Answer
$[2(cos~\frac{\pi}{2}+i~sin~\frac{\pi}{2})]^8=256$
Work Step by Step
DeMoivre's Theorem:
If $z=r(cos θ+i~sin θ)$, then
$z^n=r^n(cos~nθ+i~sin~nθ)$
$z=2(cos~\frac{\pi}{2}+i~sin~\frac{\pi}{2})$
$z^8=2^8[cos~(8·\frac{\pi}{2})+i~sin~(8·\frac{\pi}{2})]$
$z^8=256(cos~4\pi+i~sin~4\pi)$
$z^8=256(1+0i)$
$z^8=256$
