Answer
$\displaystyle \frac{1}{2048}-\frac{\sqrt{3}}{2048}i$
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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$z=\sqrt{3}-i$
$\theta$ terminates in quadrant IV
$r=\sqrt{3+1}=2$
$\displaystyle \tan\theta=\frac{-1}{\sqrt{3}} \Rightarrow \displaystyle \theta=\frac{11\pi}{6}$.
$z=2(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
$z^{-10}=(2)^{-10}(\displaystyle \cos(-10\cdot\frac{11\pi}{6})+i\sin(-10\cdot\frac{11\pi}{6}))$
$=(2)^{-10}(\displaystyle \cos\frac{-55\pi}{3}+i\sin\frac{-55\pi}{3})$
With
$\displaystyle \cos(\frac{-55\pi}{3})=\cos(\frac{55\pi}{3})=\cos(\frac{\pi}{3}+18\pi)=\frac{1}{2},$
$\displaystyle \sin(\frac{-55\pi}{3})=-\sin(\frac{55\pi}{3})=-\sin(\frac{\pi}{3}+18\pi)=-\frac{\sqrt{3}}{2},$
$z^{-10}=\displaystyle \frac{1}{1024}(\frac{1}{2}-\frac{\sqrt{3}}{2}i)$
$=\displaystyle \frac{1}{2048}-\frac{\sqrt{3}}{2048}i$
