 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2007/6/i/solution

On the unit circle, put

z = e^{i θ}

. Then

d θ = d z / (i z)

and

cos θ = (z + z^{- 1}) /2

, so

\frac{d θ}{( 2 - c o s θ ) ^{2}} = \frac{4}{i} \frac{z d z}{( z ^{2} - 4 z + 1 ) ^{2}} .

Write

r = 2 - 3

and

s = 2 + 3

. Only

r

h (z) = z / [(z - r)^{2} (z - s)^{2}]

has a double pole there. Its residue is

Res_{z = r} h = \frac{d}{d z} \frac{z}{( z - s ) ^{2}}_{z = r} = \frac{1}{( r - s ) ^{2}} - \frac{2 r}{( r - s ) ^{3}} = - \frac{r + s}{( r - s ) ^{3}} = \frac{1}{6 3} .

\int_{0}^{2 π} \frac{d θ}{( 2 - cos θ ) ^{2}} = \frac{4}{i} 2 πi \frac{1}{6 3} = \frac{4 π}{3 3} .

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