OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 3 / 3.1G / b / Solution

Partial fractions give

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

Solved by gpt-5.6-sol high.

 Ancestors (12)