The Identity Theorem

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The fact that any complex differentiable function is in fact analytic has some very surprising consequences – the most striking of which is perhaps captured by the "Identity theorem". This says that if $f, g$ are two holomorphic functions defined on a domain $U$ and we let $S=\{z \in U: f(z)=g(z)\}$ be the locus on which they are equal, then if $S$ has a limit point in $U$ it must actually be all of $U$. Thus for example if there is a disk $B(a, r) \subseteq U$ on which $f$ and $g$ agree (not matter how small $r$ is), then in fact they are equal on all of $U$ ! The key to the proof of the Identity theorem is the following result on the zeros of a holomorphic function:

Proposition 8.1. Let $U$ be an open set and suppose that $g: U \rightarrow \mathbb{C}$ is holomorphic on $U$. Let $S=\{z \in U: g(z)=0\}$. If $z_0 \in S$ then either $z_0$ is isolated in $S$ (so that $g$ is non-zero in some disk about $z_0$ except at $z_0$ itself) or $g=0$ on a neighbourhood of $z_0$. In the former case there is a unique integer $k>0$ and holomorphic function $g_1$ such that $g(z)=\left(z-z_0\right)^k g_1(z)$ where $g_1\left(z_0\right) \neq 0$.

Proof. Pick any $z_0 \in U$ with $g\left(z_0\right)=0$. Since $g$ is analytic at $z_0$, if we pick $r>0$ such that $\bar{B}\left(z_0, r\right) \subseteq U$, then we may write
\[
g(z)=\sum_{k=0}^{\infty} c_k\left(z-z_0\right)^k
\]
for all $z \in B\left(z_0, r\right) \subseteq U$, where the coefficients $c_k$ are given as in Theorem 7.11. Now if $c_k=0$ for all $k$, it follows that $g(z)=0$ for all $z \in B(0, r)$. Otherwise, we set $k=\min \left\{n \in \mathbb{N}: c_n \neq 0\right\}$ (where since $g\left(z_0\right)=0$ we have $c_0=0$ so that $k \geq 1$ ). Then if we let $g_1(z)=\left(z-z_0\right)^{-k} g(z)$, clearly $g_1(z)$ is holomorphic on $U \backslash\left\{z_0\right\}$, but since in $B\left(z_0, r\right)$ we have $g_1(z)=\sum_{n=0}^{\infty} c_{k+n}\left(z-z_0\right)^n$, it follows if we set $g_1\left(z_0\right)=c_k \neq 0$ then $g_1$ becomes a holomorphic function on all of $U$. Since $g_1$ is continuous at $z_0$ and $g_1\left(z_0\right) \neq 0$, there is an $\epsilon>0$ such that $g_1(z) \neq 0$ for all $z \in B\left(z_0, \epsilon\right)$. But $\left(z-z_0\right)^k$ vanishes only at $z_0$, hence it follows that $g(z)=\left(z-z_0\right)^k g_1(z)$ is non-zero on $B(a, \epsilon) \backslash\left\{z_0\right\}$, so that $z_0$ is isolated.

Finally, to see that $k$ is unique, suppose that $g(z)=\left(z-z_0\right)^k g_1(z)=\left(z-z_0\right)^l g_2(z)$ say with $g_1\left(z_0\right)$ and $g_2\left(z_0\right)$ both nonzero. If $k<l$ then $g(z) /(z-\left.z_0\right)^k=\left(z-z_0\right)^{l-k} g_2(z)$ for all $z \neq z_0$, hence as $z \rightarrow z_0$ we have $g(z) /(z-z_0)^k \rightarrow 0$, which contradicts the assumption that $g_1(z) \neq 0$. By symmetry we also cannot have $k>l$ so $k=l$ as required.   $\square$

Remark 8.2. The integer $k$ in the previous proposition is called the multiplicity of the zero of $g$ at $z=z_0$ (or sometimes the order of vanishing).

Theorem 8.3. (Identity theorem): Let $U$ be a domain and suppose that $f_1, f_2$ are holomorphic functions defined on $U$. Then if $S=\left\{z \in U: f_1(z)=f_2(z)\right\}$ has a limit point in $U$, we must have $S=U$, that is $f_1(z)=f_2(z)$ for all $z \in U$.
Proof. Let $g=f_1-f_2$, so that $S=g^{-1}(\{0\})$. We must show that if $S$ has a limit point then $S=U$. Since $g$ is clearly holomorphic in $U$, by Proposition 8.1 we see that if $z_0 \in S$ then either $z_0$ is an isolated point of $S$ or it lies in an open ball contained in $S$. It follows that $S=V \cup T$ where $T=\{z \in S: z$ is isolated$\}$ and $V=\operatorname{int}(S)$ is open. But since $g$ is continuous, $S=g^{-1}(\{0\})$ is closed in $U$, thus $V \cup T$ is closed, and so $\mathrm{Cl}_U(V)$, the closure¹¹ of $V$ in $U$, lies in $V \cup T$. However, by definition, no limit point of $V$ can lie in $T$ so that $\mathrm{Cl}_U(V)=V$, and thus $V$ is open and closed in $U$. Since $U$ is connected, it follows that $V=\emptyset$ or $V=U$. In the former case, all the zeros of $g$ are isolated so that $S^{\prime}=T^{\prime}=\emptyset$ and $S$ has no limit points. In the latter case, $V=S=U$ as required.   $\square$

Remark 8.4. The requirement in the theorem that $S$ have a limit point lying in $U$ is essential: For example take $U=\mathbb{C} \backslash\{0\}$ and $f_1=\sin (1 / z)$ and $f_2=0$.
Now the zeros of $f_1$ have a limit point at $0 \notin U$ since $f_1(1 /(\pi n))=0$ for all $n \in \mathbb{N}$, but certainly $f_1$ is not identically zero on $U$!
The requirement in the theorem that $U$ is connected is essential: For example take $U=U_1\cup U_2$ where $U_1$ and $U_2$ are disjoint open sets in $\Bbb C$. Define $f\upharpoonright U_1=0$ and $f\upharpoonright U_2=1$, then $f$ is holomorphic and the zeros of $f$ have limit points in $U_1$ but $f$ is not constant.
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¹¹ I use the notation $\mathrm{Cl}_U(V)$, as opposed to $\bar{V}$, to emphasize that I mean the closure of $V$ in $U$, not in $\mathbb{C}$, that is, $\mathrm{Cl}_U(V)$ is equal to the union of $V$ with the limits points of $V$ which lie in $U$.

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