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. A fractional linear transformation is determined by its values at three distinct points.
a Möbius map \(z \mapsto \frac{az+b}{cz+d}\) has three degrees of freedom (up to scaling), so prescribing three images determines it uniquely.
2. The reflection of \(z_2\) across the circle \(C\) is a point \(z_2^*\) (the inverse point of \(z_2\) with respect to \(C\)), and similarly \(z_2'^*\) for \(z_2'\) across \(C'\).
definition of the inverse point with respect to a circle.
3. A Möbius map sends \(C\) to \(C'\) iff it sends the pair \(\{z_2, z_2^*\}\) to the pair \(\{z_2', z_2'^*\}\).
a Möbius map preserves the relation of being inverse points with respect to a circle (it preserves circles and angles, hence symmetric points).
4. Hence \(f\) is determined by the three conditions \(f(z_1) = z_1'\), \(f(z_2) = z_2'\), \(f(z_2^*) = z_2'^*\).
3 and the fact that \(z_1 \in C\) maps to \(z_1' \in C'\).
5. These three conditions determine \(f\) uniquely.
1 (three points determine a Möbius map).
6. Q.E.D.
5.