Answer
$$z=4 \left ( \cos \frac{3\pi }{2} +i \sin \frac{3\pi }{2} \right )$$
Work Step by Step
We plot the complex number $z=-4i$ the same way we plot $(0,-4)$ in the rectangular coordinate system.
Please note that the complex number $z=a+bi$ is written in polar form as$$z=r(\cos \theta +i \sin \theta ),$$where $a=r\cos \theta$, $b=r\sin \theta$, $r=\sqrt{a^2+b^2}$, and $\tan \theta = \frac{b}{a}$, $0 \le \theta \lt 2\pi $.
So, for the complex number $z=-4i$, we have$$r=\sqrt{0^2+(-4)^2}=\sqrt{16}=4, \\ \sin \theta =\frac{b}{r}=\frac{-4}{4}=-1.$$ So$$\theta =\frac{3\pi }{2} $$Thus, the complex number has the polar form$$z=4 \left ( \cos \frac{3\pi }{2} +i \sin \frac{3\pi }{2} \right ).$$
