Proof that for every $z \in \mathbb C$: $z=\frac{2}{n}\sum_{k=0}^{n-1}(Re(u_{k}\overline z)u_{k})$

Question

Let $n \in \mathbb N$, $n>2$ and $u_{1},u_{2},...,u_{n-1}$ are all root of unity for n.
Proof that for every $z \in \mathbb C$: $$z=\frac{2}{n}\sum_{k=0}^{n-1}(Re(u_{k}\overline z)u_{k})$$

I know that $Re(u_{0})=Re(u_{1})$, $Re(u_{2})=Re(u_{3})$ etc. However then I have a problem with $n$ because I don't know anything about parity of $n$. Moreover I don't know what to use this idea even if I consider two cases: $n=2k$ and $n=2k+1$.
I thought also about $\sum _{{k=0}}^{{n-1}}e^{{\frac {2\pi ik}{n}}}=0$ but I still do not know how to do it.

Are you have any idea?

Answer 1

Note that $$\frac{2}{n}\sum_{k=0}^{n-1}(Re(u_{k}\overline z)u_{k}) = \frac{2}{n}\sum_{k=0}^{n-1}\frac{u_{k}\overline z+\overline{u_{k}}z}{2}u_{k} =\frac 1n (\overline z\sum_{k=0}^{n-1}u_k^2+z\sum_{k=0}^{n-1}|u_k|^2)$$

Using a geometric argument, a geometric sum, or Newton's identities it's easy to show $\sum_{k=0}^{n-1}u_k^2=0$. Besides, $|u_k|=1$, so the result follows.