 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-1/7/d/solution

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

. Let

w = z - 1

. Each fraction has a geometric expansion, so

h (z) = j = 0 \sum \infty (- 1)^{j} [\frac{1}{( 1 - i ) ^{j + 1}} + \frac{1}{( 1 + i ) ^{j + 1}}] (z - 1)^{j} .

Equivalently, its coefficient is

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

. The first terms are

h (z) = 1 - \frac{1}{2} w^{2} + \frac{1}{2} w^{3} - \frac{1}{4} w^{4} + O (w^{6}) .

The linear and fifth-order coefficients vanish. Both geometric series converge for

∣ w ∣ < 2

, and

h

has genuine simple poles at

z = \pm i

2

from the centre. Thus the Taylor series has

R = 2 .

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