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Let $G$ be an open and connected set of the complex plane $\mathbb{C}$ and $f:G\rightarrow \mathbb{C}$ be analytic, such that for each $z\in G$ there is $n=n(z)\in \mathbb{N}$ such that $f^{(n)}(z)=0$. Prove that $f$ is a polynomial.

Attempt It seemed to me like a Baire-type application, since $\displaystyle G=\bigcup_{n=1}^{\infty}\{z\in G:~f^{(n)}(z)=0\}$, but there is no completeness on $G$ to work it that way.

Edit. Motivated by Entire function with vanishing derivatives?, let $n:G\rightarrow \mathbb{C}:~n(z)=\min\{k:~f^{(k)}(z)=0\}$ which is well defined and $\displaystyle G=\displaystyle \bigcup_{k=1}^{\infty}n^{-1}(\{k\}).$ Since $G$ is open, it is uncountable and at least one $n^{-1}(\{k\})$ is uncountable, so posseses an accumulation point $z_0 (\in G?)$. Therefore by the identity theorem $f^{(k)}$ is zero on $G$ and $f$ is a polynomial. My question is why $z_0\in G$?

Thanks a lot in advance for the help!

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Nikolaos Skout
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    This is a well-known problem; see here. (One small change is that any analytic function that's a polynomial on a disk is a polynomial everywhere.) – anomaly May 18 '16 at 14:57
  • Re your edit: Note that http://math.stackexchange.com/a/995129/42969 defines the function $n$ on the compact interval $[0, 1]$. In your case, an arbitrary line segment in $G$ should work. As a compact set, it contains all accumulation points. – Martin R May 18 '16 at 15:22
  • @Martin R Thanks for the comment. Why is there an issue to work on the whole $G$ - or do you mean that a line segment is enough (in order for the compactness argument to work)? – Nikolaos Skout May 18 '16 at 15:34
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    A line segment is sufficient. – As I understand it (I am not an expert in topology) you can also apply the Baire Category Theorem (as in this answer http://math.stackexchange.com/a/995120/42969), either to a compact subset of $G$, or because BCT is valid in a locally compact Hausdorff space. – Martin R May 18 '16 at 18:02
  • Thanks a lot for the explanation! – Nikolaos Skout May 18 '16 at 18:03
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    An uncountable subset of a domain has an accumulation point in that domain, by the same argument that is used to show the existence of such a point in $\mathbb{C}$. –  May 18 '16 at 23:26

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