Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$. 
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\frac{\pi}{8}$
Since $z^n=r^n\cdot e^{in\theta}$, $z^4$ can now be rewritten as $z^4=e^{i\cdot4\cdot\frac{\pi}{8}} \iff z=e^{i\frac{\pi}{8}}$ However, my answer file says this is wrong. Can anyone give me a hint on how to find $z$?

We have $z^4=e^{3\pi i/2+2k\pi}$.
So $z=(e^{3\pi i/2+2k\pi})^{1/4}=e^{3\pi i/8+k\pi/2}$
As $e^{i\theta}=e^{i(\theta+2\pi)}$, the $4$ values of $z$ are $e^{3\pi i/8}, e^{7\pi i/8}, e^{11\pi i/8}, e^{15\pi i/8}$.
