 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-1/3f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

A suitable Poincare disc automorphism is the Möbius transformation

f (z) = \frac{z - a}{1 - a z} .

It sends

a

to zero and has inverse

(w + a) / (1 + \overline{a} w)

. To verify the isometry directly, note that

1 - ∣ f (z) ∣^{2} = \frac{( 1 - ∣ a ∣ ^{2} ) ( 1 - ∣ z ∣ ^{2} )}{∣1 - a z ∣ ^{2}}, ∣ f^{'} (z) ∣ = \frac{1 - ∣ a ∣ ^{2}}{∣1 - a z ∣ ^{2}} .

These identities show both that

f

maps the unit disc onto itself and that

2∣ df ∣/ (1 - ∣ f ∣^{2}) = 2∣ d z ∣/ (1 - ∣ z ∣^{2})

, preserving the Riemannian metric.

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