Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 1 / 2D / i

Table of contents
- Solution i

Solution

The zeroes of

cosh z

occur at

z = iπ (n + \frac{1}{2})

,

n \in Z

. They are simple because

sinh z \neq = 0

there. Squaring the denominator therefore gives

double poles at z = iπ (n + \frac{1}{2}), n \in Z .

There are no other finite singularities; this follows from the local zero-order criterion for a pole.

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