 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2021/s01/1/d/solution

Parametrize the positively oriented unit circle by

z = e^{i θ}

. Then

d z = i z d θ

and

\frac{z ^{4} - 10 z ^{2} + 1}{z ^{2}} = 2 cos (2 θ) - 10 = - 4 (2 + sin^{2} θ) .

Consequently the contour integral already evaluated satisfies

I = - \frac{i}{4} \int_{0}^{2 π} \frac{d θ}{2 + s i n ^{2} θ}, \int_{0}^{2 π} \frac{d θ}{2 + sin ^{2} θ} = 4 i I = \frac{2 π}{6} .

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