If $\Omega$ is a symmetric domain of $\mathbb{C}$ and $f$ an isomorphism with certain conditions then $f(\overline{z})=\overline{f(z)}$

Question

If $\Omega$ is a symmetric domain of $\mathbb{C}$ respect to $\mathbb{R}$ and $f: \Omega\longrightarrow D(0,1)$ is an isomorphism. If $\exists a\in \Omega \cap \mathbb{R}$ such that $f(a)=0$ and $f'(a)\in \mathbb{R}$ then $f(\overline{z})=\overline{f(z)}$ .

I am trying to apply a result that says that if a Möbius transformation sends $\mathbb{R}$ to $\mathbb{R}$ then $\varphi(\overline{z})=\overline{\varphi(z)}$. Any idea?

Answer 1

By symmetry of domain and range the function $g(z) = \overline{f(\overline{z})}$ is well defined and also an holomorphic isomorphism. Moreover, $g(a)=0$ and $g'(a) = f'(a)$. The composition $f\circ g^{-1}$ maps the unit circle onto itself, fixes the origin, and has derivative $1$ at the origin. Now apply Schwarz' lemma.