Showing a Chebyshev set

Question

I want to show that $\{1,e^{ix},...,e^{(n-1)x} \}$ is a Chebyshev Set on $(0,2\pi]$. Now I know that $\{1,x,...,x^n \}$ is one and that $e^{ix}$ is injective on $(0,2\pi]$. But how do I show that if I have a map $$ m:(0,2\pi] \rightarrow \mathbb{C} , x\rightarrow e^{ix} $$ that the Polynomial $$ a_0+a_1\cdot e^x + ... + a_n \cdot e^{ix^n} $$ has max. the same number of zeroes than $$ a_0+a_1\cdot x + ... + a_n \cdot x^n $$?

Answer 1

A non-zero polynomial $\sum_{k=0}^n a_k z^k$ has at most $n$ roots on any subset of $z\in \mathbb C$ including the circle $z=e^{ix}, x\in(0,2\pi]$ corresponding to at most $n$ distinct values of $x$ that produce such roots. That is, the polynomial $\sum_{k=0}^n a_k e^{ixk}$ has at most $n$ roots on the interval $x\in(0,2\pi]$.