Finding a conformal map from this domain into the unit disc Problem: Find a conformal map $f$ from $ A = \left\{ z \in \mathbb{C} \mid \text{Im}(z) &gt; 0, |z| &gt; 1 \right\}$ into the unit disk.

Attempt: I started off with a map $F_1: z \mapsto z + \frac{1}{z}$. I was trying to visualize what this mapping actually does to the region in the complex plane. I know that the semi-circle in the upper half plane will get mapped to the interval $[-2, 2]$ on the real line. Also, a point like $2i$ gets mapped to $3i/2$. 

I wish to map the region $A$ into the whole upper half plane (if that is possible), then I can easily find a map into the unit disc. Do I need to use a dilatation to rescale the $|z| &gt; 1$ condition? 

Help is appreciated.
Hint: With conformal mapping 
$$w=\left(\dfrac{z+1}{z-1}\right)^2$$
you map region $A=\{z\in\mathbb{C}:{\bf Im}\ z&gt;0,\ |z|&gt;1\}$ to lower half plane $\{z\in\mathbb{C}:{\bf Im}\ z&lt;0\}$.
It seems that your region $A$ is the limit as $R \to + \infty$ of
$$
A_R=\{z \in \mathbb C, \Im(z)&gt;0, 1&lt;|z|&lt;R \}=\{r\exp(i\theta), \theta \in (0,\pi), r \in (1,R) \}.
$$
I think that such a "half-annulus" can be conformally mapped into the upper plane for large $R$. Then you may follow the lines of Conformal maps from the upper half-plane to the unit disc has the form