Let \(C\) and \(C'\) be two circles and let \(z_1 \in C\), \(z_2 \notin C\), \(z'_1 \in C'\), \(z'_2 \notin C'\). Show that there is a unique fractional linear transformation \(f\) with \(f(C) = C'\) and \(f(z_1) = z'_1\), \(f(z_2) = z'_2\).
Solution
1. For a circle \(C\) and a point \(z_2\notin C\), there is a fractional linear transformation \(F\) with \(F(C)=S^1\) and \(F(z_2)=0\).
If \(C\) has center \(c\) and radius \(\rho\), then \(f_1(z)=(z-c)/\rho\) sends \(C\) to \(S^1\). If \(\abs{f_1(z_2)}>1\), let \(f_2(z)=1/z\); otherwise let \(f_2=\id\). In either case \(f_2\) preserves \(S^1\) and \(a\coloneqq f_2(f_1(z_2))\in\DD\). The Blaschke factor \(\psi_a(z)=(a-z)/(1-\bar a z)\) preserves \(S^1\) and sends \(a\) to \(0\). Put \(F\coloneqq\psi_a\circ f_2\circ f_1\).
2. There is a fractional linear transformation \(f\) with \(f(C)=C'\), \(f(z_1)=z_1'\) and \(f(z_2)=z_2'\).
Let \(F\) and \(G\) be the maps of step 1 for \((C,z_2)\) and \((C',z_2')\). Then \(F(z_1)\) and \(G(z_1')\) lie on \(S^1\), so \(\lambda\coloneqq G(z_1')/F(z_1)\) has \(\abs\lambda=1\) and \(h(z)=\lambda z\) preserves \(S^1\). Put \(f\coloneqq \inverseof{G}\circ h\circ F\). Then \(f(C)=\inverseof{G}(S^1)=C'\), \(f(z_2)=\inverseof{G}(0)=z_2'\) and \(f(z_1)=\inverseof{G}(G(z_1'))=z_1'\).
3. If \(f\) and \(\tilde f\) both satisfy the conditions, then \(f=\tilde f\).
Let \(\varphi\coloneqq F\circ\inverseof{\tilde f}\circ f\circ \inverseof{F}\). It is a fractional linear transformation with \(\varphi(S^1)=S^1\), \(\varphi(0)=0\) and \(\varphi(F(z_1))=F(z_1)\). A fractional linear transformation maps the two components of \(\widehat\CC\setminus S^1\) onto the two components, and \(\varphi(0)=0\), so \(\varphi(\DD)=\DD\). By the Schwarz lemma, an automorphism of \(\DD\) fixing \(0\) is a rotation \(z\mapsto\mu z\). It fixes \(F(z_1)\ne0\), so \(\mu=1\) and \(\varphi=\id\), that is, \(f=\tilde f\).
4. Q.E.D.
Steps 2 and 3 give existence and uniqueness.