holomorphic function on the complex plane let $ f(z) \in Hol(\mathbb{C}) $ holomorphic function such that for each $ z_0 \in \mathbb{C} $ there exists $ N(z_0) $ such that $ f^{(N(z_0))}(z_0) = 0 $
Prove: that f is a polynom
 A: Modifying Davide's argument but avoiding Baire's category theorem: Let $\overline{\mathbb E}$ be the closed unit ball and
$$F_n := \{ z \in \overline{\mathbb E} : f^{(n)}(z) = 0 \}.$$
Then, by assumption, $\overline{\mathbb E} = \bigcup_n F_n$. Since $\overline{\mathbb E}$ is uncountable, not all $F_n$ can be finite. Let $n \in \mathbb N$ s.t. $F_n$ is infinite. Since $\overline{\mathbb E}$ is compact, $F_n$ has a limit point in $\overline{\mathbb E}$, i.e. there is a $z_0 \in \overline{\mathbb E}$ and a sequence $(z_k)$ in $F_n \setminus \{z_0\}$ that converges to $z_0$. We have $f^{(n)}(z_k) = 0$ for all $k$, hence $f^{(n)} = 0$ by the identity theorem, so $f$ is a polynomial.
A: Let $F_n:=\{z\in\Bbb C, f^{(n)}(z)=0\}$. Since $f^{(n)}$ is continuous, $F_n$ is closed and by hypothesis, $\Bbb C=\bigcup_{n\in\Bbb N}F_n$. By Baire's theorem, there exists a $n_0$ such that $F_{n_0}$ has a non-empty interior. Hence $f^{(n_0)}$ vanishes on a ball, and it's necessarily the null function. This proves that $f$ is a polynomial.
Note that there is a similar result for $C^{\infty}$ functions from $\Bbb R$ to $\Bbb R$, but it's harder to show. It has been discussed here.
