 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-4/11e/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

D = {z : ∣ z ∣ < 1}, g_{D} = \frac{4∣ d z ∣ ^{2}}{( 1 - ∣ z ∣ ^{2} ) ^{2}},

h = {x + i y : y > 0}, g_{h} = \frac{d x ^{2} + d y ^{2}}{y ^{2}} .

The Cayley map

C (z) = i \frac{1 + z}{1 - z}

D

bijectively to

h

. Since

C^{'} (z) = 2 i / (1 - z)^{2}

and

Im C (z) = (1 - ∣ z ∣^{2}) /∣1 - z ∣^{2}

, direct substitution gives

C^{*} g_{h} = g_{D}

Representing

C

K = (i - 1 i 1)

, the disc isometry corresponding to

g = (a c b d)

is represented, up to a nonzero scalar, by

K^{- 1} g K = \frac{1}{2} (a + d + i (b - c) a - d + i (b + c) a - d - i (b + c) a + d - i (b - c)) .

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