Answer
$$z=4 \left ( \cos \frac{11\pi }{6} +i \sin \frac{11\pi }{6} \right )$$
Work Step by Step
We plot the complex number $z=2\sqrt{3}-2i$ the same way we plot $(2\sqrt{3}, -2)$ 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 $z=2\sqrt{3}-2i$, we have$$r=\sqrt{(2\sqrt{3})^2+(-2)^2}=\sqrt{16}=4, \\ \tan \theta =\frac{-2}{2\sqrt{3}}=-\frac{\sqrt{3}}{3}.$$Since $\cos \theta =\frac{a}{r} = \frac{2\sqrt{3}}{\sqrt{16}} \gt 0$, the argument, $\theta$, must lie in quadrant $IV$. So$$\theta =\frac{5\pi }{6}+\pi = \frac{11\pi }{6}$$Thus, the complex number has the polar form$$z=4 \left ( \cos \frac{11\pi }{6} +i \sin \frac{11\pi }{6} \right ).$$
