 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/4/c/solution

The exponential numerator is entire and never zero. The denominator vanishes precisely when

e^{2 z} = - 1

z_{n} = iπ (n + \frac{1}{2}), n \in Z .

At these points

sinh z_{n} = i (- 1)^{n} \neq = 0

, so every zero of the denominator is simple and hence every singularity is a simple pole. Using the residue formula in (b),

Res (f, z_{n}) = \frac{e ^{- ik z_{n}}}{sinh z _{n}} = \frac{( - 1 ) ^{n}}{i} exp [kπ (n + \frac{1}{2})] .

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