Answer
$[3(cos~60°+i~sin~60°)]^4=-\frac{81}{2}-\frac{81\sqrt 3}{2}i$
Work Step by Step
DeMoivre's Theorem:
If $z=r(cos θ+i~sin θ)$, then
$z^n=r^n(cos~nθ+i~sin~nθ)$
$z=3(cos~60°+i~sin~60°)$
$z^4=3^4[cos~(4 ·60°)+i~sin~(4 ·60°)]$
$z^4=81(cos~240°+i~sin~240°)$
$z^4=81[-\frac{1}{2}+i(-\frac{\sqrt 3}{2})]$
$z^4=-\frac{81}{2}-\frac{81\sqrt 3}{2}i$
