Answer
$\color{blue}{4\ \text{cis}\ 150^\circ,\ 4\ \text{cis}\ 5\pi/6}$
Work Step by Step
$z=-2\sqrt{3} +2i = x+iy \implies x=-2\sqrt{3}, y=2$
$\Huge\cdot$ modulus: $\quad r = \sqrt{x^2+y^2} = \sqrt{(-2\sqrt{3})^2 + 2^2} = \sqrt{12+4} = \sqrt{16} = 4$
$\Huge\cdot$ argument: $\quad \tan\theta=y/x=2/(-2\sqrt{3})= -1/\sqrt{3} = -\sqrt{3}/3 \implies \theta = 150^\circ \equiv 5\pi/6$ (smallest positive real angle $\theta$ from $+x$-axis to graph of $z$)
$\begin{array}{|c|c|c|} \hline
\text{Standard} & \text{Trigonometric} & \text{Trigonometric} \\
\text{Form} & \text{Form (deg)} & \text{Form (rad)} \\ \hline
-2\sqrt{3} +2i & 4\ \text{cis}\ 150^\circ & 4\ \text{cis}\ 5\pi/6 \\ \hline
\end{array}$