 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/2012/s1/1/c/solution

Parametrize the positively oriented unit circle by

z = e^{i θ}

. Then

e^{- 1/ z} = e^{- c o s θ + i s i n θ}

and

d z = i e^{i θ} d θ

, so

\oint_{C} e^{- 1/ z} d z = i \int_{0}^{2 π} e^{- c o s θ} cos (θ + sin θ) d θ - \int_{0}^{2 π} e^{- c o s θ} sin (θ + sin θ) d θ .

Equating the real part and imaginary part with

- 2 πi

gives

\int_{0}^{2 π} e^{- c o s θ} cos (θ + sin θ) d θ = - 2 π, \int_{0}^{2 π} e^{- c o s θ} sin (θ + sin θ) d θ = 0.

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