<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] Remember, that $-i = e^{i\frac{3\pi}{2}}$, then
$$ z^4 = -i = e^{i\frac{3\pi}{2}}$$
The n$^{\text{th}}$ root of a complex number is
$$z^n = re^{i\varphi} \implies z_k = r^{1/n}e^{i\left(\frac{\varphi}{n} + \frac{2\pi k}{n} \right)} \qquad \text{where} \quad k = 0, \dots, n-1$$
In your case:
$$z_0 = e^{i\left(\frac{3\pi}{8} + 0 \right)} = (r, \varphi_0) = \left(1, \frac{3 \pi}{8}\right)$$
$$z_1 = e^{i\left(\frac{3\pi}{8} + \frac{2\pi}{4} \right)} = (r, \varphi_1) = \left(1, \frac{7 \pi}{8}\right)$$
$$z_2 = e^{i\left(\frac{3\pi}{8} + \frac{4\pi}{4} \right)} = (r, \varphi_2) = \left(1, \frac{11 \pi}{8}\right)$$
$$z_3 = e^{i\left(\frac{3\pi}{8} + \frac{6\pi}{4} \right)} = (r, \varphi_3) = \left(1, \frac{15 \pi}{8}\right)$$
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