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I am reviewing some complex analysis and I have just gotten to the portion on analytic continuation. My question is about the proof of the following theorem:

Theorem. Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. The $f$ is identically zero.

Proof. Suppose $z_0 \in \Omega$ is a limit point for the sequence $\{w_k\}$ and $f(w_k) = 0$. Chose a disc $D \subset \Omega$ centered on $z_0$, and consider the power series expansion of $f$ in $D$ $$f(z) = \sum a_n(z-z_0)^n$$ If $f$ is not identically 0, there exists a smallest $m$ such that $a_m\neq 0$. Then, $f$ can be rewritten as $$f(z) = a_m(z-z_0)^m(1+g(z-z_0))$$ where $g(z-z_0) \to 0$ as $z \to z_0$. Taking $z = w_k$, we obtain a contradiction since $a_m(w_k-z_0)^m \neq 0$ and $g(w_k-z_0) \neq 0$, but $f(w_k) = 0$ ... $\blacksquare$

The proof concludes the argument by using the connectedness of $\Omega$. However, my question is why must $1+g(w_k-z_0) \neq 0$.

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  • $\begingroup$ I would have just written $f(z) = (z-z_0)^mh(z),$ where $h$ is continuous and $h(z_0)\ne 0.$ $\endgroup$ – zhw. Jul 8 '17 at 16:13
  • $\begingroup$ That is an alternative of course. I was just going through the proof in Stein and Shakarchi's Complex Analysis book where it was written this way. $\endgroup$ – Jonathan Davidson Jul 8 '17 at 20:16
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I think there is a typo here, it should be $1+g(w_k-z_0)\neq 0$. This is true because $g(z-z_0)\to 0$ as $z\to z_0$.

Then we have $$0 = f(w_k) = a_m(w_k-z_0)^m(1+g(w_k-z_0))$$ for sufficiently large $k$, all the terms on the right are not zero, so a contradiction.

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  • $\begingroup$ Oh, so if i choose $w_k$ close enough to $z_0$, that term should be really close to 1 by continuity of $g$. Thanks!!! $\endgroup$ – Jonathan Davidson Jul 8 '17 at 5:02

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