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

$$\begin{aligned} & \text { Let } \quad z=x+i y \\ & |z|=\sqrt{x^2+y^2} \\ & \because \quad \sqrt{x^2+y^2}=2 \\ & \Rightarrow \quad x^2+y^2=4 \quad \text{.... (i)} \end{aligned}$$

$$\begin{aligned} \text { and } \quad & \operatorname{Re}(z) =x \\ \text { Also, } \quad & z =x+i y \\ \Rightarrow \quad & z^2 =x^2+2 i x y-y^2 \\ \Rightarrow \quad & z^2 =\left(x^2-y^2\right)+2 i x y \\ \Rightarrow \quad & \operatorname{Re}\left(z^2\right) =x^2-y^2 \quad [\because Re(z^2)=0]\\ \Rightarrow \quad & x^2-y^2 =0\quad \text{.... (ii)} \end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned} & x^2+x^2=4 \\ & \Rightarrow \quad 2 x^2=4 \Rightarrow x^2=2 \\ & \Rightarrow \quad x= \pm \sqrt{2} \\ & \therefore \quad y= \pm \sqrt{2} \\ & \because \quad z=x+i y \\ & \Rightarrow \quad z=\sqrt{2} \pm i \sqrt{2},-\sqrt{2} \pm i \sqrt{2} \end{aligned}$$

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

Let $$z=x+i y$$

$$\begin{array}{l} \Rightarrow \quad x+i y+\sqrt{2}|x+i y+1|+i =0 \\ \Rightarrow \quad x+i(1+y)+\sqrt{2}\left[\sqrt{(x+1)^2+y^2}\right] =0 \\ \Rightarrow \quad x+i(1+y)+\sqrt{2} \sqrt{\left(x^2+2 x+1+y^2\right)} =0 \\ \Rightarrow \quad x+\sqrt{2} \sqrt{x^2+2 x+1+y^2} =0 \\ \Rightarrow \quad x^2 =2\left(x^2+2 x+1+y^2\right. \\ \Rightarrow \quad x^2+4 x+2 y^2+2 =0 \quad \text{... (ii)}\\ \Rightarrow \quad 1+y =0 \\ \Rightarrow \quad y =-1 \end{array}$$

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

$$\begin{aligned} \Rightarrow \quad & x^2+4 x+4 =0 \Rightarrow(x+2)^2=0 \\ \Rightarrow \quad & x+2 =0 \Rightarrow x=-2 \\ \therefore \quad & z =x+i y=-2-i \end{aligned}$$

Given that, $$ \begin{aligned} z & =\frac{1-i}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}}=\frac{-\sqrt{2}\left[\frac{-1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right]}{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}} \\ & =\frac{-\sqrt{2}[\cos (\pi-\pi / 4)+i \sin (\pi-\pi / 4)]}{\cos \pi / 3+i \sin \pi / 3} \\ & =\frac{-\sqrt{2}[\cos 3 \pi / 4+i \sin 3 \pi / 4]}{\cos \pi / 3+i \sin \pi / 3} \\ & =-\sqrt{2}\left[\cos \left(\frac{3 \pi}{4}-\frac{\pi}{3}\right)+i \sin \left(\frac{3 \pi}{4}-\frac{\pi}{3}\right)\right] \\ & =-\sqrt{2}\left[\cos \frac{5 \pi}{12}+i \sin \frac{5 \pi}{12}\right] \end{aligned}$$

$$\begin{aligned} & \text { Let } \quad z=r_1\left(\cos \theta_1+i \sin \theta_1\right) \text { and } w=r_2\left(\cos \theta_2+i \sin \theta_2\right) \\ & \text { Also, } \quad|z w|=|z||w|=r_1 r_2=1 \\ & \therefore \quad r_1 r_2=1 \\ & \text { Further, } \quad \arg (z)=\theta_1 \text { and } \arg (w)=\theta_2 \end{aligned}$$

$$\begin{aligned} & \text { But } \quad \arg (z)-\arg (w)=\frac{\pi}{2} \\ & \Rightarrow \quad \theta_1-\theta_2=\frac{\pi}{2} \\ & \Rightarrow \quad \arg \left(\frac{z}{w}\right)=\frac{\pi}{2} \\ & \text { Now, to prove } \bar{z} W=-i \\ & \text { LHS }=\bar{z} W \\ & =r_1\left(\cos \theta_1-i \sin \theta_1\right) r_2\left(\cos \theta_2+i \sin \theta_2\right) \\ & =r_1 r_2\left[\cos \left(\theta_2-\theta_1\right)+i \sin \left(\theta_2-\theta_1\right)\right] \\ & =r_1 r_2[\cos (-\pi / 2)+i \sin (-\pi / 2)] \\ & =1[0-i] \\ & =-i=\mathrm{RHS}\quad \text{Hence proved.} \end{aligned}$$

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

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

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

(iv) $i+i^2+i^3+\ldots$ upto 1000 terms $=i+i^2+i^3+i^4+\ldots i^{1000}=0$ $\left[\because i^n+i^{n+1}+i^{n+2}+i^{n+3}=0\right.$, where $n \in$ Ni.e., $\left.\sum_{n=1}^{1000} i^n=0\right]$

(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 $z_2=x_2+i y_2$

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

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

$$\begin{array}{ll} \Rightarrow & y_1=-y_2 \\ \because & z_2=x_2-i y_1 \\ \Rightarrow & z_2=\bar{z}_1\quad [\text{when } x_1=x_2] \end{array}$$

(vii) $$\begin{aligned} & \arg (z)+\arg (\bar{z}),(\bar{z} \neq 0) \\ & \Rightarrow \quad \theta+(-\theta)=0 \end{aligned}$$

(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.

$$\begin{aligned} & \text { (ix) Given that, } \\ & \left|\frac{z-2}{z+2}\right|=\frac{\pi}{6} \\ & \Rightarrow \quad \frac{|x+i y-2|}{|x+i y+2|}=\frac{\pi}{6} \Rightarrow \frac{|x-2+i y|}{|x+2+i y|}=\frac{\pi}{6} \\ & \Rightarrow \quad 6|x-2+i y|=\pi|x+2+i y| \\ & \Rightarrow \quad 6 \sqrt{(x-2)^2+y^2}=\pi \sqrt{(x+2)^2+y^2} \\ & \Rightarrow \quad 36\left[x^2+4-4 x+y^2\right]=\pi^2\left[x^2+4 x+4+y^2\right] \\ & \Rightarrow\left(36-\pi^2\right) x^2+\left(36-\pi^2\right) y^2-\left(144+4 \pi^2\right) x+144+4 \pi^2=0 \text {, which is a circle. } \end{aligned}$$

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

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