Find a conformal map from the region bounded by \(\abs{z - {i\over 2}} = {1\over 2}\) and \(\abs{z-i} = 1\) to \(\DD\).
Solution
The two circles are internally tangent at \(z=0\), not at \(z=i\). The desired region is the part inside \(|z-i|<1\) and outside \(|z-i/2|<1/2\).
Since both boundary circles pass through \(0\), inversion sends them to parallel lines. If \[ w={1\over z}, \] then the circle \[ |z-ia|=a \] has equation \(|z|^2=2a\Im z\), hence on its image \[ \Im w=-{\Im z\over |z|^2}=-{1\over 2a}. \] Therefore \[ |z-i|=1\longmapsto \Im w=-{1\over2}, \qquad |z-i/2|={1\over2}\longmapsto \Im w=-1. \] The lune maps to the horizontal strip \[ -1<\Im w<-{1\over2}. \]
Now set \[ \zeta=2\pi(w+i). \] Then \(0<\Im\zeta<\pi\), so \(u=e^\zeta\) maps the strip biholomorphically onto \(\HH\). Finally the Cayley map \[ C(u)={u-i\over u+i} \] sends \(\HH\) to \(\DD\).
Thus one conformal map is \[ F(z) =\frac{\exp\!\left(2\pi\left(z^{-1}+i\right)\right)-i} {\exp\!\left(2\pi\left(z^{-1}+i\right)\right)+i}. \]