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Let $f$ be an entire function. Consider $A=\{z \in \Bbb{C} : f^{(n)}(z)=0\; \text{for some}\; n \in \Bbb{N}\}$. Then how to prove if $A=\Bbb{C}$, then $f$ is a polynomial ?

This is same as proving if $f$ is not a polynomial then $A$ is not all of $\Bbb{C}$.

I show the above statement with a particular example, like $f(z)=\sin z$

How to prove generally ? Any ideas ?

JasonM
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    $A=\bigcup_n A_n$, with $A_n={z:\ f^{(n)}(z)=0}$. Since $\mathbb{C}$ is uncountable, then at least one of the $A_n$ is uncountable. Such $A_n$ must accumulate at some point different from $\infty$. Therefore, $f^{(n)}$ is constant equal to zero. –  May 22 '18 at 02:09
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    @arugula I think yours is the best solution so far as it doesn't rely on Baire. Why not write it as an answer. – zhw. May 22 '18 at 02:39
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    Another one: https://math.stackexchange.com/q/1153578/42969 – Martin R May 22 '18 at 09:47

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Suppose $A=\mathbb{C}$, we want to show that $f$ is a polynomial. Let $X_n=\{x\in \mathbb{C}: f^{(n)}(x)=0\}$. If we know one of the $X_n$ contains a open ball then by the property of entire function $f^{(n)}=0$ everywhere. Now assume $X_n$ are nowhere dense so $\mathbb{C}=\cup_{i=0}^{\infty}X_i$ but that's not possible by Baire category theorem.

Ben
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