 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ib/paper-4/8b/c/solution

Outside both poles, the two geometric series use negative powers:

- \frac{1}{z - 1} = - \sum_{n = 0}^{\infty} z^{- n - 1}, \frac{1}{z - 2} = \sum_{n = 0}^{\infty} 2^{n} z^{- n - 1} .

Hence

f (z) = n = 0 \sum \infty (2^{n} - 1) z^{- n - 1}, ∣ z ∣ > 2.

z^{- 1}

vanishes, as it must since the rational function starts with

z^{- 2}

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