I got pretty far into this question, but the further I got the more convoluted my answer was becoming to the point that I figured I was doing something wrong. I converted the equation into polar form using de moivre, and ended up getting $z= 4^{1/4} e^{-i(π/24)}$, and then I attempted to convert this to Cartesian by using $x=r\cos θ$, and $y=r\sin θ$. I worked out the $x$, by using the identity $\cos 2x=2\cos^2 x-1$, but it was a really ugly square and got the sense that I've done something wrong. Plus this is a past exam question, and I don't think the question should take as long as it is taking me (so this adds to me feeling like I am doing something wrong).
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If you're raising it to the power of 4, then you multiply the exponent, not divide! Same goes for the 4. – snowfall512 Mar 12 '17 at 03:47
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Algebraic alternative:
$$ \begin{align} (2\sqrt{3} − 2 i)^4 & = 2^4(\sqrt{3}-i)^4 = 16(2-2\sqrt{3}i)^2=64(1-\sqrt{3}i)^2=64(-2-2\sqrt{3}i) \\[3px] & =-128(1+\sqrt{3}i) \end{align} $$
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You had the original inside expression right but you have to multiply the exponent in the e by 4, and raise 4 to the power of... 4!
So it's $4^4e^{-i\frac{2\pi}{3}}=\boxed{-128-128\sqrt{3}i}.$
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1Oh woops! The next question was find the solutions of z^4=2√ 3 − 2 i so I totally brain farted thinking ahead. Thank you! – Skylineblue Mar 12 '17 at 03:56