Show that there is no bijective conformal map from \(A\da \ts{0< \abs{z} < 1}\) to \(B\da \ts{1<\abs{z} < 2}\).
Solution
Suppose toward a contradiction that \(f: A\to B\) is a bijective conformal map. Since \(f(A) \subseteq B\), which is bounded, the isolated singularity of \(f\) at \(0\) is removable. So \(f\) extends holomorphically to \(F: \DD\to \bar{B}\). As \(F\) is nonconstant, it is an open map, so \(F(\DD)\) is open and contained in \(\bar B\), hence \(F(\DD) \subseteq B\). Write \(w_0 \da F(0) \in B\). By surjectivity there is a \(z_0\in A\) such that \(f(z_0) = w_0\). Choose disjoint open disks \(U\ni z_0\) and \(V\ni0\) in \(\DD\) with \(0\notin U\). Since \(F\) is open, \(F(U)\intersect F(V)\) is an open set containing \(w_0\), so it contains a point \(w\neq w_0\). Then \(w=F(v)\) for some \(v\in V\) with \(v\neq0\) (as \(F(0)=w_0\)), and \(w=F(u)\) for some \(u\in U\). Both \(u\) and \(v\) lie in \(A\) and \(u\neq v\) since \(U\intersect V\) is empty, so \(f(u)=f(v)\) contradicts injectivity of \(f\).