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Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 12B / b / iii / Solution

Write

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

. Cauchy's coefficient formula on

∣ 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

is constant. This proves the entire function under a horizontal inverse-square-root bound result.

Solved by gpt-5.6-sol high.

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