Skip to main content
Logo image

A First Course in Complex Analysis

Section 8.2 Classification of Zeros and the Identity Principle

When we proved the Fundamental Theorem of Algebra (Theorem 5.3.2; see also Exercise 5.4.11), we remarked that, if a polynomial \(p(z)\) of degree \(d>0\) has a zero at \(a\) (that is, \(p(a) = 0\)), then \(p(z)\) has \(z-a\) as a factor. That is, we can write \(p(z)=(z-a) \, q(z)\) where \(q(z)\) is a polynomial of degree \(d-1\text{.}\) We can then ask whether \(q(z)\) itself has a zero at \(a\) and, if so, we can factor out another \((z-a)\text{.}\) Continuing in this way, we see that we can factor \(p(z)\) as
\begin{equation*} p(z) \ = \ (z-a)^m \, g(z) \end{equation*}
where \(m\) is a positive integer \(\le d\) and \(g(z)\) is a polynomial that does not have a zero at \(a\text{.}\) The integer \(m\) is called the multiplicity of the zero \(a\) of \(p(z)\text{.}\) Almost exactly the same thing happens for holomorphic functions.

Proof.

By Theorem 8.1.8, there exists \(R>0\) such that we can expand
\begin{equation*} f(z) \ = \ \sum_{k\ge0}c_k (z-a)^k \qquad \text{ for } z \in D[a,R] \, \text{,} \end{equation*}
and \(c_0 = f(a) = 0\text{.}\) There are now exactly two possibilities: either
  1. \(c_k=0\) for all \(k \ge 0\text{;}\) or
  2. there is some positive integer \(m\) so that \(c_k=0\) for all \(k\lt m\) but \(c_m\ne0\text{.}\)
The first case gives \(f(z) =0\) for all \(z \in D[a,R]\text{.}\) So now consider the second case. We note that for \(z \in D[a,R]\text{,}\)
\begin{align*} f(z) \amp \ = \ c_m(z-a)^m + c_{m+1}(z-a)^{m+1} + \cdots \ = \ (z-a)^m\left( c_m+c_{m+1}(z-a) + \cdots\right)\\ \amp \ = \ (z-a)^m\sum_{k\ge0}c_{k+m} \, (z-a)^k\text{.} \end{align*}
Thus we can define a function \(g: G \to \C\) through
\begin{equation*} g(z) := \begin{cases}\ds \sum_{k\ge0}c_{k+m}(z-a)^k \amp \text{ if } z \in D[a,R] \, , \\ \dfrac{f(z)}{(z-a)^m} \amp \text{ if } z\in G\setminus \listset a . \end{cases} \end{equation*}
(According to our calculations above, the two definitions give the same value when \(z \in D[a,R] \setminus \listset a\text{.}\)) The function \(g\) is holomorphic in \(D[a,R]\) by the first definition, and \(g\) is holomorphic in \(G \setminus \listset a\) by the second definition. Note that \(g(a)=c_m\ne0\) and, by construction,
\begin{equation*} f(z) \ = \ (z-a)^m \, g(z) \qquad \text{ for all } z \in G \,\text{.} \end{equation*}
Since \(g(a)\ne0\) there is, by continuity, \(r>0\) so that \(g(z)\ne0\) for all \(z\in D[a,r]\text{,}\) so \(D[a,r]\) contains no other zero of \(f\text{.}\) The integer \(m\) is unique, since it is defined in terms of the power series expansion of \(f\) at \(a\text{,}\) which is unique by Corollary 8.1.6.
Theorem 8.2.1 gives rise to the following result, which is sometimes called the identity principle or the uniqueness theorem.
Applying this theorem to the difference of two functions immediately gives the following variant.

Proof.

