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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?)

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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
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up vote 2 down vote accepted

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.

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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
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