 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/7/b/i/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Put

w = z - iπ

. Since

sinh (w + iπ) = - sinh w

and

sinh w = w (1 + w^{2} /6 + O (w^{4}))

, the local Laurent series is

\frac{1}{s i n h ^{3} z} = - w^{- 3} (1 + w^{2} /6 + O (w^{4}))^{- 3} = - w^{- 3} + \frac{1}{2} w^{- 1} + O (w) .

- \frac{1}{( z - iπ ) ^{3}} + \frac{1}{2 ( z - iπ )} .

There is a third-order pole, with no second-order term.

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