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This exercise has no assumption on the boundedness of $D$, hence $\bar{D}$ is not guaranteed to be compact. Would this be essential for the assertion to be true?

Suppose the nonconstant function $f(z)$ is analytic in a domain $D$ and continuous on its closure. If $|f(z)|$ is constant on the boundary of $D$, prove that $f(z)$ has a zero in $D$.

The obvious approach is to assume by contradiction that $f$ has no zero in $D$. Then we can define a nonconstant analytic function $1/f$ in $D$, which gives us by the Maximum Modulus Theorem that $1/|f|$ does not attain a maximum in $D$. But how can I use this to reach a contradiction? I am stuck here and I would greatly appreciate any help.

nomadicmathematician
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  • @MyGlasses Is the assertion false if $D$ is not bounded? – nomadicmathematician May 13 '17 at 01:59
  • As far as I know the boundedness of $D$ is necessary for applying Maximum Modulus Theorem. – Nosrati May 13 '17 at 02:21
  • Both $|f|$ and ${1 \over |f|}$ attain their $\max$ on the boundary, hence $f$ is a constant. So $f$ must have a zero in the interior. – copper.hat May 13 '17 at 02:34
  • I think this proposition is not true in general. If $D={z\in\mathbf C;|z|>1},$ and define for each $z\in D$ that $f(z)=1/z$, then $|f(z)|=1$ on $\partial D={z\in\mathbf C;|z|=1}$, and $f$ is analytic in $D$. – user88544 May 13 '17 at 02:45

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The bounded case is the application of the maximum modulus principle mentioned in the comments, or in previous questions of this type. The assertion is untrue for unbounded $D$. If $D$ is the half plane and $f(z)=e^{iz}$, then $|f|$ is constant on $\partial D=\mathbb R$ but $f$ has no zeroes in $D$.

Jason
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