Solution:
(i) $1-\sin \alpha+i \cos \alpha$
Given that $Z=1-\sin \alpha+i \cos a$
Using the formulas,
$\operatorname{Sin}^{2} \theta+\cos ^{2} \theta=1$
$\operatorname{Sin} 2 \theta=2 \sin \theta \cos \theta$
$\operatorname{Cos} 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta$
Therefore,
$\begin{array}{l}
\mathrm{Z}=\left(\sin ^{2}(\alpha / 2)+\cos ^{2}(\alpha / 2)-2 \sin (\alpha / 2) \cos (\alpha / 2)\right)+i\left(\cos ^{2}(\alpha / 2)-\sin ^{2}(\alpha / 2)\right) \\
=(\cos (\alpha / 2)-\sin (\alpha / 2))^{2}+i\left(\cos ^{2}(\alpha / 2)-\sin ^{2}(\alpha / 2)\right)
\end{array}$
It is known to us that the polar form of a complex number $Z=x+$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$
In which,
$\begin{array}{l}
|Z|=\text { modulus of complex number }=\sqrt{\left(x^{2}+y^{2}\right)} \\
{\theta} {=\arg (z)=\text { argument of complex number }=\tan ^{-1}(|y| /|x|)}
\end{array}$
Now,
$\begin{aligned}
|z| =\sqrt{(1-\sin \alpha)^{2}+\cos ^{2} \alpha} \\
=\sqrt{1+\sin ^{2} \alpha-2 \sin \alpha+\cos ^{2} \alpha} \\
=\sqrt{1+1-2 \sin \alpha} \\
=\sqrt{(2)(1-\sin \alpha)} \\
=\sqrt{(2)\left(\sin ^{2}\left(\frac{\alpha}{2}\right)+\cos ^{2}\left(\frac{\alpha}{2}\right)-2 \sin \left(\frac{\alpha}{2}\right) \cos \left(\frac{\alpha}{2}\right)\right)} \\
=\sqrt{(2)\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)^{2}} \\
=\left|\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)\right| \\
=\tan ^{-1}\left(\frac{\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)\left(\cos \left(\frac{\alpha}{2}\right)+\sin \left(\frac{\alpha}{2}\right)\right)}{\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)^{2}}\right) \\
=\tan ^{-1}\left(\frac{\cos ^{2}\left(\frac{\alpha}{2}\right)-\sin ^{2}\left(\frac{\alpha}{2}\right)}{(\cos )}\right.\\
=\tan ^{-1}\left(\frac{\tan \left(\frac{\pi}{4}\right)+\tan \left(\frac{\alpha}{2}\right)}{1-\tan \left(\frac{\pi}{4}\right) \tan \left(\frac{\alpha}{2}\right)}\right)
\end{aligned}$
$=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right)$
It is known to us that sine and cosine functions are periodic with period $2 \pi$
Here we have 3 intervals:
$\begin{array}{l}
0 \leq a \leq \pi / 2 \\
\pi / 2 \leq a \leq 3 \pi / 2 \\
3 \pi / 2 \leq a \leq 2 \pi
\end{array}$
Consider case 1:
In the interval $0 \leq \alpha \leq \pi / 2$
$\operatorname{Cos}(\alpha / 2)>\sin (\alpha / 2)$ and also $0<\pi / 4+\alpha / 2<\pi / 2$
Therefore,
$\begin{aligned}
|z| =\left|\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)\right| \\
=\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right) \\
\theta =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right) \\
=\pi / 4+\alpha / 2\left[\text { since, } \theta \text { lies in } 1^{\text {st }} \text { quadrant }\right]
\end{aligned}$
As a result, Polar form is $Z=\sqrt{2}(\cos (\alpha / 2)-\sin (\alpha / 2))(\cos (\pi / 4+\alpha / 2)+i \sin (\pi / 4+\alpha / 2))$
Consider case 2:
In the interval $\pi / 2 \leq \alpha \leq 3 \pi / 2$
$\cos (\alpha / 2)<\sin (\alpha / 2)$ and also $\pi / 2<\pi / 4+\alpha / 2<\pi$
Therefore,
$\begin{aligned}
|z| =\left|\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)\right| \\
=-\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)
\end{aligned}$
$\begin{array}{l}
