Answer
The required polar form is $ z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$
Work Step by Step
Using
$ r=\sqrt{{{a}^{2}}+{{b}^{2}}}$
and,
$\begin{align}
& \tan \theta =\frac{b}{a} \\
& =\frac{1}{-\sqrt{3}} \\
& \tan \theta =-\frac{1}{\sqrt{3}}
\end{align}$
So,
$\begin{align}
& r=\sqrt{{{a}^{2}}+{{b}^{2}}} \\
& =\sqrt{{{\left( -\sqrt{3} \right)}^{2}}+{{1}^{2}}} \\
& =\sqrt{3+1} \\
& =2
\end{align}$
Also, $\tan 30{}^\circ =\frac{1}{\sqrt{3}}$ and $\theta $ lies in the second quadrant.
So,
$\begin{align}
& \theta =180{}^\circ -30{}^\circ \\
& =150{}^\circ
\end{align}$
Hence the polar form of $ z=-\sqrt{3}+i $ is:
$\begin{align}
& z=r\left( \cos \theta +i\sin \theta \right) \\
& =r\left( \cos 150{}^\circ +i\sin 150{}^\circ \right) \\
& z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)
\end{align}$
So, $ z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$
Therefore, the required polar form is $ z=2\left( \cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6} \right)$.
