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) $1-\mathrm{i}$
Given that $Z=1-i$
Therefore now,
$\begin{array}{l}
|Z|=\sqrt{\left(x^{2}+y^{2}\right)} \\
=\sqrt{\left(1^{2}+(-1)^{2}\right)} \\
{=} &{\sqrt{(} 1+1)} \\
{=} &{\sqrt{2}} \\
{\theta} &{=\tan ^{-1}(|y| /|x|)} \\
{=\tan ^{-1}(1 / 1)} \\
{=\tan ^{-1} 1}
\end{array}$
As $x>0, y<0$ complex number lies in $4^{\text {th }}$ quadrant and the value of $\theta$ is $-90^{0} \leq \theta \leq 0^{0}$.
$\begin{array}{l}
\theta=-\pi / 4 \\
Z=\sqrt{2}(\cos (-\pi / 4)+i \sin (-\pi / 4)) \\
=\sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4))
\end{array}$
As a result, Polar form of $(1-i)$ is $\sqrt{2}(\cos (\pi / 4)-i \sin (\pi / 4))$
(ii) $(1-i) /(1+i)$
Given that $Z=(1-i) /(1+i)$
Let’s multiply and divide by $(1-i)$, we get
$\begin{aligned}
Z =\frac{1-i}{1+i} \times \frac{1-i}{1-i} \\
=\frac{(1-i)^{2}}{1^{2}-i^{2}} \\
=\frac{1^{2}+i^{2}-2(1)(i)}{1-(-1)} \\
=\frac{1+(-1)-2 i}{2}
\end{aligned}$
$\begin{array}{c}
=\frac{-2 i}{2} \\
=0-i
\end{array}$
Therefore now,
$\begin{array}{l}
\left.|Z|=\sqrt{(} x^{2}+y^{2}\right) \\
\left.=\sqrt{(} 0^{2}+(-1)^{2}\right) \\
=\sqrt{(0+1)} \\
{ } &{=\sqrt{1}} \\
{\theta} &{=\tan ^{-1}(|y| /|x|)} \\
{=\tan ^{-1}(1 / 0)} \\
{=\tan ^{-1} \infty}
\end{array}$
As $x \geq 0, y<0$ complex number lies in $4^{\text {th }}$ quadrant and the value of $\theta$ is $-90^{0} \leq \theta \leq 0^{0}$
$\begin{array}{l}
\theta=-\pi / 2 \\
Z=1(\cos (-\pi / 2)+i \sin (-\pi / 2)) \\
=1(\cos (\pi / 2)-i \sin (\pi / 2))
\end{array}$
As a result, polar form of $(1-i) /(1+i)$ is $1(\cos (\pi / 2)-i \sin (\pi / 2))$