Find a conformal map from \(\theset{z\in \CC \suchthat \abs{z} < 1,\, \abs{z - {1\over 2}} > {1\over 2} }\) to \(\DD\).
Solution
Let \[ \Omega=\left\{z:|z|<1,\ |z-1/2|>1/2\right\}. \] Both boundary circles pass through \(1\). The Möbius map \[ T(z)=\frac1{1-z} \] sends that common boundary point to infinity. If \(w=T(z)\), then \[ |z|=1 \iff |w-1|=|w| \iff \operatorname{Re}w=\frac12, \] and \[ |z-1/2|=1/2 \iff |w-2|=|w| \iff \operatorname{Re}w=1. \] Testing, for example, \(z=-1/2\in\Omega\) shows that \[ T(\Omega)=\left\{\frac12<\operatorname{Re}w<1\right\}. \]
Now \[ E(w)=\exp\bigl(2\pi i(w-1/2)\bigr) \] maps this vertical strip biholomorphically onto the upper half-plane: its argument ranges from \(0\) to \(\pi\), and its real period is \(1\), larger than the strip width \(1/2\). Finally the Cayley map \[ C(\zeta)=\frac{\zeta-i}{\zeta+i} \] maps the upper half-plane biholomorphically onto \(\mathbb D\).
Therefore one required conformal map is \[ \boxed{ z\longmapsto \frac{ \exp\!\left(2\pi i\left(\frac1{1-z}-\frac12\right)\right)-i }{ \exp\!\left(2\pi i\left(\frac1{1-z}-\frac12\right)\right)+i }.} \]