Consider the following two subsets of \(G\text{:}\)
\begin{align*} X \amp := \left\{ a \in G : \text{ there exists } r \text{ such that } f(z) = 0 \text{ for all } z \in D[a,r] \right\}\\ Y \amp := \left\{ a \in G : \text{ there exists } r \text{ such that } f(z) \ne 0 \text{ for all } z \in D[a,r] \setminus \{ a \} \right\} \text{.} \end{align*}
If \(f(a) \ne 0\) then, by continuity of \(f\text{,}\) there exists a disk centered at \(a\) in which \(f\) is nonzero, and so \(a \in Y\text{.}\)
If \(f(a) = 0\text{,}\) then Theorem 8.2.1 says that either \(a \in X\) or \(a\) is an isolated zero of \(f\text{,}\) so \(a\in Y\text{.}\)
We have thus proved that \(G\) is the disjoint union of \(X\) and \(Y\text{.}\) Exercise 8.4.11 proves that \(X\) and \(Y\) are open, and so (because \(G\) is a region) either \(X\) or \(Y\) has to be empty. The conditions of Theorem 8.2.2 say that \(\lim_{ n \to \infty } a_n\) is not in \(Y\text{,}\) and thus it has to be in \(X\text{.}\) Thus \(G = X\) and the statement of Theorem 8.2.2 follows.
The identity principle yields the strengthenings of Theorem 6.2.3 and Corollary 6.2.4 promised in Chapter 6. We recall that that we say the function \(u: G \to \R\) has a weak relative maximum \(w\) if there exists a disk \(D[w,r] \subseteq G\) such that all \(z \in D[w,r]\) satisfy \(u(z) \leq u(w)\text{.}\)
There are many reformulations of this theorem, such as:

Proof.

This is trivial if \(f\) is constant, so we assume \(f\) is non-constant. By the Extreme Value Theorem A.0.1 there is a point \(z_0\in\overline G\) so that \(\max_{z\in\overline G}|f(z)| = |f(z_0)|\text{.}\) Clearly \(\sup_{ z \in G } |f(z)| \le \max_{ z \in\overline G } |f(z)|\text{,}\) and this is easily seen to be an equality using continuity at \(z_0\text{,}\) since there are points of \(G\) arbitrarily close to \(z_0\text{.}\) Finally, Theorem 8.2.4 implies \(z_0\not\in G\text{,}\) so \(z_0\) must be in \(\partial G\text{.}\)
Theorem 8.2.4 has other important consequences; we give two here, whose proofs we leave for Exercise 8.4.12 and Exercise 8.4.13.
Note that (6.1) in Chapter 6 follows from Corollary 8.2.7 using essentially the same argument as in the proof of Corollary 8.2.5.

Proof.

Suppose there exist \(a \in G\) and \(R > 0\) such that \(|f(a)| \ge |f(z)|\) for all \(z \in D[a,R]\text{.}\) We will show that then \(f\) is constant.
If \(f(a)=0\) then \(f(z)=0\) for all \(z \in D[a,R]\text{,}\) so \(f\) is identically zero by Theorem 8.2.2.
Now assume \(f(a)\ne0\text{,}\) which allows us to define the holomorphic function \(g: G \to \C\) via \(g(z) := \frac{ f(z) }{ f(a) }\text{.}\) This function satisfies
\begin{equation*} \abs{g(z)} \ \le \ \abs{g(a)} \ = \ 1 \qquad \text{ for all } z \in D[a,R] \, \text{,} \end{equation*}
Also \(g(a)=1\text{,}\) so, by continuity of \(g\text{,}\) we can find \(r \le R\) such that \(\Re(g(z)) > 0\) for \(z \in D[a,r]\text{.}\) This allows us, in turn, to define the holomorphic function \(h: D[a,r] \to \C\) through \(h(z) := \Log(g(z))\text{,}\) which satisfies
\begin{equation*} h(a) \ = \ \Log(g(a)) \ = \ \Log(1) \ = \ 0 \end{equation*}
and
\begin{equation*} \Re(h(z)) \ = \ \Re(\Log(g(z))) \ = \ \ln(\abs{g(z)}) \ \le \ \ln(1) \ = \ 0 \, \text{.} \end{equation*}
Exercise 8.4.35 now implies that \(h\) must be identically zero in \(D[a,r]\text{.}\) Hence \(g(z)=\exp(h(z))\) must be equal to \(\exp(0)=1\) for all \(z \in D[a,r]\text{,}\) and so \(f(z)=f(a) \, g(z)\) must have the constant value \(f(a)\) for all \(z \in D[a,r]\text{.}\) Corollary 8.2.3 then implies that \(f\) is constant in \(G\text{.}\)