<s>[INST] 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 $z$?
 [/INST] the last step, you need $$-i = \cos(3\pi/2) + \sin (3\pi/2) = \cos (4 \theta) + i \sin(4 \theta) $$  which will give you $$4\theta = 3\pi/2, 3\pi/2 + 2\pi,   3\pi/2 + 4\pi,  3\pi/2 + 6\pi$$ now you can solve for $\theta.$
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