Answer
$-8+8i\sqrt 3$
Work Step by Step
$\sqrt {3}+i$ in trigonometric form is
$\sqrt {3}+i=2(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(\sqrt {3}+i)^{4}=[2(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})]^{4}=(2)^{4}(\cos 4\cdot\frac{\pi}{6}+i\sin 4\cdot\frac{\pi}{6})$
$=16(\cos \frac{2\pi}{3}+i\sin\frac{2\pi}{3})$
In standard form, our result is
$=16(-\frac{1}{2}+i\cdot \frac{\sqrt 3}{2})$
$=-8+8i\sqrt 3$
