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

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

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

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

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

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

$$\begin{array}{lrl} \text { Given that, } & & z=(1+i \sqrt{3})^2 \\ \Rightarrow & & z=1-3+2 i \sqrt{3} \Rightarrow z=-2+i 2 \sqrt{3} \\ \Rightarrow & & \tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right|=|-\sqrt{3}|=\sqrt{3} \left[\because \tan \alpha=\left|\frac{\operatorname{lm}(z)}{\operatorname{Re}(z)}\right|\right]\\ \Rightarrow & & \tan \alpha=\tan \frac{\pi}{3} \Rightarrow \alpha=\frac{\pi}{3} \\ \because & & \operatorname{Re}(z)<0 \text { and } \operatorname{lm}(z)>0 \\ \Rightarrow & & \arg (z)=\pi-\frac{\pi}{3} \Rightarrow=\frac{2 \pi}{3} \end{array}$$

Let $$z=x+i y$$

Given that, $\quad\left|\frac{z-5 i}{z+5 i}\right|=\left|\frac{x+i y-5 i}{x+i y+5 i}\right|$

$$\begin{array}{ll} \Rightarrow & \left|\frac{z-5 i}{z+5 i}\right|=\frac{|x+i(y-5)|}{|x+i(y+5)|} \\ \Rightarrow & \left|\frac{z-5 i}{z+5 i}\right|=\frac{\sqrt{x^2+(y-5)^2}}{\sqrt{x^2+(y+5)^2}} \quad \left[\because\left|\frac{z-5 i}{z+5 i}\right|=1\right] \end{array}$$

On squaring both sides, we get

$$\begin{array}{rlrl} & x^2+(y-5)^2 =x^2+(y+5)^2 \\ \Rightarrow & -10 y =+10 y \\ \Rightarrow & 20 y =0 \\ \therefore & y =0 \end{array}$$

So, $z$ lies on real axis.