Write –3i in polar form.

Let, z = -3i 

Let $0=r \cos \theta$ and $-3=r \sin \theta$

By squaring and adding, we get

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

$\Rightarrow 0+9=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$

$\Rightarrow 9=r^{2}$

$\Rightarrow r=3$

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

Since, θ lies in fourth quadrant, we have

$\theta=\frac{3 \pi}{2}$

Thus, the required polar form is $3\left(\cos \left(\frac{3 \pi}{2}\right)+\operatorname{isin}\left(\frac{3 \pi}{2}\right)\right)$