Finding Complex Roots: Showing 1 Has Three Cube Roots

[icantadd]

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]

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]

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))?