Answer
$-72+72\sqrt{3}i$
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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$z=3+\sqrt{3}i$
$\theta$ terminates in quadrant I,
$r=\sqrt{9+3}=\sqrt{12}=2\sqrt{3},$
$\displaystyle \tan\theta=\frac{\sqrt{3}}{3} \Rightarrow \displaystyle \theta=\frac{\pi}{6}$.
$z=2\displaystyle \sqrt{3}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
$z^{4}=(2\displaystyle \sqrt{3})^{4}(\cos(4\cdot\frac{\pi}{6})+i\sin(4\cdot\frac{\pi}{6}))$
$=2^{4}3^{2}(\displaystyle \cos(\frac{2\pi}{3})+i\sin(\frac{2\pi}{3}))$
$=16\displaystyle \cdot 9(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)$
$=-72+72\sqrt{3}i$
