Section8.2Classification 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
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.
(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.
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 \(G\text{.}\)
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 \(G\text{.}\)
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*}
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 \(u(z) \leq u(w)\text{.}\)
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 \(\partial G\text{.}\)
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 \(f(a)=0\text{.}\)
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.
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
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 \(G\text{.}\)