Find the principal argument of

Let, z = -2i

Let $0=r \cos \theta$ and $-2=r \sin \theta$

By squaring and adding, we get

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

$\Rightarrow 0+4=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow 4=r^{2}$

$\Rightarrow r=2$

$\therefore \cos \theta=0$ and $\sin \theta=-1$

Since, θ lies in fourth quadrant, we have

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

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