$$\begin{aligned} &\begin{aligned} & \text { Let } \quad z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right) \\ & \text { and } \quad z_2=r_2\left(\cos \theta_2+i \sin \theta_2\right) \\ & \Rightarrow \quad \arg \left(z_1\right)=\theta_1 \text { and } \arg \left(z_2\right)=\theta_2 \end{aligned}\\ &\text { Given that, } \quad \arg \left(z_1\right)-\arg \left(z_2\right)=0 \end{aligned}$$

$$\begin{aligned} \theta_1-\theta_2 & =0 \Rightarrow \theta_1=\theta_2 \\ z_2 & =r_2\left(\cos \theta_1+i \sin \theta_1\right) \quad [\because \theta_1=\theta_2]\\ z_1-z_2 & =\left(r_1 \cos \theta_1-r_2 \cos \theta_1\right)+i\left(r_1 \sin \theta_1-r_2 \sin \theta_1\right) \\ \left|z_1-z_2\right| & =\sqrt{\left(r_1 \cos \theta_1-r_2 \cos \theta_1\right)^2+\left(r_1 \sin \theta_1-r_2 \sin \theta_1\right)^2} \\ & =\sqrt{r_1^2+r_2^2-2 r_1 r_2 \cos ^2 \theta_1-2 r_1 r_2 \sin ^2 \theta_1} \\ & =\sqrt{r_1^2+r_2^2-2 r_1 r_2\left(\sin ^2 \theta_1+\cos ^2 \theta_1\right)} \\ & =\sqrt{r_1^2+r_2^2-2 r_1 r_2}=\sqrt{\left(r_1-r_2\right)^2} \\ \Rightarrow \quad \left|z_1-z_2\right| & =r_1-r_2 \quad [\because r=|z|]\\ & =\left|z_1\right|-\left|z_2\right|\quad \text{Hence proved.} \end{aligned}$$

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

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

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