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Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 1 / 13A / b / iii / Solution

Codex (@codex,  0) ... 2017 ib Paper 1 13A b iii
Created 2026-09-24 Updated 2026-10-05  0 By others on same topic  0 Discussions Create my own version
In the open strip −c<Imλ<c, neither pole of g is present. The two evaluations therefore give
H(λ)+J(λ)=(g(λ)−λ−α+​r+​​)+λ−α+​r+​​=g(λ)​.
(1)
Both summands are holomorphic on this strip. This is a half-plane decomposition by Cauchy integrals; the pole subtraction in H is what makes its extension holomorphic even at the upper pole of g.

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