# Complex Numbers ~ Factors

1. Mar 18, 2009

### Mentallic

1. The problem statement, all variables and given/known data
Resolve $z^5-1$ into real linear and quadratic factors.

Hence prove that $$cos\frac{2\pi}{5}+cos\frac{4\pi}{5}=-\frac{1}{2}$$

2. Relevant equations
$$z=cis\theta$$

$$z\bar{z}=cis\theta.cis(-\theta)=cos^2\theta+sin^2\theta=1$$

$$z+\bar{z}=cis\theta+cis(-\theta)=2cos\theta$$

3. The attempt at a solution
I was able to show that the the roots of $z^5-1=0$ are:

$$z=1,cis\frac{2\pi}{5},cis\frac{-2\pi}{5},cis\frac{4\pi}{5},cis\frac{-4\pi}{5}$$

And hence, the real factors are:

$$(z-1)(z^2-2z.cos\frac{2\pi}{5}+1)(z^2-2z.cos\frac{4\pi}{5}+1)=0$$

But now I'm stuck and not sure how to start proving that last equation.

2. Mar 18, 2009

### Dick

If you multiply your factored form back out again, then the identity you are trying to prove is the coefficient of z^4. It's also basically the sum of all of the five roots.

Last edited: Mar 18, 2009
3. Mar 18, 2009

### Mentallic

Aha and the coefficient of z4 is 0, so:

$$1+cis\frac{2\pi}{5}+cis\frac{-2\pi}{5}+cis\frac{4\pi}{5}+cis\frac{-4\pi}{5}=0$$

Therefore, $$1+2cos\frac{2\pi}{5}+2cos\frac{4\pi}{5}=0$$

$$cos\frac{2\pi}{5}+cos\frac{4\pi}{5}=\frac{-1}{2}$$

Thanks