Conformal Mappings: Find a conformal bijection from $D_1$ onto $D_2$. I am struggling a bit when it comes to conformal mappings. I have the general idea: preserves angle, analytic, non-zero derivative, etc . . . but the specifics of actually coming up with mapping are eluding me.
Problem
Let $D_1=\{z \in \mathbb{C}: |z-1|>1 \text{ and } |z+1|>1\}$ and $D_2=\{z \in \mathbb{C}: |z|<1\}$
Find a conformal bijection from $D_1$ onto $D_2$.
Attempt
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So the image set is the unit disc, I am having some trouble visualizing but I believe we are looking at the set $D_1=\{z \in \mathbb{C}: -2\Re(z) < |z|^2 < 2\Re(z)\}$
I know the map $\frac{z-1}{z+1}$ maps the unit disc onto the half-plane but I don't have a general strategy for coming up with conformal mappings.
 A: We consider several transformations $\varphi , \phi, \psi$ and $g$. So
\begin{align}
\zeta&=\varphi (z)=\frac{1}{z},\quad \varphi : D_1\to G_1=\{\zeta:-\frac{1}{2}<\operatorname{Re}\,\zeta<\frac{1}{2}\},\\
\xi&=\phi(\zeta)=i\pi \zeta, \quad \phi : G_1\to G_2=\{\xi : -\frac{\pi}{2}<\operatorname{Im}\,\xi<\frac{\pi}{2}\},\\
\eta&=\psi (\xi)=e^\xi,\quad \psi : G_2\to G_3=\{\eta : \operatorname{Re}\, \eta>0\}\\
 w&=g(\eta)=\frac{\eta-1}{\eta+1},  \quad g : G_3\to D_2.
\end{align}
See the diadram below. Then we get $$
w=g\circ \psi\circ\phi\circ\varphi (z)=\frac{\exp\left(\displaystyle\frac{i\pi}{z}\right)-1}{\exp\left(\displaystyle\frac{i\pi}{z}\right)+1}$$
as a conformal bijection from $D_1$ onto $D_2$.

Addendum: The function $\zeta=\varphi (z)=\frac{1}{z}$ maps $\{z : |z-1|>1\}$ onto $\{\zeta : \operatorname{Re}\, \zeta<\frac{1}{2}\}$ and maps $\{z : |z+1|>1\}$ onto $\{\zeta : \operatorname{Re}\, \zeta>-\frac{1}{2}\}$. Therefore $\varphi $ maps $D_1$ onto $\{\zeta : -\frac{1}{2}<\operatorname{Re}\, \zeta<\frac{1}{2}\}$.
