 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-331/4/b/v/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

At the formal value

R e = - 10

q = - 1

and

A = - i ω_{1} I + (0 - 1 1 - i)

r_{\pm} = (- 1 \pm 5) /2

. Its two eigenvalues and normalized eigenvectors are

λ_{\pm} = - i ω_{1} + i r_{\pm}, v_{\pm} = \frac{( 1 , i r _{\pm} ) ^{T}}{1 + r _{\pm}^{2}} .

Substitution uses

r_{\pm}^{2} + r_{\pm} - 1 = 0

and verifies

A v_{\pm} = λ_{\pm} v_{\pm}

. Their Hermitian inner product is

v_{+}^{†} v_{-} = \frac{1 + r _{+} r _{-}}{( 1 + r _{+}^{2} ) ( 1 + r _{-}^{2} )} = 0,

since

r_{+} r_{-} = - 1

. This provides the requested orthogonality in the formal continuation. There is no corresponding physical

R e_{N} > 0

for the printed system.

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