Solve for $r$ and $\theta$ of $z^6$

Question

$z=(-1+\sqrt(3)i)$ Find r and theta for polar form when $z^6$.

I am unsure how to solve for this. I have got $r=2$ and $\theta=\frac {\pi}3$ for regular $z$, but I am unsure how to deal with the exponent.

Answer 1

Your calculation of angle is wrong.

$$z=2e^{i2\pi/3}$$ Then $$z^6=2^6e^{6(i2\pi/3)}=64e^{i4\pi}=64$$

Answer 2

Well, if you are claiming that $z=2e^{i\pi/3}$, then $$ z^6=(2e^{i\pi/3})^6=... $$