 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-4/8b/a/solution

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

f (z) = - 1/ (z - 1) + 1/ (z - 2)

. In the interior disk, expand both terms as geometric series:

- \frac{1}{z - 1} = \sum_{n = 0}^{\infty} z^{n}, \frac{1}{z - 2} = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 1}} .

f (z) = n = 0 \sum \infty (1 - 2^{- n - 1}) z^{n}, ∣ z ∣ < 1.

The nearer pole at

z = 1

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