# Factoring z^5-1

• November 28th 2012, 07:07 AM
ImConfused
Factoring z^5-1
I have be asked to factorize z^5-1

Attempt at a solution:

z-1 is a factor because 1^5-1=0

So (z-1)(z^4+z^3+z^2+z+1)

I'm not sure how to factor (z^4+z^3+z^2+z+1)

• November 28th 2012, 07:38 AM
Plato
Re: Factoring z^5-1
Quote:

Originally Posted by ImConfused
I have be asked to factorize z^5-1

This amounts to finding the five fifth roots of one.

Let's agree that $r\exp(i\theta)=r(\cos(\theta)=i\sin(\theta)$.

Then $\rho _k = \exp \left( {\frac{{2k\pi }}{5}i} \right),\;k = 0,1,2,3,4$ are the roots.

Thus $z^5 - 1 = \prod\limits_{k = 0}^4 {\left( {z - \rho _k } \right)}$.
• November 28th 2012, 09:42 AM
zhandele
Re: Factoring z^5-1
You might want to locate the five roots on the unit circle. Remember DeMoivre's theorem? That might help you get a more intuitive grasp of it.