Let $$z_1=x+i y$$

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

$$\begin{aligned} \text{Now,}\quad z_2 & =\frac{z_1-1}{z_1+1}=\frac{x+i y-1}{x+i y+1} \\ & =\frac{x-1+i y}{x+1+i y}=\frac{(x-1+i y)(x+1-i y)}{(x+1+i y)(x+1-i y)} \\ & =\frac{x^2-1+i y(x+1)-i y(x-1)-i^2 y^2}{(x+1)^2-i^2 y^2} \\ & =\frac{x^2-1+i x y+i y-i x y+i y+y^2}{(x+1)^2+y^2} \end{aligned}$$

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

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

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

$$\begin{aligned} \text{Then,}\quad & z_2=\bar{z}_1=r_1\left(\cos \theta_1-i \sin \theta_1\right)=r_1\left[\cos \left(-\theta_1\right)+\sin \left(-\theta_1\right)\right] \\ \text { Also, }\quad & \text { let } z_3=r_2\left(\cos \theta_2+i \sin \theta_2\right) \text {, } \\ \text{Then,}\quad & z_4=\bar{z}_3=r_2\left(\cos \theta_2-i \sin \theta_2\right) \\ & \arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right) \\ & =\theta_1-\left(-\theta_2\right)+\left(-\theta_1\right)-\theta_2 \quad [\because \arg(z)=\theta]\\ & =\theta_1+\theta_2-\theta_1-\theta_2=0 \end{aligned}$$

$$\begin{array}{ll} \text { Given that, } & \left|z_1\right|=\left|z_2\right|=\cdots=\left|z_n\right|=1 \\ \Rightarrow & \left|z_1\right|^2=\left|z_2\right|^2=\cdots=\left|z_n\right|^2=1 \\ \Rightarrow & z_1 \bar{z}_1=z_2 \bar{z}_2=z_3 \bar{z}_3=\cdots=z_n \bar{z}_n=1 \\ \Rightarrow & z_1=\frac{1}{\bar{z}_1}, z_2=\frac{1}{\bar{z}_2}=\cdots=z_n=\frac{1}{\bar{z}_n} \end{array}$$

Now, $$ \begin{aligned} & \left|z_1+z_2+z_3+z_4+\cdots+z_n\right| \\ & =\left|\frac{z_1 \bar{z}_1}{\bar{z}_1}+\frac{z_2 \bar{z}_2}{\bar{z}_2}+\frac{z_3 \bar{z}_3}{\bar{z}_3}+\cdots+\frac{z_n \bar{z}_n}{\bar{z}_n}\right| \quad\left[\because z_1 \bar{z}=1 \text {, where } z_1=\frac{1}{z}, z_1=\frac{\bar{z}}{\bar{z}-\bar{z}}, z_1=\bar{z}\right] \\ & =\left|\frac{\left|z_1\right|^2}{\bar{z}_1}+\frac{\left|z_2\right|^2}{\bar{z}_2}+\frac{\left|z_3\right|^2}{z_3}+\cdots+\frac{\left|z_n\right|^2}{\bar{z}_n}\right| \\ & =\left|\frac{1}{\bar{z}_1}+\frac{1}{\bar{z}_2}+\frac{1}{\bar{z}_3}+\cdots+\frac{1}{\bar{z}_n}\right|=\sqrt{\frac{1}{\bar{z}_1}+\frac{1}{\bar{z}_2}+\frac{1}{\bar{z}_3}} \\ & =\left|\frac{1}{z_1}+\frac{1}{z_2}+\cdots+\frac{1}{z_n}\right| \quad \text { Hence proved. } \end{aligned}$$

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

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

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),

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