 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ib/paper-1/12f/c/i/solution

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

Since

tan z = sin z / cos z

a_{k} = \frac{( 2 k + 1 ) π}{2}, k \in Z .

At each one,

sin a_{k} \neq = 0

and

(cos z)^{'} ∣_{a_{k}} = - sin a_{k} \neq = 0

, so all are simple poles. There are no other finite singularities:

tan z is meromorphic with simple poles at a_{k} .

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