How to address the problem 3 in section Riemann Mapping Theorem in Conway´s book 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 $a \in G \cap R$ and suppose that $f : G \rightarrow D = \{z : |z| < 1\}$ is a one-one analytic function
with $f (a) = 0, f^{'}(a) > 0$ and $f (G) = D$. Let $G+ = \{z \in G : Im z > 0\}$. Show that $f(G+)$ must lie entirely above or entirely below the real axis.
 A: Recall the uniqeness of mapping in the Riemann mapping theorem:
Let $G$ be a simply connected region which is not the whole plane. 
Then there is an analytic one-one function $f:G→D=\{z:|z|<1\}$. 
For each point $a\in D,$
there is a unique such map $f$ such that $
f(a)=0, f^\prime(a)\in \mathbb{R}^+.$  
Let $g(z)=\overline{f(\bar{z})}.$
By the simmetry of $G$ (i.e., $\bar{z}\in G$ whenever $z\in G$), $g(z)$ is defined in $G$ and $g(G)=D,$$ g(a)=0, g^\prime (a)>0.$
Then we see $g(z)=f(z)$ by the uniqeness of mapping. This shows that $f(\bar{z})=\overline{f(z)}$.  
Suppose that $f(G+)$ does not lie entirely above or entirely below the real axis. Then there is a small disc $
K=\{z: |z-b|<r\} \subset f(G+)$, where $b$  is real.
Take two points $z_1, z_2 \in K$ with $\overline{z_2}=z_1$ and let $\zeta_i=f^{-1}(z_i)$, $i=1,2$.
Ofcourse $\overline{\zeta_2}\ne \zeta_1$, since both $\zeta_1$ and $\zeta_2$ belong to $G+$.
Now we see $$f(\overline{\zeta_2})=\overline{f(\zeta_2)}=\overline{z_2}=z_1.$$
This contradicts the one-one property of $f$, since $f(\zeta_1)=f(\overline{\zeta_2})=z_1$. 
Thus  $f(G+)$ must lie entirely above or entirely below the real axis.
