Let \(\Omega\) be the region inside the unit circle \(\abs{z} = 1\) and outside the circle \(\abs{z-{1\over 4}} = {1\over 4}\).
Find an injective conformal map from \(\Omega\) onto some annulus \(\theset{r < \abs{z} < 1}\) for constant \(r\).
Solution
The two boundary circles are coaxial. Let \[ \alpha=2-\sqrt3, \qquad \beta=2+\sqrt3. \] Then \[ \alpha\beta=1, \qquad \alpha+\beta=4. \] These are the limiting points of the coaxial family: for the outer circle \(|z|=1\) we have \(\alpha\beta=1\), while for the inner circle \(|z-1/4|=1/4\), \[ (\alpha-1/4)(\beta-1/4)=1/16. \] Therefore the Möbius transformation \[ T(z)=\frac{z-\alpha}{z-\beta} \] maps both boundary circles to circles centered at the origin.
To determine their radii, evaluate at convenient boundary points. On \(|z|=1\), using \(z=1\), \[ |T(1)| =\frac{\sqrt3-1}{\sqrt3+1} =2-\sqrt3 =\alpha. \] On \(|z-1/4|=1/4\), using \(z=1/2\), \[ |T(1/2)| =\frac{1/2-\alpha}{\beta-1/2} =\alpha^2. \] Hence \[ T(\Omega)=\{\alpha^2<|w|<\alpha\}. \] After dividing by \(\alpha\), we obtain \[ \boxed{ \Phi(z)=\frac1\alpha\frac{z-\alpha}{z-\beta}, \qquad \alpha=2-\sqrt3,} \] and \[ \Phi(\Omega)=\{\alpha<|w|<1\}. \] Thus the required annulus has inner radius \[ \boxed{r=2-\sqrt3}. \] Since \(\beta>1\) lies outside \(\Omega\), the Möbius map is holomorphic and injective on \(\Omega\).