Answer
$-8+8i$
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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$z=-1-i$
$\theta$ terminates in quadrant III,
$r=\sqrt{1+1}=\sqrt{2},$
$\displaystyle \tan\theta=\frac{-1}{-1} =1 \Rightarrow \displaystyle \theta=\frac{5\pi}{4}$.
$z=\displaystyle \sqrt{2}(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$
$z^{7}=(2^{1/2})^{7}(\displaystyle \cos(7\cdot\frac{5\pi}{4})+i\sin(7\cdot\frac{5\pi}{4}))$
$=2^{7/2}(\displaystyle \cos(\frac{35\pi}{4})+i\sin(\frac{35\pi}{4}))$
With
$\displaystyle \cos\frac{35\pi}{4}=\cos((\frac{3\pi}{4}+8\pi)=-\frac{\sqrt{2}}{2}$
$\displaystyle \sin\frac{35\pi}{4}=\sin(\frac{3\pi}{4}+8\pi)=\frac{\sqrt{2}}{2}$
$2^{7/2}=2^{3}2^{1/2}=8\sqrt{2}$
$z^{7}=8\displaystyle \sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)$
$=-8+8i$
