 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2021/4/b/iii/solution

For

z \neq = 0

, the denominator vanishes exactly when

1/ z = (2 k + 1) πi

. Thus set

z_{k} = \frac{1}{( 2 k + 1 ) πi}, k \in Z .

Each denominator zero is simple, since

(e^{1/ z} + 1)^{'}_{z = z_{k}} = - \frac{e ^{1/ z_{k}}}{z _{k}^{2}} = \frac{1}{z _{k}^{2}} \neq = 0.

Every

z_{k}

is purely imaginary and nonzero. The identity

sin (i y) = i sinh y

for real

y

shows

sin z_{k} \neq = 0

, so there is no cancellation. Each

z_{k}

is consequently a simple pole, with residue

z_{k}^{2} sin z_{k}

These poles accumulate at zero. Therefore

z_{k} are simple poles, and 0 is a non-isolated singularity .

In particular, zero has no punctured neighbourhood on which the function is holomorphic, so it cannot be classified as a removable singularity, a pole, or an isolated essential singularity. There are no other finite singularities.

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