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

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.


Suppose $\Omega$ is a region, the set

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

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:

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 $N$.


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 $D(a;r)$:


It’s clear that $g \in H(D(a;r))$ since we have

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

By hypothesis, $0 \in \Gamma$. We shall prove that $1 \in \Gamma$. Let

There exists a sequence $\{t_n\}\subset\Gamma$ such that $t_n \to s$. The continuity of $f^{(k)}$ and $\gamma$ shows that

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.