 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/41c/b/ii/solution

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

H (θ) = 1 - μ^{2} (1 - cos θ) - i μ sin θ .

A direct calculation gives

∣ H (θ) ∣^{2} = 1 - 4 μ^{2} (1 - μ^{2}) sin^{4} \frac{θ}{2} .

This is at most one for every

θ

exactly when

0 \leq μ \leq 1 .

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