 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2021/3/a/iii/solution

The denominator vanishes exactly when

e^{z} = - 1

, namely at

z = (2 k + 1) πi

k \in Z

. Its derivative there is

e^{z} = - 1 \neq = 0

, so each such zero is simple and the numerator does not cancel it. Hence

\frac{1}{1 + e ^{z}} is holomorphic on C ∖ {(2 k + 1) πi : k \in Z},

and every omitted point is a simple pole.

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