 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/12b/b/iii/solution

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

Write

h (z) = \sum_{n = 0}^{\infty} a_{n} z^{n}

∣ z ∣ = R

gives

a_{n} = \frac{1}{2 π R ^{n}} \int_{0}^{2 π} h (R e^{i θ}) e^{- in θ} d θ .

Except at the two measure-zero points where

cos θ = 0

, the hypothesis gives

∣ h (R e^{i θ}) ∣ \leq R^{- 1/2} ∣ cos θ ∣^{- 1/2} .

Since

∣ cos θ ∣^{- 1/2}

is integrable,

∣ a_{n} ∣ \leq C R^{- n - 1/2}

for a constant

C

independent of

R

. Letting

R \to \infty

shows that every

a_{n} = 0

. Hence

h \equiv 0

, in particular

h

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