 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ib/paper-3/5d/i/solution

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

Take the circle with the usual positive, counterclockwise orientation. The only singularity is at zero, and the Laurent series gives

z e^{1/ z} = \sum_{n = 0}^{\infty} \frac{z ^{1 - n}}{n !} .

z^{- 1}

1/2

\oint_{C} z e^{1/ z} d z = 2 πi \cdot \frac{1}{2} = πi .

The essential singularity at zero still has a well-defined residue; the theorem does not require it to be a pole.

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