Extension of Liouville's Theorem?

Liouville's Theorem states that if a function is bounded and holomorphic on the complex plane (i.e. bounded and entire), then it is a constant function.

What if we consider the following, slightly modified scenario:

Suppose a function $f$ is holomorphic and has constant modulus on a bounded domain $D$ (e.g. a small disk).

Can we use Liouville's Theorem to somehow conclude that $f$ is a constant function? (either on $D$ or on the whole of the complex plane?)

• Maybe not Liouvill, but maximum modulus principle will work. – Davide Giraudo Dec 5 '12 at 21:27
• Thanks Davide - indeed, this makes it much simpler. – Conan Wong Dec 5 '12 at 21:33
• @DavideGiraudo, can you say more on how to use the maximum modulus principle here? – user27126 Dec 5 '12 at 22:13
• A constant modulus on the closure of a domain gives that the function is constant. – Davide Giraudo Dec 5 '12 at 22:22

If you assume that $f$ is entire, use Cauchy-Riemann's equation on $|f|^2 = u^2 + v^2$ to show that both $u$ and $v$ must be constant on $D$. After that it follows from the uniqueness theorem that $f$ is constant everywhere.
• Thanks mrf. Just to clarify - instead of "f is entire," the assumption (as in the original scenario) that f is holomorphic on $D$ will suffice right? – Conan Wong Dec 5 '12 at 21:34
• Yes. (But of course then you only conclude that $f$ is constant on $D$.) – mrf Dec 5 '12 at 21:35