 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ib/paper-2/12b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For

∣ z ∣ > 2

, expand directly in inverse powers:

f (z) = \frac{1}{z ^{2}} \frac{1}{1 - 2/ z} .

Therefore the outer Laurent series is

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

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