The Coincidence Principle

Recall from the Orders of Roots of Analytic Complex Functions page that if $A \subseteq \mathbb{C}$ is open, $f : A \to \mathbb{C}$ is analytic on $A$ and $z_0 \in A$ is a root of $f$ then $z_0$ is said to be a root of order $k$ if we can write $f(z) = (z - z_0)^k g(z)$ where $g(z_0) \neq 0$ and $g$ is analytic on $A$.

We are about to state a very important result known as the Coincidence Principle. It tells us that if $f$ is a nonzero analytic function then the roots of $f$ are isolated points to the set of all roots of $f$ in $A$.

Theorem 1: Let $A \subseteq \mathbb{C}$ be open and let $f : A \to \mathbb{C}$ be analytic on $A$. Let $Z = \{ z \in A : f(z) = 0 \}$ (the set of roots of $f$ in $A$). If $f$ is also a nonzero function then the roots of $f$ in $A$ are isolated points of $Z$.

• Proof: Let $f$ be a nonzero analytic function on $A$ and suppose instead that there exists a $z_0 \in Z$ that is not an isolated point in $Z$.

• Since $z_0$ is not an isolated point in $Z$ there exists a sequence of points $(z_n)_{n=1}^{\infty}$ in $Z$ that converges to $z_0$.

• since $z_0 \in Z$, $z_0$ is a root of $f$ in $A$ and so there exists a $k \in \mathbb{N}$ for which $f(z) = (z - z_0)^k g(z)$ where $g(z_0) \neq 0$ and $g$ is analytic on $A$ (here, $k$ is the order of the root $z_0$). For all $n \in \mathbb{N}$ we also have that since $z_n \in Z$ that $z_n$ is a root of $f$ in $A$ so $f(z_n) = 0$. Hence:

(1)

\begin{align} \quad f(z_n) = (z_n - z_0)^k g(z_n) = 0 \end{align}

• Since $(z - z_0)^k \neq 0$ we must have that $g(z_n) = 0$ for all $n \in \mathbb{N}$.

• We know that $g$ is analytic on $A$ and so $g$ is continuous on $A$. Therefore:

(2)

\begin{align} \quad 0 = \lim_{n \to \infty} g(z_n) = g ( \lim_{n \to \infty} z_n) = g(z_0) \end{align}

• So $g(z_0) = 0$ which is a contradiction. Therefore the assumption that $z_0$ was not an isolated point in $Z$ was false. Therefore the roots of any nonzero analytic function are isolated points to the set $Z = \{ z \in A : f(z) = 0 \}$. $\blacksquare$

Corollary 2: Let $A \subseteq \mathbb{C}$ be open and let $f, g : A \to \mathbb{C}$ be analytic on $A$. If there exists a sequence of complex numbers $(z_n)_{n=1}^{\infty}$ in $A$ that converges to $z_0 \in A$ and $f(z_n) = g(z_n)$ for all $n \in \mathbb{N}$ then $f(z) = g(z)$ for all $z \in A$.

• Proof: Define a new function $h(z) = f(z) - g(z)$. Then for all $n \in \mathbb{N}$ we have that:

(3)

\begin{align} \quad h(z_n) = f(z_n) - g(z_n) = 0 \end{align}

• In other words, each $z_n$ is a root of $h$. Let $Z$ be the set of all roots of $h$. Then $\{ z_n : n \in \mathbb{N} \} \subset Z$. Furthermore, $h(z_0) = 0$, so $z_0 \in Z$. Therefore the sequence $(z_n)_{n=1}^{\infty}$ is a sequence in $Z \subset A$ that converges to $z_0 \in A$. But then $z_0$ is not an isolated point of $Z$. Since $h$ is analytic, this implies that $h$ must be identically zero on $A$, i.e., for all $z \in A$:

(4)

\begin{align} \quad 0 = f(z) - g(z) \end{align}

• Therefore $f(z) = g(z)$ for all $z \in A$. $\blacksquare$