# Complex root finding

1. Jan 9, 2009

1. The problem statement, all variables and given/known data

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

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

3. 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.

2. Jan 9, 2009

### Dick

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.

3. Jan 9, 2009

### HallsofIvy

Staff Emeritus
i2= -1 so (-1)i2= (-1)(-1)= 1. That's not a new root. 1(i2)= -1 and (-1)3= -1, not 1 so that is NOT a root.

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 $1(cos(4\pi/3)+ i sin(\4pi/3))$?