqrt{x^{2}+y^{2}}]} \end{matrix} $

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.

$ \begin{aligned} \text{Given that, } \qquad |z_1| & =|z_2| \\ \text{and } \qquad \arg (z_1)+\arg (z_2) & =\pi \\ \text{If } \qquad |z_1| & =|z_2| \\ \Rightarrow r_1 & =r_2 \\ \text{and if } \qquad \arg (z_1)+\arg (z_2) & =\pi \\ \Rightarrow \theta_1+\theta_2 & =\pi \\ \Rightarrow \theta_1 & =\pi-\theta_2 \end{aligned} $

$ \text{Now } \qquad z_1=r_1(\cos \theta_1+i \sin \theta_1) $

$ \begin{matrix} \Rightarrow z_1=r_2(\cos (\pi-\theta_2)+i \sin (\pi-\theta_2)] & [\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)] & {[\because \bar{z}_2=r_2(\cos \theta_2-i \sin \theta_2)]} \\ \Rightarrow z_1=-\bar{z}_2 & \end{matrix} $

Solution

Let

$ \begin{aligned} z_1 & =x+i y \\ \Rightarrow |z_1| & =\sqrt{x^{2}+y^{2}}=1 \qquad [\text{because} |z_1| = 1, \text{given}] \text{…(i)}\\ \text{Now, } z_2 & =\frac{z_1-1}{z_1+1}=\frac{x+i y-1}{x+i y+1} \\ & =\frac{x-1+i y}{x+1+i y}=\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)} \\ & =\frac{x^{2}-1+i y(x+1)-i y(x-1)-i^{2} y^{2}}{(x+1)^{2}-i^{2} y^{2}} \\ & =\frac{x^{2}-1+i x y+i y-i x y+i y+y^{2}}{(x+1)^{2}+y^{2}} \end{aligned} $

$ \begin{aligned} & =\frac{x^{2}+y^{2}-1+2 i y}{(x+1)^{2}+y^{2}}=\frac{1-1+2 i y}{(x+1)^{2}+y^{2}} \quad[\because x^{2}+y^{2}=1] \\ & =0+\frac{2 y i}{(x+1)^{2}+y^{2}} \end{aligned} $

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

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, $\quad z_4=\bar{z}_3=r_2(\cos \theta_2-i \sin \theta_2)$

$ \begin{aligned} \arg \frac{z_1}{z_4}+\arg \frac{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,

$\Rightarrow |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 $

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

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

$ \begin{aligned} & =|\frac{z_1 \bar z_1}{\bar z_1}+\frac{z_2 \bar z_2}{\bar z_2}+\frac{z_3 \bar z_3}{\bar z_3}+\cdots+\frac{z_{n} \bar z_n}{\bar z_n}| \quad \because z_1 \bar z=1, \text { where } z_1=\frac{1}{z}, z_1=\frac{\bar z}{\bar z-\bar z}, z_1=\bar z \\ & =|\frac{|z_1|^{2}}{\bar z_1}+\frac{|z_2|^{2}}{\bar z_2}+\frac{|z_3|^{2}}{z_3}+\cdots+\frac{|z_{n}|^{2}}{\bar z_n}| \\ & =|\frac{1}{\bar z_1}+\frac{1}{\bar z_2}+\frac{1}{\bar z_3}+\cdots+\frac{1}{\bar z_n}|=\sqrt{\frac{1}{\bar z_1}+\frac{1}{\bar z_2}+\frac{1}{\bar z_3}} \end{aligned} $

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

Hence, proved.

Solution

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

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

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

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

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

Let $z=x+iy$

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

and $\qquad Re(z)=x$

Also, $\qquad z=x+i y$

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

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

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

$\Rightarrow x^{2}-y^{2}=0 \qquad$ …(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 $\quad z+\sqrt{2}|(z+1)|+i=0$

Let $\qquad z=x+iy$

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

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

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

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

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

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

$ 1+y=0 $

$\Rightarrow y=-1$

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, $z=\frac{1-i}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}=\frac{-\sqrt{2} [\frac{-1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}]}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}$

$=\frac{-\sqrt{2}[\cos (\pi-\pi / 4)+i \sin (\pi-\pi / 4)]}{\cos \pi / 3+i \sin \pi / 3}$

$=\frac{-\sqrt{2}[\cos 3 \pi / 4+i \sin 3 \pi / 4]}{\cos \pi / 3+i \sin \pi / 3}$

