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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ib/paper-1/3f/solution

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Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 3F / Solution by

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

The partial fraction decomposition is

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

. Put

w = z - 1

and expand each reciprocal by a geometric series:

\frac{1}{1 \mp i + w} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n} w ^{n}}{( 1 \mp i ) ^{n + 1}}, ∣ w ∣ < 2 .

Adding gives the Taylor series

f (z) = n = 0 \sum \infty (- 1)^{n} [(1 - i)^{- n - 1} + (1 + i)^{- n - 1}] (z - 1)^{n} .

Equivalently, since

1 \pm i = 2 e^{\pm iπ /4}

, its coefficient is

(- 1)^{n} 2^{(1 - n) /2} cos ((n + 1) π /4)

. The first terms are

1 - \frac{1}{2} w^{2} + \frac{1}{2} w^{3} - \frac{1}{4} w^{4} + \dots

.

The nearest nonremovable poles are

i

and

- i

, both at distance

2

from the center. The function is holomorphic throughout the open disk of that radius and cannot be holomorphic across either pole. Therefore the largest radius of convergence is $r = 2$ .

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