Let $U$ be some non-empty open subset of $\mathbb C$. Suppose I have an holomorphic function $f$ defined by the series $f(z)=\sum_{n=0}^\infty a_ne^{2\pi i nz}$ where $a_n\in\mathbb C $ for every $n$ and $z\in U$.
Now say that $f(z)=0$ for every $z$. Is it true that we have $a_n=0$ for every $n$?
I ask this since we know that if we take a finite set of exponentials then they are $\mathbb C$-linearly independent functions, but passing to the infinite case seems not so clear to me.
On
Suppose the series converges to $0$ on $U$. Then in particular the terms are uniformly bounded on $U$. Since $\left|e^{2\pi i n z}\right| = e^{-2\pi n \text{Im}(z)}$ for $n \ge 0$, that says that if there is $z \in U$ with $\text{Im}(z) = r$, there is a constant $C$ such that $|a_n| \le C e^{2\pi n r}$ for all $n$. Using this, we find that the series converges uniformly on the half-plane $H_s = \{z: \text{Im}(z) \ge s\}$ for all $s > r$, and the sum $f(z)$ is analytic there. By the identity theorem for analytic functions, $f(z) = 0$ on all of $H_s$.
Now suppose $a_m$ is the first nonzero term. For $\text{Im}(z)$ sufficiently large, easy estimates give $\left| \sum_{n=m+1}^\infty a_n e^{2\pi i n z} \right| < \left|a_m e^{2\pi i m z}\right|$, implying that $f(z) \ne 0$, contradiction.
Let $V=\{e^{2\pi i z} : z \in U\}$. Then V is an open set (by Open mapping Theorem). Let $g:V \to \mathbb C$ be defined by $g(z)=\sum_0 ^{\infty}a_nz^{n}$. Then g is the sum of a convergent power series in V and $f(z)=g(e^{2\pi i z})$ If f=0 on U then g=0 on V and hence the coefficients in the power series expansion are all 0.