 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/7/c/i/solution

The zeros of

sin z

are

z = nπ

for

n \in Z

, all simple, and

cos (nπ) = (- 1)^{n} \neq = 0

. Hence the finite singularities of

cot z

are exactly

z = nπ, n \in Z, all simple poles .

Locally

cot z \sim 1/ (z - nπ)

, so each pole has residue

1

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