 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-1/3e/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Here

a_{2 n} = (- 1)^{n}

and

a_{2 n + 1} = 0

, so

\sum a_{n} z^{n} = \sum (- z^{2})^{n} = (1 + z^{2})^{- 1} = ((z - i) (z + i))^{- 1}

for

∣ z ∣ < 1

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