De Moivre's Theorem to calculate the fourth roots of 8 I was given this question:
Use de Moivre's theorem to derive a formula for the $4^{th}$ roots of 8.
As far as my understanding of this theorem goes, it is only applicaple to complex numbers. How am I supposed to use it for 8?
My initial thought was use 8 to make $z = 8 + i0$. Determining the polar form:
$$z = 8(\cos \pi + i \sin \pi )$$
from the polar form a I then get that 
$$z^4 = 8^4 (\cos (4\pi) + i \sin(4\pi))$$
This is as far as I can go. I only recently started working with de Moivre's theorem so I'm not sure about my calculations and would appreciate any clarification on how to go about answering this question.

Note that $z = 8+0i$ is equivalent to $z = 8(\cos(2k\pi) + i\sin(2k\pi))$
And then for the fourth roots:
$$z^{\frac{1}{4}}=8^\frac{1}{4}\left(\cos\left(\frac{k\pi}{2}\right) + i\sin\left(\frac{k\pi}{2}\right)\right)$$
For $k=0$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(0) + i\sin(0)) = 8^\frac{1}{4}$
For $k=1$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})) = 8^\frac{1}{4}i$
For $k=2$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\pi) + i\sin(\pi)) = -8^\frac{1}{4}$
For $k=3$: $z^{\frac{1}{4}} = 8^\frac{1}{4}(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2})) = -8^\frac{1}{4}i$
