 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-2/41c/d/solution

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

cos (π x) = \frac{1}{2} e^{iπ x} + \frac{1}{2} e^{- iπ x},

- \frac{1}{2} sin (π x) = \frac{i}{4} e^{iπ x} - \frac{i}{4} e^{- iπ x} .

Thus

c_{0} = 2, c_{1} = \frac{1}{2} + \frac{i}{4}, c_{- 1} = \frac{1}{2} - \frac{i}{4} .

Ordering the modes as

(- 1, 0, 1)

, let

a = 1/2 + i /4

. Then

C = 2 a 0 \overline{a} 2 a 0 \overline{a} 2, D = - 1 00 000001 .

Consequently

B = - C D = 2 \frac{1}{2} + \frac{i}{4} 0 000 0 - \frac{1}{2} + \frac{i}{4} - 2 .

This matrix is triangular, so its eigenvalues are its diagonal entries:

spec (B) = {- 2, 0, 2} .

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