 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-2/23f/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For

∣ z ∣ < 1

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

Therefore

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

This rational continuation is holomorphic across every point of the unit circle except

z = - 1

, where it has a double pole. The set of singular points is consequently

{- 1} .

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