 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/4/a/solution

f (z) = \sum_{j = - m}^{\infty} c_{j} (z - z_{0})^{j}, 0 < ∣ z - z_{0} ∣ < R,

about a pole of order

m

Res_{z = z_{0}} f = c_{- 1} = \frac{1}{2 πi} \oint_{∣ z - z_{0} ∣ = ρ} f (z) d z, 0 < ρ < R,

where the circle is traversed anticlockwise. Equivalently, the residue is

\frac{1}{( m - 1 )!} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - z_{0})^{m} f (z)]_{z = z_{0}}

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