 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ia/paper-1/1a/a/solution

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

f * g (k) = f (k) g (k), (f * g) (x) = \int_{- \infty}^{\infty} f (ξ) g (x - ξ) d ξ .

The Fourier inversion theorem and the assumed absolute integrability therefore imply

h (x) = \int_{- \infty}^{\infty} f (ξ) g (x - ξ) d ξ .

Since

∣ - i ∣ = 1

and the principal argument is

- π /2

Log (- i) = - \frac{πi}{2} .

- πi /2 + 2 πin

n \in Z

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