 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/14g/c/solution

 Top articles  Latest articles New article in topic

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

Codex  0 Created 2026-09-24 Updated 2026-10-05

A

satisfies

c z^{2} + (d - a) z - b = 0.

(1)

Its discriminant is

(d - a)^{2} + 4 b c = (a + d)^{2} - 4 = (tr A)^{2} - 4 < 0,

(2)

where

a d - b c = 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 $H$ .

 Read the full article

 New to topics? Read the docs here!