# Thread: Roots of a complex number

1. ## Roots of a complex number

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

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

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

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

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

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

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

Switching back to rectangular coordinates gives:

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

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?

2. First, the radius should always be positive. Therefore $16^{1/4}=2.$
Second, the angle you should have gotten is in the third quadrant, since both a and b are negative. This gives an original angle of $\frac{4\pi}{3}$.
This should help get you the correct answer.