 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/iii/solution

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

The exponential power series gives

e^{- 1/ z} = \sum_{m = 0}^{\infty} (- 1)^{m} / (m! z^{m})

for

z \neq = 0

. Hence all singular terms are

m = 1 \sum \infty \frac{( - 1 ) ^{m}}{m ! z ^{m}} = - \frac{1}{z} + \frac{1}{2 z ^{2}} - \frac{1}{6 z ^{3}} + \dots .

Infinitely many negative powers occur, so zero is an essential singularity.

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