 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-2/13a/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

r < ∣ z - z_{0} ∣ < R

f (z) = \sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n} .

For any positively oriented simple closed contour

C

in the annulus winding once around

z_{0}

a_{n} = \frac{1}{2 πi} \oint_{C} \frac{f ( ζ )}{( ζ - z _{0} ) ^{n + 1}} d ζ .

The formula follows from the Cauchy integral formula applied term by term, and it is independent of the chosen contour by the Cauchy integral theorem.

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