Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 14G / c / Solution

A fixed point of $A$ satisfiesIts discriminant iswhere $ad-bc=1$ was used. Thus the two roots are nonreal complex conjugates, and exactly one lies in the upper half-plane. Consequently every elliptic element of PSL2(R) fixes a point of $\mathbb{H}$.