A fixed point of $A$ satisfies
$$
cz^2+(d-a)z-b=0.
$$
Its discriminant is
$$
(d-a)^2+4bc=(a+d)^2-4=(\operatorname{tr}A)^2-4<0,
$$
where $ad-bc=1$ was used. Thus the two roots are nonreal complex conjugates, and exactly one lies in the upper half-plane. Consequently \b[every <elliptic element of PSL2(R)> fixes a point of $\mathbb H$.]

Solution

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

{title2=$|\operatorname{tr}A|<2$}

= Elliptic Möbius transformation
{synonym}

A real <Möbius transformation> represented by $A\in\mathrm{SL}_2(\mathbb R)$ with $|\operatorname{tr}A|<2$. It fixes a point of the <Poincare half-plane model>; an integral matrix of this type has <trace> $0$ or $\pm1$. In the <Poincare disc model>, moving its fixed point to zero makes it a rotation about zero.

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