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?
Best Answer
(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 $z_0$.