how do I write -8i in polar form?
Solving $\displaystyle z^3 = -8i$...so u want to solve for z right?
If you write z as $\displaystyle z = re^{i\theta}$ and substitute this into $\displaystyle z^3 = -8i$ and solve.
You get r = 2 and $\displaystyle 3\theta = \frac{3\pi}{2} + 2k\pi , (k=0,1)$.
Therfore $\displaystyle \theta = \frac{\pi}{2} + \frac{2k\pi}{3} , (k=0,1,2)$