I've been trying to prove the following: A trigonometric polynomial $P$ with degree $N>0$ on $\mathbb{T}$ has at most $2N$ zeroes.
So if $P(x)=\sum_{n=-N}^{N}c_ne^{inx}$, my idea was to somehow use that $\cos(nx)$ and $\sin(nx)$ have $2n$ roots in order to do this, but I'm kinda stuck on the way to write it down. Am I actually even allowed to use it like that, or should I prove that first? The more I think about it the less it wants to make sense though, since I can't write it down properly. I've read on Wikipedia that this was in a Numerical Analysis book form Powell (1981), haven't been able to access it though.
You can think of this as a complex function
$$P(z) = \sum_{n = -N}^N c_n z^n$$
on the circle, where $z = e^{ix}$. On the other hand, we can think of this as the restriction of a function on $\mathbb{C}$ with the same definition. Now $z^NP(z)$ is a complex polynomial of degree $\le 2N$, and therefore has at most $2N$ zeros on all of $\mathbb{C}$, some of which may be in $\mathbb{T}$. Dividing by $z^N$ gives the claim.