Solution:
It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$
In which,
$\begin{array}{l}
\left.|Z|=\text { modulus of complex number }=\sqrt{(} x^{2}+y^{2}\right) \\
\theta=\arg (z)=\text { argument of complex number }=\tan ^{-1}(|y| /|x|)
\end{array}$
(i) $1 /(1+\mathrm{i})$
Given that $Z=1 /(1+i)$
Let’s multiply and divide by $(1-\mathrm{i})$, we get
$\begin{aligned}
Z &=\frac{1}{1+i} \times \frac{1-i}{1-i} \\
=\frac{1-i}{1^{2}-i^{2}} \\
=\frac{1-i}{1-(-1)} \\
=\frac{1-i}{2}
\end{aligned}$
Therefore now,
$|Z|=\sqrt{\left(x^{2}+y^{2}\right)}$
$=\sqrt{\left((1 / 2)^{2}+(-1 / 2)^{2}\right)}$
$=\sqrt{(1 / 4+1 / 4)}$
$=\sqrt{(2 / 4)}$
$=1 / \sqrt{2}$
$\theta=\tan ^{-1}(|y| /|x|)$
$=\tan ^{-1}((1 / 2) /(1 / 2))$
$=\tan ^{-1} 1$
As $x>0, y<0$ complex number lies in $4^{\text {th }}$ quadrant and the value of $\theta$ is $-90^{0} \leq \theta \leq 0^{\circ}$.
$\begin{array}{l}
\theta=-\pi / 4 \\
Z=1 / \sqrt{2}(\cos (-\pi / 4)+i \sin (-\pi / 4)) \\
=1 / \sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4))
\end{array}$
As a result, polar form of $1 /(1+i)$ is $1 / \sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4))$
(ii) $(1+2 i) /(1-3 i)$
Given that $Z=(1+2 i) /(1-3 i)$
Let’s multiply and divide by $(1+3 i)$, we get
$\begin{aligned}
Z =\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i} \\
=\frac{1(1+3 i)+2 i(1+3 i)}{1^{2}-(3 i)^{2}} \\
=\frac{1+3 i+2 i+6 i^{2}}{1-9 i^{2}} \\
=\frac{1+5 i+6(-1)}{1-9(-1)} \\
=\frac{-5+5 i}{10} \\
=\frac{-1+i}{2}
\end{aligned}$
Therefore now,
$\begin{array}{l}
|\mathrm{Z}|=\sqrt{\left(x^{2}+y^{2}\right)} \\
\left.=\sqrt{(}(-1 / 2)^{2}+(1 / 2)^{2}\right) \\
=\sqrt{(1 / 4+1 / 4)} \\
=\sqrt{(2 / 4)} \\
{=} &{1 / \sqrt{2}} \\
{\theta} &{=\tan ^{-1}(|\mathrm{y}| /|\mathrm{x}|)} \\
{=\tan ^{-1}((1 / 2) /(1 / 2))} \\
{=\tan ^{-1} 1} \\
=
\end{array}$
As $x<0, y>0$ complex number lies in $2^{\text {nd }}$ quadrant and the value of $\theta$ is $90^{0} \leq \theta \leq 180^{\circ}$
$\begin{array}{l}
\theta=3 \pi / 4 \\
Z=1 / \sqrt{2}(\cos (3 \pi / 4)+i \sin (3 \pi / 4))
\end{array}$
As a result, polar form of $(1+2 i) /(1-3 i)$ is $1 / \sqrt{2}(\cos (3 \pi / 4)+i \sin (3 \pi / 4))$