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Let $D \subset \mathbb{C}$ be a bounded domain and $f$ a function holomorphic in $D$ and continuous in its closure. Suppose that $|f(z)|$ is constant on the boundary of $D$ and that $f$ does not have zeroes in $D$. Prove that $f$ is a constant function.

I think that if I can prove that $f$ attains both its maximum and minimum values on the boundary, then the result follows from the maximum principle. But I've been unable to show this. Is this the right way to approach this problem? If so, how do I show this result? Thanks in advance!

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    The problem is incorrect as stated. For example the domain could be an annulus \{z:1<|z|<2\}, in which case, the identity function satisfies the hypotheses. For the function to be constant, you additionally need that the boundary of the domain $D$ is connected.2018-10-09

2 Answers 2

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Consider $\frac{1}{f(z)}$. $\phantom{}$

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    @user123 assume $f$ is nonconstant, holomorphic on $D$ and continuous on $\bar{D}$, with $f$ nonvanishing on $D$, ... then use the minimum modulus principle we effectively just showed in the previous problem to get a contradiction2015-12-12
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By the maximum modulus principle, $f$ takes its maximum modulus on the boundary. By the minimum modulus principle (which is just the maximum modulus principle applied to $1/f$, which requires that $f$ have no zeros), $f$ also takes its minimum modulus on the boundary.

If the modulus is constant on the boundary, then the minimum modulus and the maximum modulus, both lying on the boundary, must be equal. Hence the modulus is constant on all of $D$ including the interior.

And if $|f|$ is constant on all of $D$, say $|f|(D)=\{K\}$, then the image of $D$ under $f$ lies inside the circle $\{e^{iθ}K\}.$ A circle which has empty interior in $\mathbb{C},$ so is not open.

But the open mapping theorem states that if a function $f$ is not constant, it must be an open map, i.e. it must send any open subset of $\mathbb{C}$ to an open subset.

Finally, by contraposition, since $f(D)\subseteq \{e^{iθ}K\}$ is not open, $f$ must be constant.

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    Although this years old question already has an answer which was accepted also years ago, I perceived there to be some confusion about the answer among more recent askers, even after they were linked here as a duplicate. Hence I give a slightly more detailed answer.2017-12-21