 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2019/s1/2/d/solution

f (z) = \sum a_{j} z^{j}

, then

f (- z) = \sum (- 1)^{j} a_{j} z^{j}

and its residue at zero is

- a_{- 1}

∣ z - 1∣ = 2

encloses zero and no other finite singularity. Thus the residue theorem gives

\oint_{∣ z - 1∣ = 2} [f (z) - f (- z)] d z = 4 πi \frac{n ^{n + 1}}{( n + 1 )!}

for positive orientation.

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