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?)

  • 2
    $\begingroup$ Maybe not Liouvill, but maximum modulus principle will work. $\endgroup$ Dec 5, 2012 at 21:27
  • $\begingroup$ Thanks Davide - indeed, this makes it much simpler. $\endgroup$
    – Conan Wong
    Dec 5, 2012 at 21:33
  • $\begingroup$ @DavideGiraudo, can you say more on how to use the maximum modulus principle here? $\endgroup$
    – user27126
    Dec 5, 2012 at 22:13
  • $\begingroup$ A constant modulus on the closure of a domain gives that the function is constant. $\endgroup$ Dec 5, 2012 at 22:22

1 Answer 1


I don't see how you could use Liouville's theorem to prove that, but it does follow from Cauchy-Riemann's equations.

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.

  • $\begingroup$ 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? $\endgroup$
    – Conan Wong
    Dec 5, 2012 at 21:34
  • $\begingroup$ Yes. (But of course then you only conclude that $f$ is constant on $D$.) $\endgroup$
    – mrf
    Dec 5, 2012 at 21:35

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