Find the principal argument of

$z=(1+i \sqrt{3})^{2}$

$=1+3 i^{2}+2 \sqrt{3} i$

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

$=-2+2 \sqrt{3} i$

Let $\beta$ be an acute angle given by $\tan \beta=\frac{|\operatorname{Im}(z)|}{|\operatorname{Re}(z)|}$. Then,

$\tan \beta=\frac{|2 \sqrt{3}|}{|2|}=|\sqrt{3}|$

$\Rightarrow \tan \beta=\left|\tan \frac{\pi}{3}\right|$

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

Clearly, $z$ lies in the second quadrant. Therefore, $\arg (z)=\pi-\frac{\pi}{3}=\frac{2 \pi}{3}$.

Hence, the principal argument of $z$ is $\frac{2 \pi}{3}$.