 Solution

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

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

Along real

z = x ↑ 1

, the modulus is

1/ (1 + x^{2})

and stays bounded, but along

z = i y

with

y ↑ 1

it is

1/ (1 - y^{2}) \to \infty

. Since

lim sup ∣ a_{n} ∣^{1/ n} = 1

R = 1

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