<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] for $z=x+iy$ you can write its Polar Representation $z=re^{i\theta‎}=r(\cos\theta+i\sin\theta)=\color{red}{r\cos\theta}+i\color{blue}{r\sin\theta}$(respectively real part and imaginary part red and blue) where $r=\sqrt{x^2+y^2}$ and $\theta=tan^{-1}\frac{y}{x} $ for $x>0$ and $\theta=tan^{-1}\frac{y}{x}+\pi $ for $x<0$.
For $w = \rho e^{i \phi}$ that $w^n = z$,   $\rho = r^{1/n}$, $\phi = \frac{\theta}{n} + \frac{2\pi k}{n}$ for $k=0, 1, ..,n-1$.
here for $z=i$, we have $r=\sqrt{0^2+1^2}=1$ and $\theta=tan^{-1}\frac{1}{0}=\frac{\pi}{2} $
thus answers will be $w=(1)^\frac{1}{4}e^{i(\frac{\frac{\pi}{2}}{4} + \frac{2\pi k}{4})}, k=0,1,2,3$
</s>