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Let $f:D(0,1)\longrightarrow \mathbb{C}$ be a holomorphic function such that $f(z)\in\mathbb{R} \Longleftrightarrow z\in \mathbb{R}$. How to prove that $f$ has at most one zero on the disk.

By hypothesis the zeros of $f(z)$ are real and $\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n$ , $a_n\in\mathbb{R}$ since $f(z)=\overline{f(\overline{z})}$.

Any help would be appreciated.

Antonio Vargas
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felipeuni
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    If there are two zeros, then the derivative is zero somewhere on the real axis, so it looks locally like $z^n$, right? Then it seems like the preimage of $\mathbb{R}$ should be more than one line through the critical point...? – Tim kinsella Sep 27 '14 at 02:44
  • see this –  Sep 27 '14 at 03:22
  • @OmranKouba This question is very similar, but the domain is not holomorphically equivalent, so I'd suggest that a priori it may have character different from that one. – Travis Willse Sep 27 '14 at 05:42
  • @Travis, But the proof is straightforwardly adaptable, because this is a local property. – Omran Kouba Sep 27 '14 at 06:22
  • @OmranKouba Yes, I agree that the answers are essentially the same, but the questions, at least a priori, are not. – Travis Willse Sep 27 '14 at 06:41

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