Map explicitly a simply connected region with at least two boundary points to unit disk In an old Qualifying exam there is an exercise that says:
"Let $G$ be a simply connected region of $\mathbb{C}$ such that the boundary $\partial G$ has at least two points $a\neq b$. Construct explicitly, in function of $a$ and $b$, a conformal map defined on $G$ such that $f(G)$ is open and bounded in $\mathbb{C}$".
I have problems with the generality of the region $G$. I mean if $G$ is the interior of the intersection of two circles I can manage to do it (Presumably, $f(G)$ will be the open unit disk), but in a general case like this I don't know how to proceed. 
It is possible to do this in the general case? Thanks in advance.
 A: By the assumption that $G$ is a simply connected region of $\mathbb{C}$ such that the boundary $\partial G$ has at least two points $a\ne b$, we may take a path $\Gamma$: $z=\gamma(t)$ $(0\le t<1)$ with $\gamma(0)=a\, (\text{or }b)$, $\gamma(1/2)=b\, (\text{or }a)$, and $\lim_{t\to 1} \gamma(t)=\infty.$
Let $H=\mathbb{C}\setminus \Gamma$, then $H$ is simply connected and $G\subset H.$ 
We assume that $\gamma(0)=a,\,\gamma(1/2)=b$. 
If we could construct explicitly, in function of $a$ and $b$, a conformal map $f$ defined on $H$
such that $f(H)$ is open and bounded in $\mathbb{C}$, $f$ is just desired one (since $G\subset H$, $f(G)$ should be bounded).
We consider a function $\varphi (z)=\sqrt{e^{-i\theta _0}(z-a)}$ on $H$, where $\theta_0=\arg (b-a),\, 0\le \theta _0<2\pi.$
The branch of $\varphi(z)$ is determined so that $\varphi (b)$ is real positive. By $\varphi $ a small circle $C_\varepsilon : z=a+\varepsilon e^{i\theta }\,( 0\le \theta <2\pi)$ is mapped to a half circle, so we may take a point $c$ such that $c\not \in f(H)$. 
Then $\phi(z)=\frac{1}{z-c}$ maps $\varphi (H)$ to a bounded region. Explicitly we may take $c=-\varphi (a+\varepsilon e^{i\pi})$ .Therefore $f(z)=\phi(\varphi (z))$, explicitly in function of $a$ and $b$, maps $H$ to a bounded region and $G$ too.  
