# Topological properties of the zeros of a holomorphic function

### What's going on

If for every \(z_0 \in \Omega\) where \(\Omega\) is a plane open set, the limit \[
\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}
\] exists, we say that \(f\) is holomorphic (a.k.a. analytic) in \(\Omega\). If \(f\) is holomorphic in the whole plane, it's called **entire**. The class of all holomorphic functions (denoted by \(H(\Omega)\)) has many interesting properties. For example it does form a ring.

But what happens if we talk about the points where \(f\) is equal to \(0\)? Is it possible to find an entire function \(g\) such that \(g(z)=0\) if and only if \(z\) is on the unit circle? The topological property we will discuss in this post answers this question negatively.

### Zeros

Suppose \(\Omega\) is a region, the set \[ Z(f)=\\{z_0\in\Omega:f(z_0)=0\\} \] is a at most countable set without limit point, as long as \(f\) is not identically equal to \(0\) on \(\Omega\).

Trivially, if \(f(\Omega)=\\{0\\}\), we have \(Z(f)=\Omega\). The set of unit circle is not at most countable and every point is a limit point. Hence if an entire function is equal to \(0\) on the unit circle, then the function equals to \(0\) on the whole plane.

Note: the connectivity of \(\Omega\) is important. For example, for two disjoint open sets \(\Omega_0\) and \(\Omega_1\), define \(f(z)=0\) on \(\Omega_0\) and \(f(z)=1\) on \(\Omega_1\), then everything fails.

### A simple application (Feat. Baire Category Theorem)

Before establishing the proof, let's see what we can do using this result.

Suppose that \(f\) is an entire function, and that in every power series \[ f(z)=\sum c_n(z-a)^n \] has at leat one coefficient is \(0\), then \(f\) is a polynomial.

Clearly we have \(n!c_n=f^{(n)}(a)\), thus for **every** \(a \in \mathbb{C}\), we can find a postivie integer \(n_0\) such that \(f^{(n_0)}(a)=0\). Thus we establish the identity: \[
\bigcup_{n=0}^{\infty} Z(f^{(n)})=\mathbb{C}
\] Notice the fact that \(f^{(n)}\) is entire. So \(Z(f^{n})\) is either an at most countable set without limit point, or simply equal to \(\mathbb{C}\). If there exists a number \(N\) such that \(Z(f^{N})=\mathbb{C}\), then naturally \(Z(f^{n})=\mathbb{C}\) holds for all \(n \geq N\). Whilst we see that \(f\)'s power series has finitely many nonzero coefficients, thus polynomial.

So the question is, is this \(N\) always exist? Being an at most countable set without limit points , \(Z(f^{(n)})\) has empty interior (nowhere dense). But according to **Baire Category Theorem**, \(\mathbb{C}\) could not be a countable union of nowhere dense sets (of the first category if you say so). This forces the existence of \(N\).

### Proof

The proof will be finished using some basic topology techniques.

Let \(A\) be the set of all limit points of \(Z(f)\) in \(\Omega\). The continuity of \(f\) shows that \(A \subset Z(f)\). We'll show that if \(A \neq \varnothing\), then \(Z(f)=\Omega\).

First we claim that if \(a \in A\), then \(a \in \bigcap_{n \geq 0}Z(f^{(n)})\). That is, \(f^{(k)}(a) = 0\) for all \(k \geq 0\). Suppose this fails, then there is a smallest positive integer \(m\) such that \(c_m \neq 0\) for the power series on the disc \(D(a;r)\): \[ f(z)=\sum_{n=1}^{\infty}c_n(z-a)^{n}. \]

Define

\[ \begin{aligned} g(z)=\begin{cases} (z-a)^{-m}f(z)\quad&(z\in\Omega-\\{a\\}) \\\ c_m\quad&(z=a) \end{cases} \end{aligned} \]

It's clear that \(g \in H(D(a;r))\) since we have \[ g(z)=\sum_{n=1}^{\infty}c_{m+n}(z-a)^{n}\quad(z\in D(a;r)) \]

But the continuity shows that \(g(a)=0\) while \(c_m \neq 0\). A contradiction.

Next fix a point \(b \in \Omega\). Choose a curve (continuous mapping) defined \(\gamma\) on \([0,1]\) such that \(\gamma(0)=a\) and \(\gamma(1)=b\). Let

\[ \Gamma=\\{t\in[0,1]:\gamma(t)\in\bigcap_{n \geq 0}Z(f^{(n)})\\} \] By hypothesis, \(0 \in \Gamma\). We shall prove that \(1 \in \Gamma\). Let \[ s = \sup\Gamma \] There exists a sequence \(\\{t_n\\}\subset\Gamma\) such that \(t_n \to s\). The continuity of \(f^{(k)}\) and \(\gamma\) shows that \[ f^{(k)}(\gamma(s))=0 \]

Hence \(s \in \Gamma\). Choose a disc \(D(\gamma(s);\delta)\subset\Omega\). On this disc, \(f\) is represented by its power series but all coefficients are \(0\). It follows that \(f(z)=0\) for all \(z \in D(\gamma(s);\delta)\). Further, \(f^{(k)}(z)=0\) for all \(z \subset D(\gamma(s);\delta)\) for all \(k \geq 0\). Therefore by the continuity of \(\gamma\), there exists \(\varepsilon>0\) such that \(\gamma(s-\varepsilon,s+\varepsilon)\subset D(\gamma(s);\delta)\), which implies that \((s-\varepsilon, s+\varepsilon)\cap[0,1]\subset\Gamma\). Since \(s=\sup\Gamma\), we have \(s=1\), therefore \(1 \in \Gamma\).

So far we showed that \(\Omega = \bigcap_{n \geq 0}Z(f^{(n)})\), which forces \(Z(f)=\Omega\). This happens when \(Z(f)\) contains limit points, which is equivalent to what we shall prove.

When \(Z(f)\) contains no limit point, all points of \(Z(f)\) are isolated points; hence in each compact subset of \(\Omega\), there are at most finitely many points in \(Z(f)\). Since \(\Omega\) is \(\sigma\)-compact, \(Z(f)\) is at most countable. \(Z(f)\) is also called a discrete set in this situation.