# Can a Herglotz-Nevanlinna function attain real values?

Let $\mathbb{H}^+=\{z \in \mathbb{C}\mid \Im(z)>0\}$. We say that an analytic $F\colon \mathbb{H}^+\to\overline{\mathbb{H}^+}$ is a Herglotz-Nevanlinna's function.

Question Can it be that $F(z)\in \mathbb{R}$ for some $z \in \mathbb{H}^+$?

I guess that the answer is no, because if this happened then we could find a small loop $\gamma$ around $z$ such that $F\circ \gamma$ slips outside $\overline{\mathbb{H}^+}$, but I'm not sure this is true and how to formalize this little argument.

Thank you.

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Of course in both this question and the answer by Davide Giraudo it is assumed that $F$ is not constant. – Giuseppe Negro May 22 '12 at 9:57

If $F$ is constant it's clear, otherwise, since $\mathbb H^+$ is connected, by the open mapping theorem, $F(\mathbb H^+)$ is open in $\mathbb C$. If $F(z)\in\Bbb R$ for some $z\in\mathbb H^+$, then we can find $r$ such that if $|y-F(z)|<2r$ then $y=F(y')$ for some $y'\in\mathbb H^+$. If we consider $F(z)-ri$, we get a contradiction.