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Suppose f is analytic function defined everywhere in $\mathbb C$ and such $z_0\in \mathbb C$ at least one coefficent in expansion
$f(z)=\sum^\infty_{n=0}c_n(z-z_0)^n$ is equal to $0$. Then prove that $f$ is polynomial

I understood that for each $z\in \mathbb C $ there exist some $n\in N$ such that $f^{(n)}(z)=0$
There is hint available for this problem says that use countability argument .
I know that zero of holomorphic function is at most countable by Identity theorem.
Here we get each $z$ some $n$. It does not say about for all $n$ some $n$ exist .
I am not able to use hint.
Any Help will be appreciated

José Carlos Santos
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Curious student
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1 Answers1

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Consider $A_n=\{z:f^{(n)}(z)=0\}$. The union of this sequence of sets is $\mathbb C$. Since $\mathbb C$ is uncountable at least one of the sets $A_n$, say $A_k$ is uncountable. This implies that $A_k$ has a limit point, because any set with no limit points is at most countable. But then $f^{(k)}$ is an entire function whose zeros have a limit point implying that it is the zero function. It follows that $f$ is a polynomial.

Kavi Rama Murthy
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