Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$ Let $\xi$ be the fifth primitive root of $1$ an let $\zeta= \xi + \frac{1}{\xi}$. Prove that $\zeta^2 + \zeta =1$
In my attempt I used $\xi = \cos(72k) + i\sin(72k)$ with $k=1,2,3,4$
Where I got $\zeta = \frac{1+\cos(144k)+i\sin(144k)}{\cos(72k)+i\sin(72k)}$
Which yielded $\zeta^2 + \zeta = 2\cos(72k)(2\cos(72k)+1)$
I don’t know what else I could do.

By definition, $\xi^5 = 1$ and we have the factorization $$\prod_{k=0}^4 (z - \xi^k) = z^5 - 1.$$  But since $(z - \xi^0) = (z - 1)$, it follows that $$\prod_{k=1}^4 (z - \xi^k) = \frac{z^5 - 1}{z-1} = 1 + z + z^2 + z^3 + z^4,$$ hence $$1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0.$$  Dividing by $\xi^2$ yields $$\begin{align*}0 
&= \xi^{-2} + \xi^{-1} + 1 + \xi + \xi^2 \\
&= (\xi^2 + 2 + \xi^{-2}) + (\xi + \xi^{-1}) - 1 \\
&= (\xi + \xi^{-1})^2 + (\xi + \xi^{-1}) - 1 \\
&= \zeta^2 + \zeta - 1. \end{align*}$$
