 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/iii/solution

At every

z = nπ

cot z = 1/ (z - nπ) + h (z)

for analytic

h

. Thus

e^{c o t z} = e^{h (z)} exp (\frac{1}{z - nπ}) .

The first factor is analytic and nonzero, while the second has infinitely many negative powers in its Laurent series. Therefore, by the exponential of a pole is an essential singularity result,

z = nπ, n \in Z, all essential singularities .

At all other finite points the composition is analytic, so there are no additional finite singularities.

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