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Past exam of the mathematics course of the University of Cambridge / 2023 / ia / Paper 1 / 1A / c / Solution

Since

F^{- 1} [e^{ika}] = δ (x + a), F^{- 1} [e^{- ika}] = δ (x - a),

the Inverse Fourier transforms of phase factors give

F^{- 1} [cos (ka)] = \frac{1}{2} [δ (x + a) + δ (x - a)]

and

F^{- 1} [sin (ka)] = \frac{1}{2 i} [δ (x + a) - δ (x - a)] .

Solved by gpt-5.6-sol high.

From

z = tan w = - i (e^{2 i w} - 1) / (e^{2 i w} + 1)

one obtains

e^{2 i w} = \frac{1 + i z}{1 - i z}, w = \frac{1}{2 i} lo g \frac{1 + i z}{1 - i z} .

For

z = (23 - 3 i) /7

the quotient is

1 + i 3 = 2 e^{iπ /3}

. Its principal logarithm is

lo g 2 + iπ /3

, so the principal value is

tan^{- 1} z = \frac{π}{6} - \frac{i}{2} lo g 2.

Solved by gpt-5.6-sol high.

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