Suppose we wish to solve the equation $$z^3 = 2+2 i. $$ This actually has three solutions, and we can find them using de Moivre's Theorem. Suppose that z = r (\cos \theta+ i \sin \theta) (where r > 0 and -\pi < \theta \le \pi). Then our equation becomes $$r^3 (\cos 3 \theta+ i \sin 3 \theta) = 2 + 2 i.$$ Now, the modulus of 2 + 2 i is 2 \sqrt{2}, and its argument is \pi/4. It follows that $$r^3 = 2 \sqrt{2}, $$ and therefore r = \sqrt{2}. Now, it might seem to follow that $$3 \theta = \pi/4. $$ However, it's a little more complicated than that. Because $$-\pi < \theta \le \pi, $$ we can say that $$-3 \pi < 3 \theta \le 3 \pi. $$ So as well as $$3 \theta = \pi/4, $$ we ought also to consider $$3 \theta = -2 \pi+ \pi/4 $$ and $$3 \theta = 2 \pi+ \pi/4: $$ that is $$3 \theta = -7\pi/4 $$ and $$3 \theta = 9\pi/4. $$ This gives us $$\theta = -7\pi/12, \pi/12, 3\pi/4, $$ and our solutions are therefore
\begin{eqnarray} z & = & \sqrt{2} (\cos 7\pi/12 - i \sin 7\pi/12), \\ z & = & \sqrt{2} (\cos \pi/12 + i \sin \pi/12), \\ z & = & \sqrt{2} (\cos 3\pi/4 + i \sin 3\pi/4) = -1 + i. \end{eqnarray}
The solutions are displayed on an Argand Diagram in the figure. Note that they're equally spaced, like the spokes on a wheel. This is always the case with complex solutions of equations of the form $$z^n = \omega. $$
Figure 1: The three cube roots of 2+2i