Solution
= Solution
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 $\mathbb H$.]