<s>[INST] Fourth complex roots of $i$ Solve $z^4 = i$. I cannot figure out why the angle of $i$ is $\frac{\pi}{2}$ and how to determine the values of $k$. If someone could show step-by-step that would be great!
Thanks.
 [/INST] $$z^4 = i \Longleftrightarrow$$
$$z^4 = |i| e^{\arg(i)i} \Longleftrightarrow$$
$$z^4 = 1 e^{\frac{\pi}{2}i} \Longleftrightarrow$$
$$z^4 = e^{\frac{\pi}{2}i} \Longleftrightarrow$$
$$z = \left(e^{\left(\frac{\pi}{2}+2\pi k\right)i}\right)^{\frac{1}{4}} \Longleftrightarrow$$
$$z = e^{{\frac{1}{4}}\left(\frac{\pi}{2}+2\pi k\right)i} \Longleftrightarrow$$
$$z = e^{\frac{\pi + 4\pi k}{8}i}$$
With $k\in\mathbb{Z}$ and $k:0-3$
So the solutions are:
$$z_0=e^{\frac{\pi + 4\pi \cdot 0}{8}i}=e^{\frac{\pi}{8}i}$$
$$z_1=e^{\frac{\pi + 4\pi \cdot 1}{8}i}=e^{\frac{5\pi}{8}i}$$
$$z_2=e^{\frac{\pi + 4\pi \cdot 2}{8}i}=e^{-\frac{7\pi}{8}i}$$
$$z_3=e^{\frac{\pi + 4\pi \cdot 3}{8}i}=e^{-\frac{3\pi}{8}i}$$
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