 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/2/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 .

cosh z + 1 = 0

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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