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Assume that $\gamma$ is an analytic simple closed curve in $\mathbb{C}$ which surrounds origin.

Is there a non constant entire holomorphic function $f$ such that $|f(z)|$ is constant on $\gamma$?

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  • $\begingroup$ For some curves - yes, for others - no. $\endgroup$ Commented Sep 27, 2017 at 13:36
  • $\begingroup$ @AlexandreEremenko For what curves there is no such entire function? $\endgroup$ Commented Sep 27, 2017 at 13:42

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Let $\phi$ be the conformal map of the unit disk onto the interior. The only thing that can be said about $\phi$ is that it is analytic and univalent in the closed disk. If your entire function $f$ exists, then $B=f\circ\phi$ is a finite Blaschke product (by symmetry principle). So $\phi=f^{-1}\circ B$. It is clear that not every univalent function analytic in the closed disk has such representation with entire $f$. For example, Iversen's theorem shows that $f^{-1}\circ B$ cannot have a singular arc.

EDIT. If you want a more elementary argument, take any polynomial (univalent in the unit disk, and other than a monomial) as $\phi$. Then $f\circ\phi=B$, a Blaschke product. It follows that $f$ itself is a polynomial, and thus $B(z)=z^n$. But then $\phi$ must be also a monomial, and this contradicts our assumption about $\phi$.

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  • $\begingroup$ Dear Prof. Eremenko Thanks for your answer and your attention to my question. $\endgroup$ Commented Sep 28, 2017 at 5:46

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