<s>[INST] 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.
 [/INST] You can solve this geometrically if you know polar coordinates.
In polar coordinates, multiplication goes $(r_1, \theta_1) \cdot (r_2, \theta_2) = (r_1 \cdot r_2, \theta_1 + \theta_2)$, so cubing goes $(r, \theta)^3 = (r^3, 3\theta)$. The cube roots of $(r, \theta)$ are $\left(\sqrt[3]{r}, \frac{\theta}{3}\right)$, $\left(\sqrt[3]{r}, \frac{\theta+2\pi}{3}\right)$ and $\left(\sqrt[3]{r}, \frac{\theta+4\pi}{3}\right)$ (recall that adding $2\pi$ to the argument doesn't change the number). In other words, to find the cubic roots of a complex number, take the cubic root of the absolute value (the radius) and divide the argument (the angle) by 3.
$i$ is at a right angle from $1$: $i = \left(1, \frac{\pi}{2}\right)$. Graphically:

A cubic root of $i$ is $A = \left(1, \frac{\pi}{6}\right)$. The other two are $B = \left(1, \frac{5\pi}{6}\right)$ and $\left(1, \frac{9\pi}{6}\right) = -i$.
Recalling basic trigonometry, the rectangular coordinates of $A$ are $\left(\cos\frac{\pi}{6}, \sin\frac{\pi}{6}\right)$ (the triangle OMA is rectangle at M). Thus, $A = \cos\frac{\pi}{6} + i \sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$.
If you don't remember the values of $\cos\frac{\pi}{6}$ and $\sin\frac{\pi}{6}$, you can find them using geometry. The triangle $OAi$ has two equal sides $OA$ and $Oi$, so it is isoceles: the angles $OiA$ and $OAi$ are equal. The sum of the angles of the triangle is $\pi$, and we know that the third angle $iOA$ is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$; therefore $OiA = OAi = \dfrac{\pi - \frac{pi}{3}}{2} = \dfrac{\pi}{3}$. So $OAi$ is an equilateral triangle, and the altitude AN is also a median, so N is the midpoint of $[Oi]$: $\sin\frac{\pi}{6} = AM = ON = \frac{1}{2}$. By the Pythagorean theorem, $OM^2 + AM^2 = OA^2 = 1$ so $\cos\frac{\pi}{6} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \dfrac{\sqrt{3}}{2}$.
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