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Let $f$ be an entire functions. suppose for every $z_0 \in \mathbb{C}$ we have that the power series expansion

$$\sum_{n=0}^\infty a_n(z-z_0)^n$$

has at least one coefficient $a_n=0$. Show that $f$ is a polynomial. I am very lost on where to ever begin. I know this power series converges for balls around $z_0$ but that's about it. Any hints. Could I write it out, like $0$ out every other term, and conclude that the terms need to form a polrynomial. Something like

For each $z_0 \in \mathbb{C}$ I have

$$0+a_1z+a_2z^2+...$$ $$a_0+0+a_2z^2+...$$ $$a_0+a_1z+0+...$$

well centered about $z_0$. Which forms a polynomial.

homosapien
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    this is a classic so should have been answered previously (in the form if for any $z$ there is $n_z \ge 0, f^{(n_z)}(z)=0$ then $f$ is a polynomial) but the idea is to let $F_n$ the closed sets for which that particular derivative (coefficient) is zero and use Baire theorem to show one has non empty interior since their union is the complex plane; from there result follows easily by the identity theorem – Conrad Feb 03 '22 at 00:56
  • woah wouldn't have even thought! thanks!! – homosapien Feb 03 '22 at 00:56

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