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I have a conjecture that I can´t prove nor disprove, any help on doing so will be very grateful.

Let $f: \{z: |z|<2\} \to \mathbb C$ be a non constant analytic function such that if $|z|=1$ then $|f(z)|=1$.

Is it true that the zeros of $f$ can not be in $\{ z: 1/2< |z| < 2 \}$ ?

I have successfully proven, by the maximum modulus theorem, that $f$ must have a zero inside $\{ z: |z|<1 \}$. However, I can not seem to prove that all the zeros must be in $\{ z:|z| < 1/2 \}$, neither to find a counter example.

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Alonso Delfín
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1 Answers1

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It is true. If a function is unimodular on the unit circle ($|f(z)|=1$), you can use Schwarz reflection principle that tells you that the zeros and poles are symmetric wrt the unit circle ($z\ \leftrightarrow\ \bar{z}^{-1}$). There are no poles inside $|z|=1$, so there are no zeros outside $|z|=1$ for the analytical continuation. All poles are outside $|z|=2$ then all zeros are inside $|z|=1/2$.

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A.Γ.
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  • Neat. I surprisingly never encountered this result before. $+1$ – Cameron L. Williams Jul 28 '15 at 02:45
  • @A.G. Thank you for your prompt answer. Then If I get you right, my conjecture follows by using that since $f$ is unimodular then it is a rational function, also by the maximum principle we get that all zeros of $f$ must be in ${z:|z|<1}$, say $Z(f)={z_1, \cdots, z_n}\subset {z:|z|<1}$ are all the zeros of $f$, then $P(f)=\left{\left(\overline{z_1}\right)^{-1}, \cdots, \left(\overline{z_n}\right)^{-1}\right} \subset {z:|z|>1}$ are the poles of $f$, however, since in this case $P(f) \subset {|z|\geq 2}$ we conclude that $Z(f)\subset {z : |z|<1/2}$?? – Alonso Delfín Jul 28 '15 at 04:21
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    @AlonsoDelfín That's right. Under the conditions the function is a finite Blaschke product. – A.Γ. Jul 28 '15 at 10:01