Question:
If $z=\frac{-2}{1+i \sqrt{3}}$, then the value of arg ( $\mathrm{z}$ ) is
(a) $\pi$
(b) $\frac{\pi}{3}$
(c) $\frac{2 \pi}{3}$
(d) $\frac{\pi}{4}$
Solution:
(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, $\arg (z)=\pi-\frac{\pi}{3}$
$=\frac{2 \pi}{3}$