3.1 Zeros

Definition 3.1

Suppose \(f\) is analytic in at \(z_0\in \mathbb {C}\). Set

\[ \operatorname{\mathrm{ord}}_{z_0} f := \sup \{ i\in \mathbb {N}: j {\lt} i\Rightarrow f^{(j)}(z_0) = 0\} . \]

We call \(\operatorname{\mathrm{ord}}_{z_0} f\) the order of vanishing of \(f\) at \(z_0\), and, we say that \(f\) has a zero of order \(\operatorname{\mathrm{ord}}_{z_0} f\) at \(z_0\).

Note that \(\operatorname{\mathrm{ord}}_{z_0} f = 0\Leftrightarrow f(z_0)\neq 0\). Suppose \(f = \sum _{i=0}^{\infty } a_i (z - z_0)^i\) is the Taylor series of \(f\) at \(z_0\). Then, since \(a_i = f^{(i)}(z_0)/i!\),

\[ \operatorname{\mathrm{ord}}_{z_0} f = \sup \{ i\in \mathbb {N}: j {\lt} i\Rightarrow a_j = 0\} . \]

If \(S\) is an open set containing the domain of two functions \(f\) and \(g\), let’s write “\(f\equiv g\) on \(S\)” to mean that \(f(z) = g(z)\) for all \(z\in S\). In this case, we say that \(f\) is identically equal to \(g\) on \(S\). We say \(f\) is identically zero on \(S\) if \(f\equiv 0\).

Lemma 3.2

Suppose \(z_0\in \mathbb {C}\), \(r {\gt} 0\), and \(f\) is analytic on \(D := D(z_0,r)\). Then

\(\operatorname{\mathrm{ord}}_{z_0} f = +\infty \) if and only if \(f(z) =0\) for all \(z\in D\). In other words, \(\operatorname{\mathrm{ord}}_{z_0} f = +\infty \) if and only if \(f\equiv 0\) on \(D\).
\(\operatorname{\mathrm{ord}}_{z_0} f = m {\lt} \infty \) if and only if there exists \(g\in A(D)\) with \(g(z_0)\neq 0\) such that \(f(z) = (z-z_0)^m g(z)\) for all \(z\in D\).

Proof ▼

(a) Suppose \(\operatorname{\mathrm{ord}}_{z_0} f = +\infty \). Since \(\operatorname{\mathrm{ord}}_{z_0} f = +\infty \), every term in the Tayor series of \(f\) at \(z_0\) is \(0\). So, since the Taylor series of \(f\) converges to \(f\) on \(D\), \(f=0\) on \(D\). The converse is obvious.

(b) Suppose \(\operatorname{\mathrm{ord}}_{z_0} f = m {\lt}+\infty \). Then the Taylor series,

\begin{equation} \label{etup} \sum _{n\geq m}^{\infty } a_n (z-z_0)^n, \end{equation}

3.3

of \(f\) at \(z_0\) converges to \(f\) on \(D\). Since 3.3 converges on \(D\), the sequence

\begin{equation} \label{etdiv} \sum _{n\geq m}^{\infty } a_n (z-z_0)^{n-m} \end{equation}

3.4

obtained by dividing each of the terms in 3.3 by \((z-z_0)^m\) converges on \(D\setminus \{ z_0\} \). But 3.4 obviously converges at \(z = z_0\) as well. So 3.4 converges on \(D\), and, as it is a power series, it converges to an analytic function \(g\) on \(D\). Moreover, 3.4 is the Taylor series of \(g\). So \(g(z_0) = a_m\neq 0\) and \(f(z) = (z-z_0)^m g(z)\) for all \(z\in D\). Again, the converse is obvious.

Corollary 3.5

Suppose \(z_0\in \mathbb {C}\), \(r {\gt} 0\), \(D = D(z_0, r)\) and \(f\in A(D)\). Suppose further that \(f\) is not identically \(0\) on \(D\). Then there exists \(\epsilon {\gt} 0\) such that

\begin{equation} \label{nzp} 0 < |z - z_0| < \epsilon \Rightarrow f(z) \neq 0. \end{equation}

3.6

Proof ▼

Suppose \(f\) is not identically \(0\) on \(D\). Then, by Lemma 3.2, \(\operatorname{\mathrm{ord}}_{z_0} f = m {\lt} \infty \), and we can find \(g\in A(D)\) with \(g(z_0)\neq 0\) such that \(f(z) = (z-z_0)^m g(z)\) for all \(z\in D\). As \(g(z_0)\neq 0\) and \(g\), being analytic, is continuous, we can find \(\epsilon {\gt} 0\), such that

\begin{equation} \label{nzg} |z - z_0| <\epsilon \Rightarrow g(z) \neq 0. \end{equation}

3.7

But then 3.7 directly implies 3.6.

Definition 3.8

Suppose \(S\subseteq \mathbb {C}\). We say \(z_0\in \mathbb {C}\) is a limit point of \(S\) if every punctured neighborhood of \(z_0\) contains a point of \(S\).

In other words, \(z_0\) is a limit point of \(S\) if, for every \(\epsilon {\gt} 0\), there exists \(z\in S\) such that \(0 {\lt} |z - z_0| {\lt} \epsilon \).

Theorem 3.9

Suppose \(D\subseteq \mathbb {C}\) is a domain and \(f\in A(D)\). If \(D\) contains a limit point of the set \(Z := \{ z\in D: f(z) = 0\} \) of zeros of \(f\), then \(D = Z\).

Proof ▼

Suppose \(z_0\in D\) is a limit point of \(Z\). Since \(D\) is open, we can find find an \(r {\gt} 0\) such that \(D(z_0,r)\subseteq D\). Then, for every \(\epsilon {\gt} 0\), there exists a \(z\in D\) such that \(0 {\lt} |z - z_0| {\lt} \epsilon \) and \(f(z) = 0\). It follows from Corollary 3.5 that \(f\) is identically \(0\) on \(D\). Thus, \(\operatorname{\mathrm{ord}}_{z_0} f = +\infty \).

Now, define a function \(\varphi :D\to \mathbb {C}\) by setting

\[ \varphi (z) = \begin{cases} 0, & \operatorname{\mathrm{ord}}_{z} f = +\infty ;\\ 1, & \text{else.} \end{cases} \]

I claim that \(\varphi \) is locally constant.

To see this, suppose \(\varphi (z) = +\infty \). Then, since \(D\) is open, there exists \(r {\gt} 0\) such that \(D(z,r) \subseteq D\). And, by Lemma 3.2, \(f\) is \(0\) on \(D(z,r)\). Therefore, \(\varphi (z) = 0\) on \(D(z,r)\) as well.

On the other hand, if \(\varphi (z) {\lt} \infty \), then \(\operatorname{\mathrm{ord}}_z f = m {\lt} +\infty \). So, by Corollary 3.5, there exists \(\epsilon {\gt} 0\), such that \(0 {\lt} |w-z| {\lt} \epsilon \Rightarrow f(w) \neq 0\). Thus \(\operatorname{\mathrm{ord}}_w f = 0\) for all \(w\) with \(0{\lt} |w-z|{\lt}\epsilon \). Hence, \(\varphi (w) = 0\) for all \(w\) with \(|w-z|{\lt}\epsilon \).

This proves the claim that \(\varphi \) is locally constant. Thus, as \(D\) is a domain, \(\varphi \) is constant. So, as \(\varphi (z_0) = 0\), \(\varphi (z) = 0\) for all \(z\in D\). Thus, \(f\equiv 0\) on \(D\).