 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2019/s1/3/d/solution

[- 1, 1]

z^{2} - 1 \sim z

. The Laurent series from the preceding solution has

z^{- 1}

3/8

. Consequently every sufficiently large positively oriented circle gives

\oint_{C_{\infty}} f (z) d z = \frac{3 πi}{4} .

Although

f

grows cubically, its other Laurent series terms have zero contour integral. Choosing the opposite branch of a multivalued function reverses the sign; neither the branch of a multivalued function nor the orientation can be omitted from the answer.

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