Solve the following

(c) $\frac{2 \pi}{3}$

$z=\frac{-2}{1+i \sqrt{3}}$

Rationalising z, we get,

$z=\frac{-2}{1+i \sqrt{3}} \times \frac{1-i \sqrt{3}}{1-i \sqrt{3}}$

$\Rightarrow z=\frac{-2+i 2 \sqrt{3}}{1+3}$

$\Rightarrow z=\frac{-1+i \sqrt{3}}{2}$

$\Rightarrow z=\frac{-1}{2}+\frac{i \sqrt{3}}{2}$

$\tan \alpha=\left|\frac{\operatorname{Im}(z)}{\operatorname{Re}(z)}\right|$

$=\sqrt{3}$

$\Rightarrow \alpha=\frac{\pi}{3}$

Since, $z$ lies in the second quadrant.

Therefore, $\arg (z)=\pi-\frac{\pi}{3}$

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