Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 2 / 16B / b

Table of contents
- Solution b

Solution

If the pole has order

k + 1

, multiplying the Laurent series by

(z - a)^{k + 1}

gives a function

h

with a holomorphic extension across

a

:

h (z) = c_{- (k + 1)} + c_{- k} (z - a) + \dots + c_{- 1} (z - a)^{k} + \dots .

Its coefficient of

(z - a)^{k}

is exactly

c_{- 1}

. The Taylor series coefficient formula gives

h^{(k)} (a) = k! c_{- 1}

. Continuity of this derivative at

a

therefore proves

res (f, a) = \frac{h ^{(k)} ( a )}{k !} = z \to a lim \frac{h ^{(k)} ( z )}{k !} .

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