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 Incoming links: Complex cosine

Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 1 / 5C / ii / Solution Created 2026-09-24 Updated 2026-10-14

Using

cos z = (e^{i z} + e^{- i z}) /2

and collecting real and imaginary terms gives the complex cosine identity

cos (x + i y) = cos x cosh y - i sin x sinh y

. Thus

Re (cos z) = cos x cosh y, Im (cos z) = - sin x sinh y .

The sign follows from the factors

e^{- y}

and

e^{y}

in the two complex exponentials.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 5 / c / Solution Created 2026-10-18 Updated 2026-10-20

Use the exponential definitions of complex sine function and complex cosine. Multiplying the equation by

2 i

gives

e^{i z} - e^{- i z} = - 2 (e^{i z} + e^{- i z})

, hence

3 e^{i z} + e^{- i z} = 0

. Since exponentials never vanish, this is equivalent to

e^{2 i z} = - 1/3

. All branches of the complex logarithm give

2 i z = - ln 3 + i (π + 2 πn), z = \frac{π}{2} + nπ + \frac{i}{2} ln 3, n \in Z .

The derivation does not divide by

cos z

, so it introduces no lost zero-denominator cases.

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