Conformal map from $\mathbb{C}\setminus\{z\in\mathbb{C} : \Re(z)=0, \, |z| \geq 1 \}$ to unit disc I want to find a conformal map from $U=\mathbb{C}\setminus\{z\in\mathbb{C} : \Re(z)=0, \, |z| \geq 1 \}$ to the unit disc. Thus far, this is what i have:

*

*$z \mapsto iz+1$ to map $U$ to $\mathbb{C}\setminus \left( (-\infty,0] \cup [2,\infty)\right)$.

*$z\mapsto \sqrt{z}$ to map to the right half plane minus $[\sqrt{2},\infty)$.

Can someone help me to find a conformal map from the latter to the unit disc?
Thank you in advance.
 A: I think it's best to imagine your set as a subset of the Riemann sphere. Visualizing it that way, it is the sphere with a straight slit going from $\mathrm i$ to $-\mathrm i$ via $\infty$. We can now rotate this slit sphere using a Möbius transformation such that the slit goes from $0$ to $\infty$ via $1$. This gives us a slit plane missing the positive real axis. To do so, use the Cayley transform
$$z\mapsto\frac{z-\mathrm i}{z+\mathrm i}.$$
Now apply $z\mapsto\sqrt z$ to transform the slit plane to the upper half plane. Follow up by applying the Cayley transform again, which transforms the upper half-plane to the unit disc.
The basic idea is to recognize that both the unit disc and the half-plane are hemispheres of the Riemann sphere, and that both your given set and any slit plane are slit Riemann spheres. Now all that remains is to transform a slit sphere to a hemisphere, which the square root does nicely if the slit starts or ends at $0$. So we just have to rotate the slit sphere so that the slit starts at $0$ (that's the first Cayley transform), transform it to a hemisphere (the upper half-plane in this case, that's the square root transformation), and then rotate back so we actually have the unit disc (that's what the second Cayley transform does).
