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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.

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