 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ib/paper-1/3/3-1g/b/solution

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

f = \frac{1}{2 z ^{2}} - \frac{3}{4 z} + \frac{1}{z + 1} - \frac{1}{4 ( z + 2 )} .

For

1 < ∣ z ∣ < 2

this becomes

\frac{1}{2 z ^{2}} - \frac{3}{4 z} + n \geq 0 \sum (- 1)^{n} z^{- n - 1} - \frac{1}{8} n \geq 0 \sum (- z /2)^{n} .

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