What is the relationship between complex numbers and trigonometric functions?

[Mentallic]

Homework Statement

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}

Homework 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

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.

 

[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.

 

[Mentallic]

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.

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