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

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

Solution

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

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

$\therefore \qquad r_1 r_2=1$

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

$ \begin{aligned} \text { But } \quad \arg (z)-\arg (w) & =\frac{\pi}{2} \\ \Rightarrow \quad \theta_1-\theta_2 & =\frac{\pi}{2} \\ \Rightarrow \quad \arg (\frac{z}{w}) & =\frac{\pi}{2} \\ \text { Now, } \quad \text { 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=RHS \end{aligned} $

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

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

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

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

(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| \\ & =|z+4-3| \leq 3+3 \\ & =|z+4-3| \leq 6 \end{aligned} $

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

For, now, 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,

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

$ \begin{matrix} \Rightarrow \frac{|x+i y-2|}{|x+i y+2|}=\frac{\pi}{6} \Rightarrow \frac{|x-2+i y|}{|x+2+i y|}=\frac{\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. } \end{matrix} $

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

$ \begin{matrix} \text { Let } z =x+i y=r(\cos \theta+i \sin \theta) \\ \Rightarrow |z| =r=4 \text { and } \arg (z)=\theta \\ \because \tan \theta =\frac{5 \pi}{6} \\ \Rightarrow z =4 (\cos \frac{\pi}{6}+i \sin \frac{\pi}{6})=4[\cos (\pi-\pi / 6)+i \sin (\pi-\pi / 6)] \\ =4[-\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}]=4 [\frac{-\sqrt{3}}{2}+i \frac{1}{2}]=-2 \sqrt{3}+2 i \end{matrix} $

Solution

(i) False

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

(ii) False

$ (x, y)\left[\begin{gathered} -1 \\ 0 \end{gathered}\right]=\left[\begin{gathered} 0 \\ -y \end{gathered}\right] \text {, which is false. } $

(iii) True

$ \begin{aligned} & \text { Let } \begin{aligned} z & =x+i y \\ \qquad|z|+|z-1| & =\sqrt{x^{2}+y^{2}}+\sqrt{(x-1)^{2}+y^{2}} \end{aligned} \end{aligned} $

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

(iv) True

$ \begin{aligned} \text{Let } \qquad z & =x+i y \\ |z-1| & =|z-i| \\ \text{Then, } \qquad |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 \text{…(i)} \end{aligned} $

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

$ \begin{aligned} & y-0=\frac{1-0}{0-1}(x-1) \\ & \Rightarrow \quad y=-(x-1) \Rightarrow x+y=1 \qquad \text{…(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., $\qquad x=0$

$\therefore \quad z=i y$

$Im(z^{2})=i^{2} y^{2}=-y^{2}$ which is real.

(vi) True

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

Let $\qquad z=x+i y$

$\therefore \begin{aligned}|x-4+i y| & <|x-2+i y| \\ \Rightarrow \sqrt{(x-4)^{2}+y^{2}} & <\sqrt{(x-2)^{2}+y^{2}}\end{aligned}$

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

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

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

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

$\Rightarrow -4 x<-12$

$\Rightarrow 4 x>12$

$\Rightarrow x>3$

(vii) False

$ \begin{aligned} \text{Let }& z_1=x_1+i y_1 \quad \text { and } \quad z_2=x_2+i y_2 \\ \text{Given that, } & |z_1+z_2|=|z_1|+|z_2| \end{aligned} $

$ \begin{aligned} |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})} \end{aligned} $

On squaring both sides, we get

$ \begin{aligned} & & (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})} \end{aligned} $

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 \frac{y_1}{x_1} =\frac{y_2}{x_2} $

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

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

(viii) True

We know that, 2 is a real number.

Since, 2 is not a complex number.

Solution

$ \begin{aligned} \text{(i) Given, } \quad 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] \\ \Rightarrow \quad \tan \alpha & =|\frac{r \sin \theta}{r \cos \theta}|=\frac{1}{\sqrt{3}} \\ \Rightarrow \quad \tan \alpha & =\frac{1}{\sqrt{3}} \Rightarrow \alpha=\frac{\pi}{6} \\ \because \quad x & >0, \quad y>0 \\ \text{and } \quad \arg (z) & =\theta=\frac{\pi}{6} \end{aligned} $

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

(ii) Given that, $\quad z=-1+\sqrt{-3}=-1+i \sqrt{3}$

$\therefore \quad \tan \alpha=|\frac{\sqrt{3}}{-1}|=\sqrt{3}$

$\Rightarrow \quad \tan \alpha=\tan \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3}$

$ \begin{matrix} \because & x<0, \quad y>0 \\ \Rightarrow & \theta=\pi-\alpha=\pi-\frac{\pi}{3}=\frac{2 \pi}{3} \end{matrix} $

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

$ \begin{matrix} \Rightarrow |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 \end{matrix} $

It is a straight line which is a perpendicular bisector of segment joining the points $(-2,0)$ and $(2,0)$

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

$ \begin{matrix} \Rightarrow & |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 \end{matrix} $

It is a straight line, which is a perpendicular bisector of segment joining $(0,-2)$ and $(0,2)$.

