Answer
$(cos~\frac{\pi}{4}+i~sin~\frac{\pi}{4})^{12}=-1$
Work Step by Step
DeMoivre's Theorem:
If $z=r(cos θ+i~sin θ)$, then
$z^n=r^n(cos~nθ+i~sin~nθ)$
$z=(cos~\frac{\pi}{4}+i~sin~\frac{\pi}{4})$
$z^{12}=1^{12}[cos~(12·\frac{\pi}{4})+i~sin~(12·\frac{\pi}{4})]$
$z^{12}=(cos~3\pi+i~sin~3\pi)$
$z^{12}=-1+0i$
$z^{12}=-1$
