Map $ \mathbb C\setminus\gamma $ conformally to a punctured disk, where $ \gamma = \{ z\in S:Re(z)\le 0 \} $ 
Let $ S $ denote the circle of radius $ \frac{1}{2} $ centered at $ \frac i 2\in\mathbb C $ and let 
  $$ \gamma = \{ z\in S:Re(z)\le 0 \} $$
  be the semicircle. 
  Find a one-to-one and onto holomorphic map from $ \mathbb C\setminus\gamma $ (the complement of $ \gamma $ in $ \mathbb C $ ) to the punctured disk
  $$ \Delta^*=\{ z\in\mathbb C:0<|z|<1 \}. $$

Can someone give me a hint? 

Edit:
Here is what I have by now:
We start from both $ \mathbb C\setminus\gamma $ and $ \Delta^* $.
On the $ \mathbb C\setminus\gamma $ side, we have:
$$ \begin{align}
&1^\circ\ \text{Translation(Move the center of $ S $ to $ 0 $):}& & z\mapsto z-\frac i 2 \\
&2^\circ\ \text{Rotation(Rotate $ \frac\pi 2 $ clockwise):}& & z\mapsto\exp(-\frac{\pi}{2}i)z \\
&3^\circ\ \text{Linear Fractional Transformation(Map the semicircle to $yi$, $ y
>0 $): }& &z\mapsto\frac{z+1}{z-1}\\
&4^\circ\ \text{Rotation(Rotate $ \frac\pi 2 $ clockwise):}&& z\mapsto\exp(-\frac{\pi}{2}i)z\\
&5^\circ\ \text{Logarithm:}& & z\mapsto \log(z)\\
&6^\circ\ \text{Scale:}& & z\mapsto\frac 1 2 z
\end{align} $$
Through the above process, we get the region $$ \Omega_1:\{ z\in\mathbb C: Im(z)\in (0, \pi ) \} .$$
On the other side, starting from $ \Delta^* $, we have:
\begin{align}
&1^\circ\ \text{Logarithm:}&& z\mapsto\log(z)\\
&2^\circ\ \text{Rotation(Rotate $ \frac\pi 2 $ clockwise):}&& z\mapsto\exp(-\frac{\pi}{2}i)z\\
&3^\circ\ \text{Scale:}& & z\mapsto\frac 1 2 z\\
&4^\circ\ \sin (z):&& z\mapsto \sin(z)\\
&5^\circ\ \text{Logarithm:}&& z\mapsto\log(z)
\end{align}
And we get the region:
$$ \Omega_2:\{z\in\mathbb C: Im(z)\in[0,\pi)\} .$$
But they still vary from the real axis. So how to move on?
 A: Take a Mobius transformation $\mu$ that maps $\gamma$ to $[-1, 1]$. It maps $\mathbb C \setminus \gamma$ to $\overline {\mathbb C} \setminus ([-1, 1] \cup \{\mu(\infty)\})$. Take the branch $J^{-1}$ of the inverse Joukowsky transformation that maps $\overline {\mathbb C} \setminus [-1, 1]$ to the open unit disk $\mathbb D$. The composition $J^{-1} \circ \mu$ maps $\mathbb C\setminus \gamma$ to $\mathbb D \setminus \{J^{-1}(\mu(\infty))\}$. Then take a Mobius transformation that maps $\mathbb D$ to itself and maps $J^{-1}(\mu(\infty))$ to $0$.
A: This is a detailed version of @Maxim's:
$ 1^\circ $ Take a Mobius transformation $ \mu $ that maps $ \gamma $ to $ [-1,1] $: Let's find such a Mobius transformation, such that 
\begin{align} &z_1=&i&\mapsto 1&=\omega_1\\
&z_2=&-\frac 1 2+\frac 1 2i&\mapsto 0&=\omega_2\\
&z_3=&0&\mapsto -1&=\omega_3
\end{align}
Since Mobius transformation preserves the corss-ratio, we have:
\begin{align}
\frac{\omega-\omega_1}{\omega-\omega_2}:\frac{\omega_3-\omega_1}{\omega_3-\omega_2}&=\frac{z-z_1}{z-z_2}:\frac{z_3-z_1}{z_3-z_2}\\
\frac{\omega-1}{\omega}\cdot\frac{-1}{-1-1}&=\frac{z-i}{z+\frac{1}{2}-\frac{1}{2}i}\cdot\frac{\frac{1}{2}-\frac 1 2 i}{-i}\\
1-\frac 1 \omega&=\frac{2z-2i}{2z+1-i}\cdot(1+i)\\
\omega&=\frac{2z+1-i}{-2iz-1+i}
\end{align}
Note that $$ \mu(\infty)=i .$$
$2^\circ$ Take the branch of inverse Joukowsky transformation:
$$ J^{-1}(z)=z-\sqrt{z^2-1} $$ 
mapping $ \overline{\mathbb C}\setminus [-1,1] $ to the unit open disk $ \mathbb D .$
Note that $$ J^{-1}(\mu(\infty))=(1-\sqrt{2})i .$$
$ 3^\circ $ Take a Mobius transformation that maps $ \mathbb D $ to itself and maps $ \mu(\infty)=i $ to $ 0 $. It is of the form:
$$ z\mapsto \omega=e^{i\theta}\frac{z-(1-\sqrt{2})i}{1-z\overline{(1-\sqrt{2})i}} $$ 
Take $ \theta=0 $ and we get
$$ z\mapsto \omega=\frac{z-(1-\sqrt 2)i}{1-z(\sqrt 2-1)i} $$
A: EDIT: the answer below is wrong, as a result of my misunderstanding of what the original region was.

