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Let $f:\Omega \to \mathbb{C}$ be non-constant and holomorphic function and $\Omega\subseteq\mathbb{C}$ be an open region. We will show that for any $z_0\in\Omega$, there is a neighborhood of $z_0$ such that for all $z\not=z_0$ in this neighborhood, $f(z)\not=f(z_0)$. To do this, we need the result that

If $f,g$ holomorphic on region $\Omega$ agree on a sequence of points with limit point in $\Omega$, they are identical on $\Omega$.

Proof: We prove the claim in the title by contradiction. Suppose that for each neighborhood of $z_0$ in $\Omega$, there is a $z$ in that neighborhood with $f(z)=f(z_0)$. Since $\Omega$ is open, there is a $\varepsilon>0$ such that $D_\varepsilon(z_0)\subseteq\Omega$. Consider the monotonically decreasing sequence $(r_n=\varepsilon/n)_n$ with $r_n\to 0$. By assumption, for each $n\in\mathbb{N}_0$ there is a $z_n\in D_{r_n}(z_0)\setminus\{z_0\}$ such that $f(z_n)=f(z_0)$. Then $z_n\to z_0$. By the result above this implies that $f=f(z_0)$ on all of $\Omega$, which is a contradiction. QED

Question: I've only now stumbled on this fact about holomorphic functions, even though I've been studying the complex analysis course for an entire semester. It's not mentioned in the book. Is it an obvious property that is much more easily seen than my proof above, perhaps?

Stijn D'hondt
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    You forgot to say $z_n \neq z_0$. – geetha290krm Jan 27 '23 at 11:17
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    I'm not sure there's a much easier way to prove it than what you did. I wouldn't say it's an "obvious" fact, but if you think long enough about why it is a corollary of the theorem you mentioned, it should appear intuitive. The identity theorem you mentioned tells you that you can't have an accumulation of points around $z_0$ such that $f(z) = f(z_0)$. So if you have no such accumulation, then surely, there's a small enough neighbourhood around $z_0$ such that the function $f$ never takes the value $f(z_0)$ again – Azur Jan 27 '23 at 15:05
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    Beware: the result quoted in the question is wrong. Consider $\Omega=\Bbb{C}\setminus\Bbb{R}$ and define $f(z)=z$ on $\Omega$; also define $g(z)=\begin{cases} z&\ \mathrm{if}\ \Im(z)>0,\ z-1&\ \mathrm{if}\ \Im(z)<0. \end{cases}$

    Both $f$ and $g$ are holomorphic, they coincide on a half-plane but $f\neq g$.

     – jp boucheron Jan 27 '23 at 22:41
  • @jp boucheron, the standard definition of "open region" is an open and connected, not just open. – Chris JS Jan 28 '23 at 00:46
  • And same of "region", it requires connected. – Chris JS Jan 28 '23 at 00:52
  • @ChrisJS Thanks for the info; I was not aware of this definition. English is not my native language and, in French, we simply say ouvert connexe (“open connected”). At least today I have learned something new and (hopefully) I won't make the same mistake again. – jp boucheron Jan 28 '23 at 01:00
  • @jp boucheron, the same thing happened to me when I first took a complex analysis class and a classmate had to correct me. – Chris JS Jan 28 '23 at 01:03

1 Answers1

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(Without making use of the assumption quoted in the question)$f$ has a Taylor expansion at $z_0$ with radius of convergence $r\in\mathopen]0\mathbin;+\infty]$. Without loss of generality we may also suppose that $f(z_0)=0$ (replacing $f$ by $f-f(z_0)$). That is $f(z)=\sum_{k=1}^{+\infty} a_k(z-z_0)^k$. If all $a_k$ were 0, $f$ would be constant; hence $\{k\in\Bbb{N} \mid a_k\neq0 \}\neq\emptyset$, and this set has a least element $n$.

Now $f(z)=(z-z_0)^n\sum_{k=n}^{+\infty} a_k(z-z_0)^{k-n} =\color{blue}{(z-z_0)^n g(z-z_0)}$, where $g(z)=\sum_{k=0}^{+\infty}a_{k+n}\,z^k$, and the radius of convergence of $g$ is also equal to $r>0$.

Since $g(0)=a_n\neq0$ and $g$ is continuous at $0$ there exists some $\varrho\in\mathopen]0\mathbin;r\mathclose[$ such that $g(z)\neq0$ for any $z\in D(0\mathbin;\varrho)$. On the disk $D(z_0\mathbin;\varrho)$, $f$ has no other zero than $z_0$.

jp boucheron
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