 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/12b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Put

w = z - 1

. Then

f (z) = \frac{1}{( w + 1 ) ( w - 1 )} = - \frac{1}{1 - w ^{2}} .

Thus

f (z) = - n = 0 \sum \infty (z - 1)^{2 n}, 0 < ∣ z - 1∣ < 1.

This is actually a Taylor series, because

z = 1

is a regular point. The point

z = 0

f (1/ w) = \frac{w ^{2}}{1 - 2 w}

extends analytically to

w = 0

f

, with extended value zero.

For the real integral, first write

\frac{2 - c o s θ}{5 - 4 c o s θ} = \frac{1}{4} + \frac{3}{4 ( 5 - 4 c o s θ )} .

Let

J = \int_{0}^{2 π} \frac{d θ}{5 - 4 c o s θ} .

With

z = e^{i θ}

J = \oint_{∣ z ∣ = 1} \frac{d z}{i ( - 2 z ^{2} + 5 z - 2 )} .

1/2

and

2

, so only

1/2

lies inside the contour. Its residue is

1/ (3 i)

, and the residue theorem gives

J = 2 πi \frac{1}{3 i} = \frac{2 π}{3} .

Consequently

\int_{0}^{2 π} \frac{2 - cos θ}{5 - 4 cos θ} d θ = \frac{π}{2} + \frac{3}{4} \frac{2 π}{3} = π .

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