 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/12b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

f (z) = - \frac{1}{2 z} + \frac{1}{2 ( z - 2 )} .

For

0 < ∣ z ∣ < 2

\frac{1}{2 ( z - 2 )} = - \frac{1}{4} \frac{1}{1 - z /2} = - \sum_{n = 0}^{\infty} \frac{z ^{n}}{2 ^{n + 2}} .

Hence

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

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