Question:
Write $z=(-1+i \sqrt{3})$ in polar form.
Solution:
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 $2\left(\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}\right)$