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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/24g/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 1 / 24G / iii / Solution by

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

z w - 2 i w - i z - 2 = (z - 2 i) (w - i) .

(1)

Thus

Y

is the union of the two complex lines

z = 2 i

and

w = i

, meeting at

(2 i, i)

. A sufficiently small neighborhood of the intersection, with that point removed, has two connected components, whereas a punctured disc is connected. The intersection therefore has no neighborhood homeomorphic to an open subset of

C

. Hence

Y

cannot be a Riemann surface.

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