Answer
$-4$
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
$(1+i)^{4}=[\sqrt {2}(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4})]^{4}=(\sqrt {2})^{4}(\cos 4\cdot\frac{\pi}{4}+i\sin 4\cdot\frac{\pi}{4})$
$=4(\cos \pi+i\sin\pi)$
In standard form, our result is
$=4(-1+i\cdot 0)$
$=-4$