 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2016/s1/3/a/solution

z = e^{i θ}

. The integral becomes

\oint_{∣ z ∣ = 1} \frac{d z}{i [( 1 - i /2 ) z ^{2} + 3 z + ( 1 + i /2 )]} .

The roots are

z = (- 3 \pm 2) / (2 - i)

, only

z_{+} = - 1/ (2 - i)

lying inside. The derivative of the quadratic at

z_{+}

is two, so the residue is

1/ (2 i)

. Consequently

\int_{0}^{2 π} d θ / (3 + 2 cos θ + sin θ) = π

. The denominator is positive since

3 > 5

, so no principal-value qualification is required.

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