Conformal map from exterior of unit circle to upper half plane

I'm trying to find a conformal map from the space $\Omega = \mathbb{H}\setminus\{z : |z-\frac{i}{2}|\leq\frac{1}{2}\}$ to the upper half plane. I think I'm most of the way there, but I wanted to check on the last part. Both $\Omega$ and $\mathbb{H}$ are simply connected, as far as I can tell, so this should be possible.

My idea was to map this first to the exterior of the unit circle, and then from there to $\mathbb{H}$. The first part can be achieved by composing $f_1(z)=2z-i$, $f_2(z)=-iz$, and $f_3(z)=z^2$ to lower/expand the disc to $\mathbb{D}$, and extend the right-side exterior to the whole exterior of $\mathbb{D}$.

Now, at this point, I think that $f_4(z)=i\frac{z-1}{z+1}$ will map $\mathbb{C}\setminus\mathbb{D} \to \mathbb{H}$, but I'm having trouble proving it. If this works, then the composition of those functions would be the transformation I'm looking for. Am I completely off-base here, or no?

Your composition unfortunately does not work out. We have

$$\Omega_1 = f_1(\Omega) = \{z : \operatorname{Im} z > -1\} \setminus \overline{\mathbb{D}},$$

and thus

$$\Omega_2 = f_2(\Omega_1) = \{z : \operatorname{Re} z > -1\} \setminus \overline{\mathbb{D}},$$

and both $f_1$ and $f_2$ are biholomorphic. But then $f_3$ is not injective on $\Omega_2$. While indeed $f_3(\Omega_2) = \mathbb{C}\setminus \overline{\mathbb{D}}$, the mapping is only locally conformal, not conformal. And $\mathbb{C}\setminus \overline{\mathbb{D}}$ is not simply connected, hence there is no conformal map from that onto the upper half-plane (you get a conformal map $\widehat{\mathbb{C}}\setminus \overline{\mathbb{D}} \to \mathbb{H}$ from $f_4$, though).

To get a conformal map $\Omega \to \mathbb{H}$, the usual way is to start by using a Möbius transformation to map the point of contact of the two boundary curves to $\infty$. That maps the two boundary curves to parallel lines (since both are circles or straight lines), and $\Omega$ to a parallel strip. Mapping a parallel strip conformally to a half-plane is a standard exercise.

Taking $g_1(z) = \frac{1}{z}$ for the Möbius transformation, the real line is mapped to itself (interchanging $0$ and $\infty$), and the circle $\left\lvert z - \frac{i}{2}\right\rvert = \frac{1}{2}$ is mapped to the line $\operatorname{Im} z = -1$, and $\Omega$ to the parallel strip $\{ z : -1 < \operatorname{Im} z < 0\}$ between these two lines.

The map $g_2(z) = \exp\left(\pi\left(z + \frac{i}{2}\right)\right)$ conformally maps the strip to the right half-plane, and a rotation then maps the right half-plane to the upper. Altogether, we obtain the conformal map

$$F \colon z \mapsto i\cdot \exp\left(\pi\left(\frac{1}{z}+\frac{i}{2}\right)\right)$$

from $\Omega$ to $\mathbb{H}$.

• Thanks for the correction! Commented Apr 2, 2014 at 16:47