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)

Help please :)

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- Nov 28th 2012, 06:07 AMImConfusedFactoring 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)

Help please :) - Nov 28th 2012, 06:38 AMPlatoRe: Factoring z^5-1

This amounts to finding the five fifth roots of one.

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

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

Thus $\displaystyle z^5 - 1 = \prod\limits_{k = 0}^4 {\left( {z - \rho _k } \right)} $. - Nov 28th 2012, 08:42 AMzhandeleRe: 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.