$$\begin{aligned} & \text { Given that, } \quad \arg (z-1)=\arg (z+3 i) \\ & \text { and let } z=x+i y \\ & \text { Now, } \quad \arg (z-1)=\arg (z+3 i) \\ & \Rightarrow \quad \arg (x+i y-1)=\arg (x+i y+3 i) \\ & \Rightarrow \quad \arg (x-1+i y)=\arg [x+i(y+3)] \\ & \Rightarrow \quad \tan ^{-1} \frac{y}{x-1}=\tan ^{-1} \frac{y+3}{x} \\ & \Rightarrow \quad \frac{y}{x-1}=\frac{y+3}{x} \Rightarrow x y=(x-1)(y+3) \\ & \Rightarrow \quad x y=x y-y+3 x-3 \Rightarrow 3 x-3=y \end{aligned}$$

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

Let $z=x+i y$

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

$$\begin{aligned} & \Rightarrow \quad\left|\frac{x+i y-2}{x+i y-3}\right|=2 \\ & \Rightarrow \quad|x-2+i y|=2|x-3+i y| \\ & \Rightarrow \quad \sqrt{(x-2)^2+y^2}=2 \sqrt{(x-3)^2+y^2} \end{aligned}$$

$$\left[\because|x+i y|=\sqrt{x^2+y^2}\right]$$

On squaring both sides, we get

$$\begin{aligned} & x^2-4 x+4+y^2 =4\left(x^2-6 x+9+y^2\right) \\ \Rightarrow \quad & 3 x^2+3 y^2-20 x+32 =0 \\ \Rightarrow \quad & x^2+y^2-\frac{20}{3} x+\frac{32}{3} =0\quad \text{... (i)} \end{aligned}$$

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

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

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

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

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

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

$$\begin{array}{ll} \Rightarrow & x^2-1+y^2=0 \Rightarrow x^2+y^2=1 \\ \Rightarrow & \sqrt{x^2+y^2}=\sqrt{1} \Rightarrow|z|=1\quad \left[\because|z|=\sqrt{x^2+y^2}\right] \end{array}$$

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

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

and $$\arg \left(z_1\right)+\arg \left(z_2\right) =\pi$$

If $\left|z_1\right| =\left|z_2\right|$

$\Rightarrow \quad r_1 =r_2\quad \text{... (i)}$

$$\begin{aligned} \text { and if } & \arg \left(z_1\right)+\arg \left(z_2\right) =\pi \\ \Rightarrow \quad & \theta_1+\theta_2 =\pi \\ \Rightarrow \quad & \theta_1 =\pi-\theta_2 \end{aligned}$$

$$\begin{array}{lll} \text { Now, } & z_1=r_1\left(\cos \theta_1+i \sin \theta_1\right) & \\ \Rightarrow & z_1=r_2\left[\cos \left(\pi-\theta_2\right)+i \sin \left(\pi-\theta_2\right)\right] & {\left[\because r_1=r_2 \text { and } \theta_1=\left(\pi-\theta_2\right)\right]} \\ \Rightarrow & z_1=r_2\left(-\cos \theta_2+i \sin \theta_2\right) & \\ \Rightarrow & z_1=-r_2\left(\cos \theta_2-i \sin \theta_2\right) & \\ \Rightarrow & z_1=-\left[r_2\left(\cos \theta_2-i \sin \theta_2\right)\right] & \\ \Rightarrow & z_1=-\bar{z}_2 & {\left[\because \bar{z}_2=r_2\left(\cos \theta_2-i \sin \theta_2\right)\right]} \end{array}$$

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.