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Assume $f$ is an entire function such that for each $z\in \mathbb{C}$, we have an $n=n(z)$ such that $f^{(n)}(z)=0$, show that $f$ is a polynomial.

I got stuck in my following argument.

First, we only need to show that we have a uniform $n$ such that $f^{(n)}=0$.

To show this, we only need to show that we have a uniform $n$ such that $f^{(n)}|_{\bar{D}}=0$, where $\bar{D}=\{z||z|\leq 1\}$.

Suppose not, we get a sequence $z_n\in \bar{D}$, such that $f^{(n)}(z_n)\neq 0$, then W.L.O.G., we can assume $z_n\rightarrow z\in \bar{D}$, suppose $f^{(m)}(z)=0$, then ???

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up vote 6 down vote accepted

If $f$ is not a polynomial, then every $f^{(n)}$ is an entire non-constant function, so the set of solutions of $f^{(n)}(z) = 0$ is discrete, hence countable for every $n$. This implies that the union of those sets over all $n$ is still countable, so it can't be the whole plane.

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Thanks! nice solution. – ougao Nov 26 '12 at 1:59

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