 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2021/5/a/i/solution

\frac{1}{( z + 1 ) ( z - 2 )} = - \frac{1}{3 ( z + 1 )} + \frac{1}{3 ( z - 2 )} .

For

1 < ∣ z ∣ < 2

, expand the first fraction in negative powers and the second in nonnegative powers using the geometric series:

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

\frac{1}{( z + 1 ) ( z - 2 )} = - \frac{1}{3} n = 0 \sum \infty (- 1)^{n} z^{- n - 1} - \frac{1}{3} n = 0 \sum \infty \frac{z ^{n}}{2 ^{n + 1}}, 1 < ∣ z ∣ < 2.

For

∣ z ∣ > 2

, both fractions expand in negative powers:

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

Combining the two geometric series gives

\frac{1}{( z + 1 ) ( z - 2 )} = \frac{1}{3} n = 0 \sum \infty (2^{n} - (- 1)^{n}) z^{- n - 1}, ∣ z ∣ > 2.

The

z^{- 1}

coefficient is zero, consistently with the original rational function being

O (z^{- 2})

at infinity. Each displayed series converges absolutely on its stated annulus and uniformly on every compact subset of that annulus.

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