Find the three cube roots of $z = -2+2i$ $n$th roots of a complex number:
For a positive integer $n$, the complex number $z=r(\cos(\theta) + i\sin(\theta))$ has exactly $n$ distinct $n$th roots given by
$$\sqrt[n]r\left(\cos\left(\frac{\theta + 2\pi k}n\right) + i\sin\left(\frac{\theta + 2\pi k}n\right)\right)$$
where $k=0, 1, 2, ..., n-1$
Find the three cube roots of $z=-2 + 2i$
$$r = \sqrt{(-2)^2 + 2^2} = \sqrt{8}$$
$$\tan(\theta) = \frac ba = \frac {2}{-2} = -1$$
$\theta = 45^o$, because $z$ is in Quad II, $\theta = 135^o$
The trigonometric form of $z$ is $z = -2 + 2i = \sqrt{8}(cos135^o + isin135^o)$
Using the formula of $n$th roots, the cube roots have the form
$$\sqrt[6]8\left(\cos\left(\frac{135^o + 360^ok}3\right) + i\sin\left(\frac{135^o + 360^ok}3\right)\right)$$
For $k = 0$ we get the root $1 + i$
For $k = 1$ we get the root $-1.3660 + 0.3660i$
For $k = 2$ we get the root $0.3660 - 1.3660i$
My question is, using the formula for $nth$ roots, why is the first part of the formula given by $$\sqrt[6]8$$ and not $$\sqrt[3]8$$
because $n=3$, so I'm confused as to where 6 is coming from. 

You have $r=\sqrt{8}$ and then you have to compute $\sqrt[3]{r}=\sqrt[3]{\sqrt{8}}=\sqrt[6]{8}$. This is the basic property that you have to use: if $a\geq 0$ and $n,m\in \mathbb{N}$, with $n,m\geq 1$, then $\sqrt[n]{\sqrt[m]{a}} = \sqrt[nm]{a}$.
