# Complex root

1. Sep 13, 2006

### Valhalla

I just bombed a quiz because it was 2 questions and this was one of them:

Find all three complex roots of the following equation (give answers in polar and rectangular form)

$$z^3+8=0$$

Looks easy enough,

$$z=2e^{-i\frac{\theta}{3}}$$

This is where I think I completely realized I wasn't sure what I was doing. My roommate suggested I look for the roots of unity which I know that:

$$r^n(cos(n\theta)+isin(n\theta))=1+i*0$$

so if I want to consider mine it should be:

$$8^{1/3}(cos(\frac{\theta}{3})+isin(\frac{\theta}{3})=-8$$

so then

$$\theta=\frac{k2\pi}{3}$$

is this the right track?

Last edited: Sep 13, 2006
2. Sep 13, 2006

### StatusX

I'm not sure what you're trying to do (eg, your second to last latex line doesn't look right). Just keep in mind that there will be a unique solution for r, and then you're left with finding the cube roots of -1 (not 1). If it helps, these are the sixth roots of 1 that aren't also cube roots of 1.