Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 7 / c

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- Solution c

Solution

Multiply the exponential power series by the two prefactors:

f (z) = \sum_{n = 0}^{\infty} \frac{z ^{1 - n}}{n !} + \sum_{n = 0}^{\infty} \frac{z ^{- 1 - n}}{n !} .

Collecting powers yields the Laurent series

f (z) = z + 1 + m = 1 \sum \infty (\frac{1}{( m + 1 )!} + \frac{1}{( m - 1 )!}) z^{- m}, 0 < ∣ z ∣ < \infty.

Its first terms are

z + 1 + 3/ (2 z) + 7/ (6 z^{2}) + 13/ (24 z^{3}) + \dots

. Infinitely many negative-power coefficients are nonzero, so

z = 0 is an essential singularity

. The residue is the coefficient of

z^{- 1}

:

Res_{z = 0} f = \frac{1}{2 !} + \frac{1}{0 !} = \frac{3}{2} .

The phrase "power series" here refers to a Laurent series; no ordinary Taylor series about an essential singularity exists.

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