 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2004/ib/paper-1/5a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

z^{2} (1 + z^{2})

. Near zero,

\frac{1}{z ^{2} ( 1 + z ^{2} )} = z^{- 2} - 1 + z^{2} - \dots,

so zero is a double pole with residue zero. At

z = \pm i

the poles are simple and the denominator derivative is

2 z + 4 z^{3} = - 2 z

. Therefore

Res_{0} = 0, Res_{i} = \frac{i}{2}, Res_{- i} = - \frac{i}{2} .

These are all the poles.

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