put the following in standard form how do I compute the following by putting it in standad form?
\begin{equation}
(2-2\sqrt{3i})^{20}
\end{equation}
what I have tried to do is use de Moivre theorem
but that requires me to put it in polar form, as in z=...
but I don't know how to do that either.
 A: Hint:..................
$$i=\frac{(1+i)^2}{2}$$
$$i=-\frac{(1-i)^2}{2}$$
A: Let $2 - 2\sqrt 3 i = 4(\frac 1 2 - \frac {\sqrt 3} 2 i)$.  
Now a neat thing happens when we raise $(\frac 1 2 - \frac {\sqrt 3} 2 i)$ to powers.
$(\frac 1 2 - \frac {\sqrt 3} 2 i)^2 = \frac 1 4 - \frac 3 4 - \frac {\sqrt 3} 2 i = - \frac 1 2 - \frac {\sqrt 3} 2 i$
So $(\frac 1 2 - \frac {\sqrt 3} 2 i)^3 = (- \frac 1 2 - \frac {\sqrt 3} 2 i)(\frac 1 2 - \frac {\sqrt 3} 2 i) = - (\frac 1 2)^2 + (\frac {\sqrt 3} 2 i)^2 = - \frac 1 4 - \frac 3 4 = -1$
$(\frac 1 2 - \frac {\sqrt 3} 2 i)^{20} = ((\frac 1 2 - \frac {\sqrt 3} 2 i)^3)^6*(\frac 1 2 - \frac {\sqrt 3} 2 i)^2 = 1*(- \frac 1 2 - \frac {\sqrt 3} 2 i) = - \frac 1 2 - \frac {\sqrt 3} 2 i$
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So $(2 - 2\sqrt 3 i)^{20} = 4^{20}(\frac 1 2 - \frac {\sqrt 3} 2 i)^{20} = 2^{40}(- \frac 1 2 - \frac {\sqrt 3} 2 i) = -2^{39}(1 + \sqrt 3 i)$
+++++++++++++++
And we all could have saved ourselves a lot of time is we recognize $(1/2, -\sqrt3 /2)$ as the height and base of a 30-60-90 triangle and thus the number is a 6th root of 1 and a cube root of -1.
