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## Every expansion has a 0

Suppose $f$ is entire such that for each $x_0 \in \mathbb{C}$ at least one of the coefficients of the expansion $f=\sum\limits_{n=0}^\infty c_n (z-z_0)^n$ is equal to 0. Prove that $f$ is a polynomial.

Solution: The statement of the problem is equivalent to $\forall z \in \mathbb{C} \exists n \in \mathbb{N}$ such that $f^{(n)}(z)=0$. Therefore, if we denote $A_n=\{ z \in \mathbb{C} : f^{(n)}(z)=0\}$ we have that $A_n$ is closed for any $n$ (because $f$ is entire and $f^{(n)}$ is continuous ) and $\displaystyle \bigcup_{n=0}^\infty A_n=\mathbb{C}$. Since $\mathbb{C}$ is a complete metric space, by Baire’s Category Theorem, one of the sets $A_n$ has non-void interior. This means that there exists $n \in \mathbb{N}$ such that $f^{(n)}(z)=0$ for $z$ in an open disk. This means that $f$ is a polynomial in that disk, and furthermore, by analytic continuation, $f$ must be a polynomial everywhere.