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 $1+g(w_k-z_0) \neq 0$.
Best Answer
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.