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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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    $\begingroup$ 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. $\endgroup$ – Anon Oct 9 '18 at 16:24
  • $\begingroup$ Apologies if I am making an error: the maximum modulus principle follows from the open mapping theorem. Provided D is bounded, the arguments in the answers below can be used. Is the problem that, on your domain, the modulus is not constant as it is 1 on part of it and 2 on another part of it. $\endgroup$ – Ethan Horsfall Jun 19 '20 at 18:58
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Consider $\frac{1}{f(z)}$. $\phantom{}$

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    $\begingroup$ Hi Kevin, thanks but could you be more specific? I see $\frac{1}{f(z)}$ is also holomorphic on $D$ and continuous on its closure, and $\left|\frac{1}{f(z)}\right|$ is constant on the boundary of the unit disk, but how does this help with the problem? Sorry I'm missing the point here. $\endgroup$ – anegligibleperson Oct 28 '12 at 6:53
  • $\begingroup$ @anegligibleperson So $|1/f(z)|$ must have a maximum on the boundary of the disc. $\endgroup$ – WimC Oct 28 '12 at 7:45
  • $\begingroup$ @anegligibleperson If you could help me? How did you approach this question? $\endgroup$ – abcdef Dec 12 '15 at 14:53
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    $\begingroup$ @user123 The maximum of $f$ on the closure of $D$ is the maximum on the boundary by the maximum modulus principle, and because $\bar{D}$ is compact. Comsidering $g=1/f$, we see $g$ has the same property, and hence $f$ attains both its maximum and minimum on the boundary. Since $f$ is constant there, by the open mapping theorem, we have $f$ is constant. $\endgroup$ – anegligibleperson Dec 12 '15 at 15:21
  • $\begingroup$ @anegligibleperson If I have the same conditions as yours, meaning |f(z)| constant on the boundary, how can I show that f(z) has a zero in D. $\endgroup$ – abcdef Dec 12 '15 at 15:31
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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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  • $\begingroup$ 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. $\endgroup$ – ziggurism Dec 21 '17 at 4:00

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