Answer
$-\displaystyle \frac{\sqrt{3}}{2048}-\frac{1}{2048}i$
Work Step by Step
See p. 606,
If $z=r(\cos\theta+i\sin\theta)$, then for any integer $n$
$z^{n}=r^{n}(\cos n\theta+i\sin n\theta)$
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$z=2\sqrt{3}+2i$
$\theta$ terminates in quadrant I,
$r=\sqrt{4\cdot 3+4}=\sqrt{16}=4,$
$\displaystyle \tan\theta=\frac{2}{2\sqrt{3}}=\frac{\sqrt{3}}{3} \Rightarrow \displaystyle \theta=\frac{\pi}{6}$.
$z=4(\displaystyle \cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
$z^{-5}=4^{-5}(\displaystyle \cos(-5\cdot\frac{\pi}{6})+i\sin(-5\cdot\frac{\pi}{6}))$
$=\displaystyle \frac{1}{1024}(\cos(\frac{5\pi}{6})-i\sin(\frac{5\pi}{6}))$
$=\displaystyle \frac{1}{1024}(-\frac{\sqrt{3}}{2}-\frac{1}{2}i)$
$=-\displaystyle \frac{\sqrt{3}}{2048}-\frac{1}{2048}i$
