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
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)} $.