Find all the roots in rectangular coordinates, exhibit them as vertices of certain squares, and point out which is the principal root.

$\displaystyle z_0=(-8-8i \sqrt3)^{1/4}=z^{1/4}$

The given answer is $\displaystyle \pm (\sqrt 3-i), \pm (1+ i\sqrt3)$, but I just do not see at all how they arrive at that answer. Here's what I got:

$\displaystyle r=\sqrt(64+192)=\sqrt(256)=16$

$\displaystyle tan(\theta)=\frac{-8\sqrt3}{-8}=\sqrt3...\theta= \frac{\pi }{3} $

$\displaystyle z=16e^{i(\frac{\pi}{3})}$

$\displaystyle z_0={16e^{i(\frac{\pi}{3})}}^{1/4}=\pm2e^{(\frac{\pi}{12}+\frac{2\pi k}{4})}=\pm2e^{(\frac{\pi}{12}+\frac{\pi k}{2})}, k=\pm1, \pm2,...$

Switching back to rectangular coordinates gives:

$\displaystyle \pm2\{cos(\frac{\pi}{12}+k\frac{\pi}{2})+isin(\fra c{\pi}{12}+k\frac{\pi}{2})\}$

This is clearly not correct, and my eyes are beginning to glaze over from looking at this problem all day. What am I doing wrong?