=\sqrt{2}\left(\sin \left(\frac{\alpha}{2}\right)-\cos \left(\frac{\alpha}{2}\right)\right) \\
\theta=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)\right) \\
=\pi-[\pi / 4+\alpha / 2]\left[\text { since, } \theta \text { lies in } 4^{\text {th }} \text { quadrant }\right] \\
=3 \pi / 4-\alpha / 2
\end{array}$
Since, $(1-\sin \alpha)>0$ and $\cos \alpha<0\left[Z\right.$ lies in $4^{\text {th }}$ quadrant]
$=a / 2-3 \pi / 4$
As a result, Polar form is $Z=-\sqrt{2}(\cos (\alpha / 2)-\sin (\alpha / 2))(\cos (\alpha / 2-3 \pi / 4)+i \sin (a / 2-3 \pi / 4))$
Consider case $3:$
In the interval $3 \pi / 2 \leq a \leq 2 \pi$
$\operatorname{Cos}(\alpha / 2)<\sin (\alpha / 2)$ and also $\pi<\pi / 4+a / 2<5 \pi / 4$
Therefore,
$\begin{array}{l}
|z|=\left|\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right)\right| \\
\quad=-\sqrt{2}\left(\cos \left(\frac{\alpha}{2}\right)-\sin \left(\frac{\alpha}{2}\right)\right) \\
\quad=\sqrt{2}\left(\sin \left(\frac{\alpha}{2}\right)-\cos \left(\frac{\alpha}{2}\right)\right) \\
\theta=\tan ^{-1}(\tan (\pi / 4+\alpha / 2))
\end{array}$
$=\Pi-(\pi / 4+\alpha / 2)$ [since, $\theta$ lies in 1 st quadrant and tan’s period is $\pi]$,
$=\alpha / 2-3 \pi / 4$
As a result, Polar form is $Z=-\sqrt{2}(\cos (\alpha / 2)-\sin (\alpha / 2))(\cos (a / 2-3 \pi / 4)+i \sin (a / 2-3 \pi / 4))$
(ii) $(1-i) /(\cos \pi / 3+i \sin \pi / 3)$
Given that $Z=(1-i) /(\cos \pi / 3+i \sin \pi / 3)$
Multiply and divide by $(1-\mathrm{i} \sqrt{3})$, we get
$Z=\frac{1-i}{\frac{1}{2}+\frac{i \sqrt{3}}{2}}$
$\begin{array}{l}
=2 \times \frac{1-i}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}} \\
=2 \times \frac{1+i^{2} \sqrt{3}-i(1+\sqrt{3})}{1-i^{2} 3} \\
=2 \times \frac{(1+(-\sqrt{3})-i(1+\sqrt{3}))}{1-(-3)} \\
=2 \times \frac{(1-\sqrt{3})-i(1+\sqrt{3})}{4} \\
=\frac{(1-\sqrt{3})-i(1+\sqrt{3})}{2}
\end{array}$
It is known to us that the polar form of a complex number $Z=x+i y$ is given by $Z=|Z|(\cos \theta+i \sin \theta)$
In which,
$|Z|=$ modulus of complex number $=\sqrt{\left(x^{2}+y^{2}\right)}$
$\theta=\arg (\mathrm{z})=$ argument of complex number $=\tan ^{-1}(|\mathrm{y}| /|\mathrm{x}|)$
Now,
$\begin{aligned}
|z| =\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^{2}+\left(\frac{-1-\sqrt{3}}{2}\right)^{2}} \\
=\sqrt{\frac{1+3-2 \sqrt{3}+1+2+2 \sqrt{3}}{4}} \\
=\sqrt{\frac{8}{4}} \\
=\sqrt{2}
\end{aligned}$
$\begin{aligned}
\theta =\tan ^{-1}\left(\left|\frac{\frac{1+\sqrt{3}}{2}}{\frac{1-\sqrt{3}}{2}}\right|\right) \\
=\tan ^{-1}\left(\left|\frac{1+\sqrt{3}}{1-\sqrt{3}}\right|\right) \\
=\tan ^{-1}\left(\left|\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}\right|\right) \\
=\tan ^{-1}\left(\left|\frac{1+3+2 \sqrt{3}}{1-3}\right|\right)
\end{aligned}$
$\begin{array}{l}
=\tan ^{-1}\left(\left|\frac{1+3+2 \sqrt{3}}{1-3}\right|\right) \\
=\tan ^{-1}\left(\frac{4+2 \sqrt{3}}{2}\right)
\end{array}$
Since $x<0, y<0$ complex number lies in $3^{\text {rd }}$ quadrant and the value of $\theta$ is $180^{0} \leq \theta \leq-90^{\circ}$.
$\begin{array}{l}
=\tan ^{-1}(2+\sqrt{3}) \\
=-7 \pi / 12 \\
Z=\sqrt{2}(\cos (-7 \pi / 12)+i \sin (-7 \pi / 12)) \\
=\sqrt{2}(\cos (7 \pi / 12)-i \sin (7 \pi / 12))
\end{array}$
As a result, Polar form of $(1-i) /(\cos \pi / 3+i \sin \pi / 3)$ is $\sqrt{2}(\cos (7 \pi / 12)-i \sin (7 \pi / 12))$