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## Homework Statement

This is a claim from a Wikipedia page about analytic functions (http://en.wikipedia.org/wiki/Analytic_function), and I can't seem to prove it.

If [tex](r_{n})[/tex] is a sequence of distinct numbers such that [tex]f(r_{n}) = 0[/tex] for all n and this sequence converges to a point r in the domain of D, then f is identically zero on the connected component of D containing r.

## Homework Equations

[tex]f(x) = \sum a_{n}x^{n}[/tex]

[tex]f(x_{n})=0[/tex]

## The Attempt at a Solution

I'm trying to prove the case when r=0. If f is analytic, then it can be represented by some power series. If you look at some sequence [tex](r_n) = 1/n[/tex], then the zeros of f bunch up near 0. It is clear to me that f(0)=0 (since f is continuous), so [tex]a_{0}=0[/tex]. It is also clear to me that f '(0)=0, and so [tex]a_{1}=0[/tex]. It seems like the way to go is to show (using induction) that each [tex]a_{n}[/tex] is 0 by showing that all derivatives at 0 are equal to 0.

Thoughts?