Answer
$-8i$
Work Step by Step
$1+i$ in trigonometric form is
$1+i=\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})$
Using de Moivre's theorem ($z^{n}=r^{n} cis \,n\theta$ where $z=r\, cis\,\theta$ and $n$ is an integer), we get
$(1+i)^{6}=[\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})]^{6}=(\sqrt {2})^{6}(\cos 6\cdot\frac{\pi}{4}+i\sin 6\cdot\frac{\pi}{4})$
$=8(\cos \frac{3\pi}{2}+i\sin\frac{3\pi}{2})$
In standard form, our result is
$=8(0+i\cdot -1)$
$=-8i$
