In Conway's Functions of One Complex Variable, the section on the Riemann Mapping Theorem has the following exercise:
Let $G$ be a simply connected region which is not the whole plane, and suppose that $\bar{z}\in G$ whenever $z\in G$. Let $a\in G\cap\mathbb{R}$ and suppose that $f:G\rightarrow D=\{z:|z|<1\}$ is a bijective analytic function with $f(a)=0,\ f'(a)>0$. Let $G_+=\{z\in G:\text{Im }z>0\}$. Show that $f(G_+)$ must lie entirely above or below the real axis.
My proof showed that $f(z)=\overline{f(\bar{z})}$ using the classification of conformal automorphisms of $D$, which then implied the desired result. Using the Riemann mapping theorem, I believe this can be extended to bijective conformal maps between any two simply connected regions $G_1$ and $G_2$ that are symmetrical across $\mathbb{R}$.
It seems plausible that the condition of being simply connected is unnecessary. Proceeding as in my proof, this seems to require some results like Schwarz's lemma bounding the derivative of conformal automorphisms of a region. Is there a proof or a counterexample for the case when the regions are not simply connected (but still connected), which preferably offers more intuition into the result?
Best Answer
If $G$ is a domain (connected open set) that is symmetric with respect to the real axis, then $G\cap \mathbb{H} = \{ z\in G : \operatorname{Im} z > 0\}$ is also connected. So if $f\colon G_1 \to G_2$ is a holomorphic map between two domains symmetric with respect to the real axis, then $f(G_1\cap\mathbb{H})$ does not lie entirely above or below the real axis if and only if $f$ attains real values in $G_1\cap \mathbb{H}$. Thus, for the case of a biholomorphic $f$, it is sufficient to have $f(G_1\cap \mathbb{R}) = G_2 \cap \mathbb{R}$ to conclude that $f(G_1\cap \mathbb{H})$ is either the part of $G_2$ in the upper half-plane, or the part of $G_2$ in the lower half-plane.
If there is an $a\in G_1 \cap\mathbb{R}$ with $f(a) \in G_2 \cap \mathbb{R}$ and $f'(a) \in \mathbb{R}$, then $g(z) = \overline{f\left(\overline{z}\right)}$ is also a biholomorphic map $G_1 \to G_2$, with $g(a) = f(a)$ and $g'(a) = f'(a)$. Thus $h = g^{-1}\circ f$ is an automorphism of $G_1$ with fixed point $a$ and $h'(a) = 1$.
We obtain the conclusion if we can show that $h = \operatorname{id}_{G_1}$. For then $g = f$ shows that $f(G_1\cap\mathbb{R}) \subset G_2 \cap \mathbb{R}$, and the same argument for $f^{-1}$ shows the equality.
For that, we don't need any symmetry or other particular properties:
Theorem: Let $G\subset \mathbb{C}$ a domain, and $a\in G$. If $h$ is an automorphism of $G$ with $h(a) = a$ and $h'(a) = 1$, then $h = \operatorname{id}_G$.
Proof: If $G = \mathbb{C}$, then the automorphisms of $G$ are precisely the polynomials of degree $1$, and $h'(a) = 1$ then implies that $h$ is a translation $z \mapsto z+b$, and since $h(a) = a$, it follows that $b = 0$.
If $G = \mathbb{C}\setminus \{p\}$, then $h$ extends to an automorphism $\tilde{h}$ of the Riemann sphere $\widehat{\mathbb{C}}$, and either $\tilde{h}(p) = p,\, \tilde{h}(\infty) =\infty$, in which case $h = \operatorname{id}_{G}$ follows as above, or $\tilde{h}(p) = \infty$ and $\tilde{h}(\infty) = p$, in which case we have
$$\tilde{h}(z) = \frac{c}{z-p}+p$$
for some $c\in\mathbb{C}$. Then $h(a) = a$ yields $c = (a-p)^2$, and we obtain
$$h'(a) = -\frac{(a-p)^2}{(a-p)^2} = -1 \neq 1,$$
so that cannot happen: $h(a) = a$ and $h'(a) = 1$ implies $h = \operatorname{id}_{G}$ also for $G = \mathbb{C}\setminus \{p\}$.
If $\mathbb{C}\setminus G$ contains at least two points, then the family $\operatorname{Hol}(U,G)$ of holomorphic functions on any domain $U$ with values in $G$ is a normal family by Montel's big theorem. In particular - picking $U = G$ - the sequence $\left(h^n\right)_{n\in\mathbb{N}}$ of iterates of $h$ contains a locally uniformly convergent subsequence $\left(h^{n_k}\right)_{k\in\mathbb{N}}$. By Weierstraß' theorem, also the sequences $\left(\frac{d^m}{dz^m}h^{n_k}\right)_{k\in\mathbb{N}}$ of $m$-th derivatives converges locally uniformly for every $m\in\mathbb{N}$. If we already know that $h^{(r)}(a) = 0$ for $2 \leqslant r < m$ (which we vacuously do for $m = 2$), the Taylor expansion of $h$ about $a$ is
$$h(z) = a + (z-a) + c_m (z-a)^m + O\left((z-a)^{m+1}\right).$$
From that we find that the $n$-fold iterate of $h$ has the Taylor expansion
$$h^n(z) = a + (z-a) + n\cdot c_m (z-a)^m + O\left((z-a)^{m+1}\right)$$
by induction:
$$\begin{align} h^{n+1}(z) &= h\left(a+(h^n(z)-a)\right)\\ &= a + \left(h^n(z)-a\right) + c_m \left(h^n(z)-a\right)^m + O\left(\left(h^n(z)-a\right)^{m+1}\right)\\ &= a + (z-a) + n\cdot c_m(z-a)^m + O\left((z-a)^{m+1}\right)\\ &\qquad + c_m \left((z-a) + O\left((z-a)^m\right)\right)^m + O\left((z-a)^{m+1}\right)\\ &= a + (z-a) + (n+1)c_m(z-a)^m + O\left((z-a)^{m+1}\right). \end{align}$$
So
$$\frac{d^m}{dz^m}\Biggl\lvert_a h^n = n\cdot \frac{d^m}{dz^m}\Biggl\lvert_a h,$$
and that converges only if $h^{(m)}(a) = 0$. By induction on $m$, it follows that all derivatives of $h$ of order greater than one vanish in $a$, and hence $h = \operatorname{id}_{G}$.