Question:
If $z=-1+\sqrt{-} \overline{3}$, then $\arg (z)=$
Solution:
$z=-1+i \sqrt{-} \overline{3}$
$x=-1$ and $y=\sqrt{3}$
$\theta=\tan ^{-1} \frac{\sqrt{3}}{-1}$
$=\tan ^{-1}|-\sqrt{3}|$
$\theta=\frac{\pi}{3}$
Since z lies in IV quadrant.
$\Rightarrow$ argument of $z$ is $\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$.
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