Answer
$(\frac{1}{2}+\frac{\sqrt 3}{2}i)^{20}=-\frac{1}{2}-\frac{\sqrt 3}{2}i$
Work Step by Step
$z=\frac{1}{2}+\frac{\sqrt 3}{2}i$
$r=|z|=\sqrt {a^2+b^2}=\sqrt {(\frac{1}{2})^2+(\frac{\sqrt 3}{2})^2}=\sqrt {\frac{1}{4}+\frac{3}{4}}=1$
$tan~θ=\frac{b}{a}=\frac{\frac{\sqrt 3}{2}}{\frac{1}{2}}=\sqrt 3$
$θ=\frac{2\pi}{3}$
$z=r(cos~θ+i~sin~θ)$
$z=1(cos\frac{2\pi}{3}+i~sin\frac{2\pi}{3})$
$z^{20}=(\frac{1}{2}+\frac{\sqrt 3}{2}i)^{20}=[1(cos\frac{2\pi}{3}+i~sin\frac{2\pi}{3})]^{20}$
$z^{20}=1^{20}[cos(20·\frac{2\pi}{3})+i~sin(20·\frac{2\pi}{3})]$
$z^{20}=cos(20·\frac{2\pi}{3})+i~sin(20·\frac{2\pi}{3})$
$z^{20}=cos\frac{40\pi}{3}+i~sin\frac{40\pi}{3}$
$\frac{40\pi}{3}=6(2\pi)+\frac{4\pi}{3}$
So:
$z^{20}=cos\frac{4\pi}{3}+i~sin\frac{4\pi}{3}$
$z^{20}=-\frac{1}{2}-\frac{\sqrt 3}{2}i$
