Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 2

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- Solution 2

Solution

The exponential definition of hyperbolic sine gives

e^{z} - e^{- z} = 0

, or

e^{2 z} = 1

. Taking every branch of the complex logarithm gives

sinh z = 0 ⟺ z = iπk, k \in Z .

For hyperbolic cosine, multiply

cosh z + 1 = 0

by

2 e^{z}

to obtain

(e^{z} + 1)^{2} = 0

. Thus

e^{z} = - 1

cosh z + 1 = 0 ⟺ z = i (2 k + 1) π, k \in Z .

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