TITLE: Prove that the zeros of an analytic function are finite and isolated
QUESTION [14 upvotes]: Let us assume that the zeros of $f = \{Z_1,\ldots,Z_n,a\}$ are infinite and converge towards $a$.
The book which I am reading says that any neighborhood of $a$ will contain infinite zeros. Since $f$ is continuous, $f$ must be identically zero in this neighborhood. .....(A)
Then as the neighborhood grows, it will be identically $0$ in that neighborhood too until it engulfs the whole of this region. Hence, the function becomes identically zero in the whole region.
I don't quite understand (A). I am confused over (A) with this analogy. Consider $f(z) = z-1$. This has a zero at $z=1$. Since $f$ is continuous in the $z$ plane, this means that as per (A) points in the neighborhood of $z =1$ should also have mapped value $= 0$.
I am confused; where could I be making a mistake?
Thanks

REPLY [3 votes]: The set of Zeros of an analytic function may be countably infinite. For example $f(z)=\sin\left(\frac1z\right)$ is analytic on $C\setminus\{0\}$. It is true that The set of Zeros of an analytic function at most countably infinite.