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

e^{- 1/ z} = \sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{n !} z^{- n} = 1 - \frac{1}{z} + \frac{1}{2 z ^{2}} - \frac{1}{6 z ^{3}} + \dots, z \neq = 0.

(1)

Its principal part contains infinitely many nonzero negative powers, so zero is an essential singularity. The first three displayed terms are not an asymptotic truncation as

z \to 0

; they are the beginning of the convergent Laurent series in powers of

1/ z

b

 0  0

Solution

 0  0

The coefficient of

z^{- 1}

in the Laurent series is

- 1

. The residue theorem therefore gives, for the standard positive orientation,

\oint_{C} e^{- 1/ z} d z = 2 πi Res_{z = 0} e^{- 1/ z} = - 2 πi .

(1)

The residue theorem applies to an isolated essential singularity as well as to a pole.

c

 0  0

Solution

 0  0

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 .

(1)

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.

(2)

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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

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