Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 3 / 3A (Complex Methods)

Table of contents
- a 3A
  - Solution a
- b 3A
  - Solution b

a

Solution

Uniqueness of Laurent coefficients applied to

f (z) = f (- z)

gives

\sum_{n} a_{n} z^{n} = \sum_{n} (- 1)^{n} a_{n} z^{n} .

Hence $a_{n} = (- 1)^{n} a_{n}$ , so every coefficient with odd $n$ vanishes.

b

Solution

The poles inside

C_{n}

are zero and

\pm kπ

for

1 \leq k \leq n

. At a nonzero pole,

Res_{kπ} \frac{1}{z ^{3} s i n z} = \frac{( - 1 ) ^{k}}{k ^{3} π ^{3}},

and the residue at

- kπ

is its negative. At zero, the Laurent series contains only even powers because the function is even, so its residue is zero. All residues cancel and therefore

I_{n} = 0 (n = 0, 1, 2, \dots) .

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