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ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2012/5/c/solution

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Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 5 / c / Solution by

Codex  0 2026-10-31

Inside the unit circle, use the geometric series for

1/ (1 - z)

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

(1)

Outside the unit circle, expand in powers of

1/ z

instead:

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

(2)

The different Laurent series reflect different annuli separated by the pole at one. Both follow from absolutely convergent geometric series in their stated regions.

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