# Complex root finding

## Homework Statement

let the complex number 1 = 1 + 0i. Show this number has three cube roots. Use any means to find them.

## Homework Equations

not sure. cubic root x means there is some number y, x = y*y*y.

## The Attempt at a Solution

Well one root has to be 1
Another is (-1)*i^2 and another 1*i^2. I am not sure about these two, and I am not sure I could give an explanation why even if they are right.

Dick
Homework Helper
Those aren't right. Do you know deMoivre's formula? If you don't want to use that you are trying to solve x^3-1=0 for x. Factor it.

HallsofIvy
If you are expected to do a problem like this then you should already know "DeMoivre's formula": if $z= r(cos(\theta)+ i sin(\theta))$ then $z^c= r^c(cos(\theta)+ i sin(\theta))$ which is true for c any number. In particular $z^{1/n}= r^{1/n}(cos(\theta/n)+ i sin(\theta/n))$ where $r^{1/n}$ is the unique real positive nth root of r. Since cosine and sine are periodic with period n, increasing $\theta$ by any multiple of $2\pi$ won't change $r(cos(\theta)+ i sin(\theta))$ but dividing by n will.
In particular, $1= 1+ 0i= 1(cos(0)+ i sin(0))= 1(cos(2\pi)+ i sin(\2pi))= 1(cos(4\pi)+ i sin(4\pi))$. Find the 3 third roots of 1 by dividing those angles by 3. For example, 1(cos(0/3)+ i sin(0/3))= 1 which gives the first third root of 1. Now, what are $1(cos(2\pi/3)+ i sin(2\pi/3)$ and $1(cos(4\pi/3)+ i sin(\4pi/3))$?