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,
$\begin{array}{l}
|\mathrm{Z}|=\sqrt{\left(x^{2}+y^{2}\right)} \\
{=} &{\left.\sqrt{(} 1^{2}+1^{2}\right)} \\
{=} &{\sqrt{(} 1+1)} \\
{=} &{\sqrt{2}} \\
{\theta=\tan ^{-1}(|\mathrm{y}| /|\mathrm{x}|)} \\
{=\tan ^{-1}(1 / 1)} \\
{=\tan ^{-1} 1}
\end{array}$
As $x>0, y>0$ complex number lies in $1^{\text {st }}$ quadrant and the value of $\theta$ is $0^{0} \leq \theta \leq 90^{0}$.
$\begin{array}{l}
\theta=\pi / 4 \\
Z=\sqrt{2}(\cos (\pi / 4)+i \sin (\pi / 4))
\end{array}$
As a result, polar form of $(1+i)$ is $\sqrt{2}(\cos (\pi / 4)+i \sin (\pi / 4))$
(ii) $\sqrt{3}+i$
Given that $Z=\sqrt{3}+i$
Therefore now,
$\begin{array}{l}
\left.|Z|=\sqrt{(} x^{2}+y^{2}\right) \\
\left.=\sqrt{(}(\sqrt{3})^{2}+1^{2}\right) \\
=\sqrt{(} 3+1) \\
=\sqrt{4} \\
=2 \\
\theta=\tan ^{-1}(|y| /|x|) \\
=\tan ^{-1}(1 / \sqrt{3})
\end{array}$
As $x>0, y>0$ complex number lies in $1^{\text {st }}$ quadrant and the value of $\theta$ is $0^{0} \leq \theta \leq 90^{0}$.
$\begin{array}{l}
\theta=\pi / 6 \\
Z=2(\cos (\pi / 6)+i \sin (\pi / 6))
\end{array}$
As a result, polar form of $(\sqrt{3}+i)$ is $2(\cos (\pi / 6)+i \sin (\pi / 6))$