How can you find the cubed roots of $i$? I am trying to figure out what the three possibilities of $z$ are such that 
$$
z^3=i
$$
but I am stuck on how to proceed. I tried algebraically but ran into rather tedious polynomials. Could you solve this geometrically? Any help would be greatly appreciated.

Taking the absolute value of both sides: $|z^3| = |i|$, gives $|z| = 1$.  So, $z = \cos (\theta) + i \sin (\theta)$ for some real $\theta$.
Using De Moivre's formula gives $z^3 = \cos(3\theta) + i \sin(3\theta)$.  Given that $z^3 = i = 0 + 1i$, this means that $\cos(3\theta) = 0$ and $\sin(3\theta) = 1$.  Solving this system gives $3\theta = \frac{\pi}{2} + 2\pi n$, or $\theta = \frac{\pi}{6} + \frac{2 \pi n}{3}$, for any $n \in \mathbb{Z}$.
Plugging in a few values for $n$ gives:


*

*$n = 0$ → $\theta = \frac{\pi}{6}$ → $z = \frac{\sqrt{3}}{2} + \frac{1}{2} i$

*$n = 1$ → $\theta = \frac{5\pi}{6}$ → $z = \frac{-\sqrt{3}}{2} + \frac{1}{2} i$ 

*$n = 2$ → $\theta = \frac{3\pi}{2}$ → $z = -i$


And we can stop there because this is a polynomial equation of degree 3, and the Fundamental Theorem of Algebra guarantees that it has at most 3 distinct roots.  The solution set is thus $z \in \{ \frac{\sqrt{3}}{2} + \frac{1}{2} i, \frac{-\sqrt{3}}{2} + \frac{1}{2} i, -i \}$.
