Constant function in $\mathbb{C}$

Question

The following five problems (Ahlfors Theorem 11 chapter 3) are trivial consequences of the Open mapping theorem.

(a) Show that a holomorphic function $f$ on a domain $\Omega$ whose derivative vanishes identically is a constant.

(b) Show that the same conclusion holds if

(I) the real part is constant. (II) the imaginary part is constant. (III) modulus of $f$ is constant. (IV) the argument of f is constant.

These can be easily done by the CR equations or by some simple tools, for example, the first one can be done by FTC for contour integrals. Could you please share a couple more simple and cool results/problems that conclude that $f$ is constant in an open set in $\mathbb{C}$? Thanks so much.

Answer 1

Here are some examples of results; hopefully they inspire someone to read more about complex analysis; it truly is a fascinating subject.

If $\Omega =\mathbb{C}$ there are a bunch of results characterizing holomorphic functions with certain behavior.

1. If $|f(z)|\leq M|z|^m$ for some $m\in \mathbb{N}\cup\{ 0\}$, then $f$ is a polynomial of degree at most $m$. The case $m=0$ is Liouville's theorem.

2. If the function $f$ misses two (or more) points, i.e. $\{ z_1,z_2\}\notin f(\mathbb{C})$, then $f$ is constant. In the case $f$ is a polynomial missing one point is enough (by the fundamental theorem of algebra), however in general for instance $f(z)=e^z$ misses $0$. This is called the Little Picard theorem.

3. If there is a sequence $\{ z_k\}$ converging to $z\in \mathbb{C}$, such that $f(z_j)= c$ for all $j$, then $f$ is constant. This is called the identity theorem (if instead of a sequence you have $f\equiv c$ in some open neighborhood of $z$, this follows immediately from the expansion of $f$ in a Taylor series).

For a general domain $\Omega$, only 3 continues to apply. For instance, there's a bunch of bounded holomorphic functions on the unit disk (and so 2 has to fail as well on this $\Omega$).

On the other hand, don't let this results deceive you; while holomorphic functions are rigid, there's still plenty of them. For instance, relating to 3, given any sequence $z_j\subset \Omega$ having no limit point in $\Omega$ (so we either tend to $\infty$ or to the boundary), then you can construct an analytic function such that $f(z_j)=0$ and this are precisely all the zeroes of it. In other words, having a limit point in $\Omega$ is the only obstruction to having a non-constant function. This is called the Weierstrass factorization theorem.