 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/7/c/solution

Put

w = z - π

. The identity

cosh^{2} (1/ w) = [1 + cosh (2/ w)] /2

gives a Laurent series containing a constant and only even negative powers

w^{- 2}, w^{- 4}, \dots

. There is no

w^{- 1}

Res_{z = π} cosh^{2} (\frac{1}{z - π}) = 0.

The singularity is an essential singularity, not a pole: infinitely many negative Laurent coefficients are nonzero. The residue is still defined as the coefficient of

w^{- 1}

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