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I was trying to prove the claim 'If $f$ is an analytic on a domain $D$ and $|f(z)|=|\sin z|$ for all $z \in D$, then $f(z)=c \sin z,~z \in D$ for some $c\in \mathbb C$'.

It is clear that $g(z)=\frac{f(z)}{\sin z}$ can be analytically extended through out $D$ since $g$ has only removable singularities. However, I couldn't apply Liouville's theorem to state $$g(z)=c,$$(for some complex constant $c$) as $D$ been just a domain. How to justify it?

Messi Lio
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    If $|g(z)|$ is constant then $g(z)$ is constant if the domain is connected. It can be proved for example by applying C-R equations. – Ryszard Szwarc Sep 12 '22 at 16:32
  • Sir, How can we pretend $|g(z)|$ is constant through out $D$? – Messi Lio Sep 12 '22 at 16:36
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    By assumption $|f(z)|=|\sin z|.$ So $|g(z)|=1$ for $z\in D.$ You do not need the Liouville theorem. Besides call me Ryszard. – Ryszard Szwarc Sep 12 '22 at 16:40
  • How to see an analytic function is constant if it's modulus is constant through a domain (open connected subset of $\Bbb C$)? I am well aware of Little Picards Theorem while $D=\Bbb C$. – Messi Lio Sep 12 '22 at 16:47
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    One route without appealing to the Cauchy-Riemann equations is through the maximum modulus principle. – csch2 Sep 12 '22 at 16:51
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    Consult https://math.stackexchange.com/questions/1588511/holomorphic-function-with-a-local-maximum-modulus-is-constant – Ryszard Szwarc Sep 12 '22 at 16:53
  • Yes...Now I got the things...Thanks all.... – Messi Lio Sep 12 '22 at 16:54
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    @csch2 You are not avoiding C-R, by applying maximum modulus principle, as C-R equations are used in the proof of the principle. First part of the proof gives that $|f|$ is constant by the mean value property. Then C-R are applied. So avoiding C-R by maximum modulus is a circular reasoning. – Ryszard Szwarc Sep 12 '22 at 18:20
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    @RyszardSzwarc true. I should have said "without directly appealing" to CR. – csch2 Sep 12 '22 at 19:22
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    @RyszardSzwarc -- it should be said that C-R equations are not necessary for proving max modulus theorem of analytic functions; you'll most likely see this with authors taking the Weierstrass approach to complex analysis. E.g. Beardon proves max modulus in chapter 8 and doesn't mention C-R until chapter 10. (He doesn't even introduce basic differentiation or integration until chapter 9.) Put differently it is not circular per se to avoid C-R by max modulus. There are multiple roads to Rome. – user8675309 Sep 13 '22 at 18:04
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    @user8675309 You are right. But Beardon proves the maximum principle for power series, i.e. analytic functions. The ratio $f(z)/\sin z$ is holomorphic. In order to prove it is analytic integration is needed. It is possible to take a long detour. In my opinion using C-R is the shortest road, and most popular. In this problem we start right from the begining from $|g(z)|=1.$ So using C-R is the simplest way in my opinion. In many textbooks C-R are introduced long before Taylor series. – Ryszard Szwarc Sep 13 '22 at 19:06
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    @RyszardSzwarc -- "The ratio ()/sin is holomorphic. In order to prove it is analytic integration is needed. It is possible to take a long detour." -- there's actually a very short integration-free approach to solving this problem at the level of Beardon chapter 8 (with no mention of differentiation, singularities, etc.). I may drop said solution in one of the duplicate links as I think the people would benefit from seeing a different approach. – user8675309 Sep 15 '22 at 17:04
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    @user8675309 Thanks. I am very interested for didactic reasons. – Ryszard Szwarc Sep 15 '22 at 19:34

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