 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/7/c/solution

Expand the exponential in inverse powers of

z

g (z) = \sum_{n = 0}^{\infty} \frac{z ^{- n}}{n !} - \sum_{n = 0}^{\infty} \frac{z ^{2 - n}}{n !} = - z^{2} - z + \frac{1}{2} + \sum_{k = 1}^{\infty} (\frac{1}{k !} - \frac{1}{( k + 2 )!}) z^{- k} .

This is a Laurent series, not a Taylor series, and converges throughout

0 < ∣ z ∣ < \infty

. Every displayed coefficient of a negative power is nonzero, so its principal part has infinitely many terms. Thus

z = 0 is an essential singularity .

z^{- 1}

Res_{z = 0} g = 1 - \frac{1}{3 !} = \frac{5}{6} .

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