simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere. 
Any help is much appreciated!

$$(-2 + 2\sqrt3i) = 4 \exp\left(\frac{2\pi}{3}i\right) = 4 \cos \left(\frac{2\pi}{3}\right) +  4 \sin \left(\frac{2\pi}{3}\right) i$$
and I would say $$\left(4 \exp\left(\frac{2\pi}{3}i\right)\right)^{\frac{3}{2}} = 4^{\frac{3}{2}} \exp\left(\frac{3}{2} \times \frac{2\pi}{3}i\right) =8 \exp(\pi i) = -8. $$ 
I think using $(-2 + 2\sqrt3i) = 4 \exp\left(\frac{8\pi}{3}i\right)$ or  $4 \exp\left(-\frac{4\pi}{3}i\right)$ here would be unconventional. 
To get an answer of $\pm 8$ you would need to believe $\sqrt{-2 + 2\sqrt3i} = -1-\sqrt3 i$ as well as $1+\sqrt3 i$ and while the square of each of them gives the intended value, I would take what I regard as the principal root giving a single answer, and so does Wolfram Alpha.
It is like saying $\sqrt{4} = 2$ alone even though $(-2)^2=4$ too.
