 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/13a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Using the Taylor series of the exponential and series division,

\frac{1}{e ^{2 z} - 1} = \frac{1}{2 z} - \frac{1}{2} + \frac{z}{6} - \frac{z ^{3}}{90} + O (z^{5}) .

Thus the first three nonzero Laurent terms are

\frac{1}{2 z} - \frac{1}{2} + \frac{z}{6} .

The singularities occur at

z = kπi

. The nearest nonzero ones are

\pm πi

, so this Laurent series is valid on $0 < ∣ z ∣ < π$ .

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