# Math Help - Cube root of a complex number

1. ## Cube root of a complex number

Just trying to confirm something if possible:

a) Complex numbers have 3 cube roots.

b) The usual method for finding them is to write in polar form, then add 2 pi and 4 pi to the angle and use De Moivre's theorem on the resulting three expressions.

Cheers

2. Originally Posted by DangerousDave
Just trying to confirm something if possible:

a) Complex numbers have 3 cube roots.

b) The usual method for finding them is to write in polar form, then add 2 pi and 4 pi to the angle and use De Moivre's theorem on the resulting three expressions.

Cheers
You are on the right track. They do have cube roots, find them using

$z^{\frac{1}{n}} = r^{\frac{1}{n}}\times \text{cis}\left( \frac{\theta+2k\pi}{n}\right), k= 0,1,\dots,n-1$