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If I have an analityc function $f(z)$ on a domain $B \subset \mathbb{C}$ and a simple and closed curve $C$ that encloses $B$, and if $|f|$ is constant over C, the $f$ is constant on $B$?

I haven’t found a counterexample but i dont know how to apply the maximum principle or apply some analityc continuation. Please some help with this.

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    $\begingroup$ $f(z) = z$ on $|z| = 1$ ? Otherwise $f(z) =c \frac{z+b}{\overline{b}z+1}$ with $|b| < 1$ on $|z|=1$ $\endgroup$ – reuns Nov 18 '18 at 4:24
  • $\begingroup$ @reuns: Can you post this as an answer so that this question can be moved out of the Unanswered queue? $\endgroup$ – aleph_two Dec 20 '18 at 4:35

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