Write the principal argument of

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

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

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

$z=0+2 \sqrt{3} i$

Let $0=r \cos \theta$ and $2 \sqrt{3}=r \sin \theta$

By squaring and adding, we get

$(0)^{2}+(2 \sqrt{3})^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$

$\Rightarrow 0+(2 \sqrt{3})^{2}=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow(2 \sqrt{3})^{2}=r^{2}$

$\Rightarrow r=2 \sqrt{3}$

∴ cosθ= 0 and sinθ=1

Since, θ lies in first quadrant, we have

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

Since, $\theta \in(-\pi, \pi]$ it is principal argument.