# Zeros of Analytic Functions

A strange property of analytic functions is that the zeros are isolated. I don’t remember the proof I originally learned of this fact, but today I saw a really interesting topological way to do it. It makes sense now.

More precise formulation: If $\Omega\subset\mathbb{C}$ is a connected open set, then $\{z: f(z)=0\}$ consists of isolated points if $f$ is analytic on $\Omega$. (Oops, I started writing this up and realized that I need to trivially throw out the case where $f\equiv 0$.

Proof: Let $U_1=\{a\in\Omega : \exists\delta>0, \ f(z)\neq 0 \ on \ 0<|z-a|<\delta\}$ and let $U_2=\{a\in\Omega : \exists\delta>0, \ f(z)\equiv 0 \ on \ 0<|z-a|<\delta \}$. Reformulating the setup we see that $U_1$ means: if f has a zero, it is isolated since f is nonzero on a punctured disk (meaning the zero must be the punctured part). Also $U_2$ is just the regions that f has non-isolated zeros.

It is straightforward to check that both $U_1$ and $U_2$ are open (just choose $\delta$‘s sufficiently small to stay inside the declared sets). Also we have that $U_1\cap U_2=\emptyset$ and I now claim $\Omega=U_1\cup U_2$.

This seems obvious, but should be pinned down in some sort of argument. Let $z_0\in\Omega$. We claim that there is a punctured disk about $z_0$ such that either $f\equiv 0$ on the disk or $f\neq 0$ anywhere on the disk. By analyticity, we have a power series convergent on some radius $r>0$ about $z_0$, i.e. $f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n$ on $|z-z_0|.

Suppose $a_k$ is the first nonzero coefficient (by not being equivalently zero, this must exist). Then $f(z)=\sum_{n=0}^\infty a_{n+k}(z-z_0)^n=(z-z_0)^{-k}\sum_{n=k}a_n(z-z_0)^n$. So since the series converges in $0<|z-z_0| and since $f$ is continuous we can choose $0<\delta small enough so that $|f(z)-f(z_0)|=|f(z)-a_k|<\frac{|a_k|}{2}$. This clearly shows that $f(z)\neq 0$ on $0<|z-z_0|<\delta$ else we’d have $|a_k|<\frac{|a_k|}{2}$. So either there is a punctured disk on which f is non-zero, or the f has no first non-zero coefficient making it zero everywhere on that first disk $|z-z_0| proving the claim.

The properties $U_1\cap U_2=\empty$ and $\Omega=U_1\cup U_2$ (along with both sets being open) combine to give that either $U_1=\emptyset$ or $U_2=\emptyset$ by the connectedness of $\Omega$. This simply means that all the zeros are isolated since we ruled out the alternative of being equivalently zero.

This goes to show how remarkably different analytic on $\mathbb{C}$ is to continuous on $\mathbb{R}$. In fact, even infinitely differentiable functions on $\mathbb{R}$. Bump functions play a crucial role in many areas of analysis and they are smooth functions with compact support meaning that outside of a bounded they are zero. An entire class of important functions violates this property that analytic functions are guaranteed to have.