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.

Write $\zeta= \xi + \dfrac{1}{\xi} = \xi + \xi^4$, because $\xi^5=1$. Then 
$$\zeta^2 + \zeta = \xi^2 + 2\xi \xi^4 + \xi^8 + \xi + \xi^4 = \xi^2 + 2 + \xi^3 + \xi + \xi^4 = 1 + 1 + \xi + \xi^2 + \xi^3 + \xi^4 = 1
$$
because $1 + \xi + \xi^2 + \xi^3 + \xi^4 = 0$, since $0=\xi^5-1=(\xi-1)(1 + \xi + \xi^2 + \xi^3 + \xi^4)$ and $\xi \ne 1$.
