# If I have a holomorphic function on the unit disc, do I know anything about the radius of convergence of its series expansion about zero?

I'm looking at a proof that assumes only that
$f : \mathbb{D} \rightarrow \mathbb{D}$ is holomorphic with $f(0) = 0$

The first step in the proof is to "expand $f$ in a power series centered at $0$ and convergent in all of $\mathbb{D}$."

In other words, $f(z) = a_0 + a_1 z + a_2 z^2 + ...$ inside the disc. Is this a valid step? How do I know that the function's series expansion is valid in the whole disc?

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I believe this might be useful. –  Tunococ May 31 '13 at 0:01

Following Tunococ's link, I found the following: "the fact that the radius of convergence is always the distance from the center a to the nearest singularity; if there are no singularities (i.e., if ƒ is an entire function), then the radius of convergence is infinite. Strictly speaking, this is not a corollary of the theorem but rather a by-product of the proof."

Since there are no singularities, then the series is valid for the whole disc!

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This is incorrect: you may not assume that $f$ has no singularity. Counter-example: $f(z)=\frac {1}{3}+\frac {1}{z-3}$. –  Georges Elencwajg May 31 '13 at 7:57
Sorry, I meant that there are no singularities within the disc. –  Mark May 31 '13 at 17:04