Solve this

We have, $z=(-1+i \sqrt{3})$

Let $-1=r \cos \theta$ and $\sqrt{3}=r \sin \theta$

By squaring and adding, we get

$(-1)^{2}+(\sqrt{3})^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$

$\Rightarrow 1+3=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow 4=r^{2}$

$\Rightarrow r=2$

$\therefore \cos \theta=\frac{-1}{2}$ and $\sin \theta=\frac{\sqrt{3}}{2}$

Since, θ lies in second quadrant, we have

$\theta=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$

Thus, the required polar form is $2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)$