Convert the following in the polar form:

(i) Here, $z=\frac{1+7 i}{(2-i)^{2}}$

$=\frac{1+7 i}{(2-i)^{2}}=\frac{1+7 i}{4+i^{2}-4 i}=\frac{1+7 i}{4-1-4 i}$

$=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}=\frac{3+4 i+21 i+28 i^{2}}{3^{2}+4^{2}}$

$=\frac{3+4 i+21 i-28}{3^{2}+4^{2}}=\frac{-25+25 i}{25}$

$=-1+i$

Let $r \cos \theta=-1$ and $r \sin \theta=1$

On squaring and adding, we obtain

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2 \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$\Rightarrow r=\sqrt{2} \quad[$ Conventionally, $r>0]$

$\therefore \sqrt{2} \cos \theta=-1$ and $\sqrt{2} \sin \theta=1$

$\Rightarrow \cos \theta=\frac{-1}{\sqrt{2}}$ and $\sin \theta=\frac{1}{\sqrt{2}}$

$\therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$  [As $\theta$ lies in II quadrant]

$\therefore z=r \cos \theta+i r \sin \theta$

$=\sqrt{2} \cos \frac{3 \pi}{4}+i \sqrt{2} \sin \frac{3 \pi}{4}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

This is the required polar form.

(ii) Here, $z=\frac{1+3 i}{1-2 i}$

$=\frac{1+3 i}{1-2 i} \times \frac{1+2 i}{1+2 i}$

$=\frac{1+2 i+3 i-6}{1+4}$

$=\frac{-5+5 i}{5}=-1+i$

Let $r \cos \theta=-1$ and $r \sin \theta=1$

On squaring and adding, we obtain

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1$

$\Rightarrow r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=2$

$\Rightarrow r^{2}=2 \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right]$

$\Rightarrow r=\sqrt{2} \quad$ [Conventionally, $r>0$ ]

$\therefore \sqrt{2} \cos \theta=-1$ and $\sqrt{2} \sin \theta=1$

$\Rightarrow \cos \theta=\frac{-1}{\sqrt{2}}$ and $\sin \theta=\frac{1}{\sqrt{2}}$

$\therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$   [As $\theta$ lies in II quadrant]

$\therefore z=r \cos \theta+i r \sin \theta$

$=\sqrt{2} \cos \frac{3 \pi}{4}+i \sqrt{2} \sin \frac{3 \pi}{4}=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$

This is the required polar form.