Question:
Find the principal argument of $(1+i \sqrt{3})^{2}$.
Solution:
$z=(1+i \sqrt{3})^{2}$
$=1+3 i^{2}+2 \sqrt{3} i$
$=1-3+2 \sqrt{3} i$
$=-2+2 \sqrt{3} i$
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 $\frac{2 \pi}{3}$.
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