Classification of Zeros and the Identity Principle

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.

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Theorem 8.2.1. Classification of Zeros.

Suppose \(f: G \to \C\) is holomorphic and \(f\) has a zero at \(a \in G\text{.}\) Then either

\(f\) is identically zero on some open disk \(D\) centered at \(a\) (that is, \(f(z)=0\) for all \(z \in D\)); or
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there exist a positive integer \(m\) and a holomorphic function \(g: G \to \C\text{,}\) such that \(g(a)\ne0\) and

\begin{equation*} f(z) \ = \ (z-a)^m \, g(z) \qquad \text{ for all } z \in G \,\text{.} \end{equation*}

In this case the zero \(a\) is isolated: there is a disk \(D[a,r]\) which contains no other zero of \(f\text{.}\)

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The integer \(m\) in the second case is uniquely determined by \(f\) and \(a\) and is called the multiplicity of the zero at \(a\text{.}\)

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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

\(c_k=0\) for all \(k \ge 0\text{;}\) or
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there is some positive integer \(m\) so that \(c_k=0\) for all \(k\lt m\) but \(c_m\ne0\text{.}\)
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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 \\ \amp \ = \ (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*}

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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.

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Theorem 8.2.1 gives rise to the following result, which is sometimes called the identity principle or the uniqueness theorem.

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Theorem 8.2.2.

Suppose \(G\) is a region, \(f: G \to \C\) is holomorphic, and \(f(a_n) = 0\) where \((a_n)\) is a sequence of distinct numbers that converges in \(G\text{.}\) Then \(f\) is the zero function on \(G\text{.}\)

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Applying this theorem to the difference of two functions immediately gives the following variant.

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Corollary 8.2.3.

Suppose \(f\) and \(g\) are holomorphic in a region \(G\) and \(f(a_k)=g(a_k)\) at a sequence that converges to \(w \in G\) with \(a_k\ne w\) for all \(k\text{.}\) Then \(f(z)=g(z)\) for all \(z\) in \(G\text{.}\)

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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*}

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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 \(a \in Y\text{.}\)

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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 \(a\in Y\text{.}\)

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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.

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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 \(u(z) \leq u(w)\text{.}\)

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Theorem 8.2.4. Maximum-Modulus Theorem.

Suppose \(f\) is holomorphic and nonconstant in a region \(G\text{.}\) Then \(|f|\) does not attain a weak relative maximum in \(G\text{.}\)

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There are many reformulations of this theorem, such as:

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Corollary 8.2.5.

Suppose \(f\) is holomorphic in a bounded region \(G\) and continuous on its closure. Then

\begin{equation*} \sup_{ z \in G } |f(z)| \ = \ \max_{ z \in \partial G } |f(z)| \,\text{.} \end{equation*}

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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 \(\partial G\text{.}\)

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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.

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Corollary 8.2.6. Minimum-Modulus Theorem.

Suppose \(f\) is holomorphic and nonconstant in a region \(G\text{.}\) Then \(|f|\) does not attain a weak relative minimum at a point \(a\) in \(G\) unless \(f(a)=0\text{.}\)

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Corollary 8.2.7.

If \(u\) is harmonic and nonconstant in a region \(G\text{,}\) then it does not have a weak relative maximum or minimum in \(G\text{.}\)

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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.

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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.

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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.

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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 \(G\text{.}\)

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