Polar form of -$2$ - $2\sqrt{2i}$ -$2$ - $2\sqrt{2i}$
So I am confused on the $\sqrt{2i}$ part, so what I did was using the rule $\sqrt{ab}$ = $\sqrt{a} \sqrt{b}$ and that gives $\sqrt{2} \sqrt{i}$
-$2$ - $2\sqrt{2i}$ = -2 -2$\sqrt{2} \sqrt{i}$ multiply by $i^\frac{9}{2}$ to get an $i$ therefore it becomes -2 -2$\sqrt{2}$i ; r = 2$\sqrt{3}$ and $\theta$ = $\tan^{-1}({\sqrt{2})}$
so the polar form is 2$\sqrt{3}$ cis($\tan^{-1}({\sqrt{2})}$)
Is this correct?

Note
\begin{align}
-2-2\sqrt{2i}&= -2-2\sqrt2 (e^{i\frac\pi2})^{1/2}\\
&= -2-2\sqrt2e^{i\frac \pi4}\\
&= -4-2i \\
&= 2\sqrt5 e^{i\tan^{-1}\frac12}
\end{align}
