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!

So this can be written as $$4^{3/2} \cdot (-\frac{1}{2} + \frac{\sqrt{3}}{2} i)^{3/2} = 8 \cdot (\cos\frac{2\pi}{3} + i\cdot \sin\frac{2\pi}{3})^{3/2}$$
and $4^{3/2} =8$. Use De moivre now. And you can also pull out $-4$ and get going. Hence your $z = \pm{8}$.
