 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/12/d/solution

Using the exponential definition of hyperbolic cosine, multiply

e^{z} + e^{- z} = - 2

by the nonzero quantity

e^{z}

. This yields

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

, hence

e^{z} = - 1

z = x + i y

, its complex modulus requires

e^{x} = 1

, so

x = 0

; its phase then requires

y = (2 k + 1) π

. Thus all complex solutions are

z = (2 k + 1) πi, k \in Z .

There are no real solutions, since real

cosh x \geq 1

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