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Let $f$ be holomorphic on the upper half plane and continuous on $\mathbb{R}$, with $|f(r)|=1$ for all $r\in\mathbb{R}$. Prove that $f$ is rational.

I was playing around with conformal maps and $\overline{f(\bar{z})}$, but I would really like a hint on how exactly "rationality" comes up. I'm guessing Schwarz Lemma is involved?

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    @quanta: if you are referring to the composition of f(z)=Rez and g(z)=$e^z$, then f(z) is not holomorphic; f(z)=U+iV=x+i.0, so that U(x,y)=x, V(x,y)=0, and $U_x$=1 , but $V_y$=0.2011-07-14

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I think you also want $\lim_{r \to +\infty} f(r)$ and $\lim_{r \to -\infty} f(r)$ to exist and be equal. Schwarz Reflection principle shows $f$ is meromorphic on $\mathbb C$ with $f(\overline{z}) = 1/\overline{f(z)}$. Same applies to $f(1/z)$. So $f$ is an analytic function from the Riemann sphere to itself, and such functions are rational.

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    @Plop: you're right, you do need th$a$t stronger condition.2011-05-10