I feel kind of stupid for not seeing how to do this sooner. A good sleep helped, I guess.
The circular disk whose right half you want to exclude is tangent to the real axis at the origin, and its maximum point is $i$. Let’s call the circle itself (not the disk) $C$, the right semicircle $C_R$, the left $C_L$. The vertical diameter, running from the origin to $i$ I’ll call $D$.
Use the map $z\mapsto1/z$ and see what happens. $C_R$ gets sent to the horizontal ray going rightwards from $-i$ and $D$ goes to the vertical ray going downwards from $-i$. ($C_L$, which is in the midst of the region of our interest, goes to the horizontal ray that runs leftward from $-i$.)
Thus the bad region, the righthand interior of the circular disk, goes to the right-angled wedge going SE from the point $-i$, while the good region is everything else, that is, $\{z:\Re(z)<0\}\cup\{z:\Im(z)>-1\}$.
From here on, it’s clear what you need to do: move up by one unit, $z\mapsto z+i$ to get your wedges with their vertex at the origin, and then apply the conformal map $z\mapsto z^{2/3}$, which has the effect of straightening out the $270^\circ$ angle to $180^\circ$. Now the good region is a half-plane, in fact the upper half-plane. Note that $\infty$, which is, so to speak, the isolated point in the exterior of our original good region, has gone first to $0$, the to $i$, and to $\frac12+i\frac{\sqrt3}2\,$. I’m going to call this number $\zeta$: it’s a primitive sixth root of unity.
Finally, map the upper half-plane to the interior of the unit circle by
$$
z\mapsto\frac{z-\zeta}{z-\bar\zeta}\,,
$$
in which you see that real numbers get sent to the unit circle, while $\zeta$ gets sent to the origin. Compose all the steps in the right order, see that $i\mapsto-i\mapsto0\mapsto0\mapsto-\frac12+i\frac{\sqrt3}2$; $\infty\mapsto0\mapsto i\mapsto\zeta\mapsto0$; $\frac i2\mapsto-2i\mapsto-i\mapsto\frac{\sqrt3}2+\frac i2\mapsto\zeta$; and $0\mapsto\infty\mapsto\infty\mapsto\infty\mapsto1$.
And there you are.