(v) Given that, $\quad |z+4 i| \geq 3 =|x+i y+4 i| \geq 3$

$ \begin{aligned} \Rightarrow \qquad &=|x+i(y+4)| \geq 3 \\ \Rightarrow \qquad &=\sqrt{x^{2}+(y+4)^{2}} \geq 3 \\ \Rightarrow \qquad &=x^{2}+(y+4)^{2} \geq 9 \\ \Rightarrow \qquad &=x^{2}+y^{2}+8 y+16 \geq 9 \\ \qquad &=x^{2}+y^{2}+8 y+7 \geq 0 \end{aligned} $

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

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

$ \begin{matrix} \Rightarrow & |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 \end{matrix} $

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

(vii) Given that,

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

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

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

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

$\therefore \quad \frac{1}{z}=\frac{1}{1-i}=\frac{1+i}{(1-i)(1+i)}=\frac{1+i}{1-i^{2}}=\frac{1}{2}(1+i)$

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

Solution

$ \begin{aligned} \text{Given that, } \quad |z_1| & =|z_2| \\ \text{Let } \quad 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| \end{aligned} $

$ \begin{aligned} \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 \end{aligned} $

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

Solution

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

$ \begin{matrix} \Rightarrow & \frac{(a^{2}+1)^{2}(2 a+i)}{(2 a-i)(2 a+i)} & =x+i y \\ \Rightarrow & \frac{(a^{2}+1)^{2}(2 a+i)}{4 a^{2}+1} & =x+i y \\ \Rightarrow x & =\frac{2 a(a^{2}+1)^{2}}{4 a^{2}+1} \text { and } y=\frac{(a^{2}+1)^{2}}{4 a^{2}+1} \\ \\ \therefore & x^{2}+y^{2} & =4 a^{2} \left[\frac{(a^{2}+1)^{2}}{4 a^{2}+1}\right]+\left[\frac{(a^{2}+1)^{2}}{4 a^{2}+1}\right] \\ \\ & =\frac{(4 a^{2}+1)(a^{2}+1)^{4}}{(4 a^{2}+1)^{2}}=\frac{(a^{2}+1)^{4}}{(4 a^{2}+1)} \end{matrix} $

Solution

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

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

Solution

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

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

Solution

Given that,

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

$ \begin{matrix} \Rightarrow & \tan \alpha=|\frac{2 \sqrt{3}}{-2}|=|-\sqrt{3}|=\sqrt{3} \qquad \left[\because \tan \alpha = \left|\frac{Im(z)}{Re(z)}\right|\right] \\ \Rightarrow & \tan \alpha=\tan \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3} \\ \because & Re(z)<0 \text { and } Im(z)>0 \\ \Rightarrow & \arg (z)=\pi-\frac{\pi}{3} \Rightarrow=\frac{2 \pi}{3} \end{matrix} $

Solution

Let $z=x+iy$

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

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

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

On squaring both sides, we get

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

$ \begin{matrix} \Rightarrow & & -10 y & =+10 y \\ \Rightarrow & 20 y & =0 \\ \therefore & y & =0 \end{matrix} $

So, $z$ lies on real axis.

Solution

(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 $\tan 2 x=1$

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

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

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

Solution

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

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

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

$ \begin{aligned} \therefore \quad \frac{-3 \sin \alpha}{1+4 \sin ^{2} \alpha} & =0 \\ \Rightarrow \quad -3 \sin \alpha & =0 \Rightarrow \sin \alpha=0 \\ \alpha & =n \pi \end{aligned} $

Solution

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

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

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

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

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

$ \begin{matrix} \therefore \frac{x^{2}-y^{2}}{x^{2}+y^{2}}<0 \text { and } \frac{-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 \end{matrix} $

Solution

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

Let $\quad z=x+i y$

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

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

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

Solution

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

$ \begin{aligned} \Rightarrow & & \left[\frac{(1+i)(1+i)}{(1-i)(1+i)}\right]^x & =1 \Rightarrow \left[{\frac{1+2 i+i^{2}}{1-i^{2}}}\right]^{x}=1 \\ \Rightarrow & \left[\frac{2 i}{1+1}\right]^{x} & =1 \Rightarrow \left[\frac{2 i}2\right]^{x}=1 & \\ \Rightarrow & i^{x} & =1 \Rightarrow i^{x}=(i^{4 n}) & \\ \Rightarrow & x & =4 n & \end{aligned} $

