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!
 A: $$(-2 + 2\sqrt3i) = 4 \exp\left(\frac{2\pi}{3}i\right) = 4 \cos \left(\frac{2\pi}{3}\right) +  4 \sin \left(\frac{2\pi}{3}\right) i$$
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.
A: 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 $z = \pm{8}$.
A: Try something like this, perhaps:
$$(-2+2\sqrt{3}i)^{3/2}=\exp \left(\frac{3}{2}\operatorname {Log} (-2+2\sqrt{3}i)\right)$$
We know the principal value of $\operatorname {Log z}$ is given by $\operatorname {Log z}=\log |z|+i\arg \theta=$ for $n \in \mathbb{Z}$:
$$\exp \left(\frac{3}{2}\operatorname {Log} (-2+2\sqrt{3}i)\right)=\exp \left(\frac{3}{2}(\log 4+i \frac{2\pi}{3})\right)=8$$
Keep in mind that because we only used the single values $\operatorname {Log z}$, we only get the positive answer.
A: $\:\sqrt{-2+2\sqrt{-3}}\:$ can be denested by a radical denesting formula that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \  \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $  
and, furthermore,  $\rm\:w\:$ has trace $\rm\: =\: w+w' =  (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\:  2\:a$

Here $ {-}2+2\sqrt{-3}\:$ has norm $= 16.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -4\ $ yields $\  2+2\sqrt{-3}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ hence\ \  \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1+\sqrt{-3}.$
Checking we have
$\ \smash[t]{\displaystyle \left(1+\sqrt{-3}\right)^2 =\ 1-3 + 2\sqrt{-3}\ =\ -2 + 2 \sqrt{-3}}$
Therefore $\quad\ \begin{eqnarray}\rm\:(-2 + 2\sqrt{-3})^{3/2} &=&\ (-2+2\sqrt{-3})\ (-2+2\sqrt{-3})^{1/2} \\  &=&\ -2\,(1-\sqrt{-3})\ (1+\sqrt{-3}) \\
 &=&\ -8\rm\ \ \ (up\ to\ sign) 
\end{eqnarray}$
See this answer for general radical denesting algorithms.
A: $$-2+2\sqrt{3}i = 4 \exp(2 \pi/3 i) = 4 \exp(2 \pi/3 i + 2 k \pi i) \text{ where }k\in \mathbb{Z}$$ Hence, $$\left(-2+2\sqrt{3}i \right)^{3/2} = \left(4 \exp(2 \pi/3 i + 2k \pi i) \right)^{3/2} = 8 \exp(\pi i + 3k \pi i) = \pm 8$$
