 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s1/2/c/solution

For a positively oriented circle of any radius

R > 0

\oint_{∣ z ∣ = R} f (z) d z = 2 πi \frac{n ^{n + 1}}{( n + 1 )!} .

Only the

z^{- 1}

coefficient in the Laurent series contributes. Reversing the orientation changes the sign.

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