Solution

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

$ \begin{matrix} \Rightarrow & \left[\frac{3-4 i x}{3+4 i x}\right]=\alpha-i \beta \\ \\ \text { Now, } & (\alpha-i \beta)=\frac{(3-4 i x)(3-4 i x)}{(3+4 i x)(3-4 i x)}=\frac{9+16 i^{2} x^{2}-24 i x}{9-16 i^{2} x^{2}} \\ \\ \Rightarrow & \alpha-i \beta=\frac{9-16 x^{2}-24 i x}{9+16 x^{2}} \\ \\ \Rightarrow & \alpha-i \beta=\frac{9-16 x^{2}}{9+16 x^{2}}-\frac{i 24 x}{9+16 x^{2}} \qquad \text{…(i)}\\ \\ \therefore & \alpha+i \beta=\frac{9-16 x^{2}}{9+16 x^{2}}+\frac{i 24 x}{9+16 x^{2}} \qquad \text{…(ii)} \end{matrix} $

$ \begin{aligned} \text { So, } \quad (\alpha-i \beta)(\alpha+i \beta) & =\left(\frac{9-16 x^{2}}{9+16 x^{2}}\right)-\left(\frac{i 24 x}{9+16 x^{2}}\right)^{2} \\ \therefore \quad \alpha^{2}+\beta^{2} & =\frac{81+256 x^{4}-288 x^{2}+576 x^{2}}{(9+16 x^{2})^{2}} \\ & =\frac{81+256 x^{4}+288 x^{2}}{(9+16 x^{2})^{2}} \\ & =\frac{(9+16 x^{2})^{2}}{(9+16 x^{2})^{2}}=1 \end{aligned} $

Solution

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

$\Rightarrow \quad|z_1|=r_1$

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

$\Rightarrow \quad|z_2|=r_2$

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)]

Solution

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

It is rotated about origin through an angle $\frac{\pi}{2}$ 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(\frac{-\pi}{2}\right)+i \sin \left(\frac{-\pi}{2}\right)\right]=(2-i)[0-i] \\ & =-2 i-1=-1-2 i \end{aligned} $

Solution

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

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

Solution

(d) Given that,

$ \begin{gathered} a+i b=c+i d \\ \Rightarrow |a+i b|=|c+i d| \\ \Rightarrow \sqrt{a^{2}+b^{2}}=\sqrt{c^{2}+d^{2}} \end{gathered} $

On squaring both sides, we get

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

Solution

(b) Given that, $\quad|\frac{i+z}{i-z}|=1$

Let $z = x+iy$

$ \begin{aligned} & \therefore \quad|\frac{x+i(y+1)}{-x-i(y-1)}|=1 \Rightarrow \frac{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).

Solution

(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 \text{…(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 \text{…(ii)} \end{aligned} $

From Eqs. (i) and (ii),

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

Solution

(c) Given that,

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

$\Rightarrow|(r_1 \cos \theta_1+r_2 \cos \theta)+i(r_1 \sin \theta_1+r_2 \sin \theta_2)|=r_1+r_2$

$\Rightarrow \sqrt{r_1^{2} \cos ^{2} \theta_1+r_2^{2} \cos ^{2} \theta_2+2 r_1 r_2 \cos \theta_1 \cos \theta_2+r_1^{2} \sin ^{2} \theta_1+r_2^{2} \sin ^{2} \theta_2}$

$\sqrt{r_1^{2}+r_2^{2}+2 r_1 r_2[\cos (\theta_1-\theta_2)}]=r_1+r_2$

$\Rightarrow \quad \sqrt{+2 r_1 r_2 \sin \theta_1 \sin \theta_2}=r_1+r_2$

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) \end{aligned} $

Solution

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

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

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

$ \begin{matrix} \Rightarrow & 3 \cos \theta & =0 \\ \Rightarrow & \cos \theta & =\cos \frac{\pi}{2} \\ \Rightarrow & \theta & =2 n \pi \pm \frac{\pi}{2} \end{matrix} $

Solution

(c) Let

$ \begin{aligned} z & =x+0 i \text { and } x<0 \\ |z| & =\sqrt{(-1)^{2}+(0^{2})}=1 \end{aligned} $

Since, the point $(x, 0)$ represent $z=x+0 i$ lies on the negative side of real axis.

$\therefore \quad$ Principal $\arg (z)=\pi$

Solution

(a)

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

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