Answer
(a) See the graph.
(b) $z=2(cos~\frac{\pi}{3}+i~sin~\frac{\pi}{3})$
(c) $z^9=-512$
Work Step by Step
(b) $z=a+bi=1+\sqrt 3i$
$r=|z|=\sqrt {a^2+b^2}=\sqrt {1^2+(\sqrt 3)^2}=\sqrt 4=2$
$tan~θ=\frac{b}{a}=\frac{\sqrt 3}{1}=\sqrt 3$
$θ=\frac{\pi}{3}$
Polar form:
$z=2(cos~\frac{\pi}{3}+i~sin~\frac{\pi}{3})$
(c)
$z^n=r^n(cos~nθ+i~sin~nθ)$
$z^9=2^9[cos~(9·\frac{\pi}{3})+i~sin~(9·\frac{\pi}{3}))$
$z^9=512(cos~3\pi+i~sin~3\pi)$
$z^9=512(-1+0i)$
$z^9=-512$
