Answer
$$z=4 \left ( \cos \frac{2\pi }{3} +i \sin \frac{2\pi }{3} \right )$$
Work Step by Step
We plot the complex number $z=-2+2\sqrt{3}i$ the same way we plot $(-2,2\sqrt{3})$ 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 $z=-2+2\sqrt{3}i$, we have$$r=\sqrt{(-2)^2+(2\sqrt{3})^2}=\sqrt{16}=4, \\ \tan \theta =\frac{2\sqrt{3}}{-2}=-\sqrt{3}.$$Since $\cos \theta =\frac{a}{r} = \frac{-2}{\sqrt{16}} \lt 0$, the argument, $\theta$, must lie in quadrant $II$. So$$\theta =\frac{2\pi }{3} $$Thus, the complex number has the polar form$$z=4 \left ( \cos \frac{2\pi }{3} +i \sin \frac{2\pi }{3} \right ).$$
