3.5

3.5 Prove that any Moebius transformation different from the identity map can have at most two fixed points.
—————————————————————————–
Proof: Assume that \(f(z)\) has 3 or more fixed points. Then for some complex numbers \(z_1,z_2,z_3\), then \(f(z_1)=z_1,f(z_2)=z_2,f(z_3)=z_3\). There exists a Moebius Transformation, call it \(g(z)\), where \(g(z_1)=0, g(z_2)=1,\) and \(g(z_3)=\infty\). So \(g(f(z_1))=0,g(f(z_2))=1, g(f(z_3))=\infty\). Then by proposition 3.16, the composition of \(g\) and \(f\), \(g(f(z))=g(z)\) for every \(z \in \mathbb{C}\). Thus \(g(f(z))=g(z)\) for every \(z \in \mathbb{C}\). Therefore, for all \(z \in \mathbb{C}\), we know \(f(z)=z\), so \(f(z)\) is the identity map.