Discussion
Let \(f(z) = \mathrm{cosec}\left(\dfrac{1}{z - 1}\right)\) for \(z \ne 1\), \(z \in \mathbf{C}\). Then at the point \(z = 1\), \(f\) has :
Answer: D
At \(z = 1\) the function has a non-isolated singularity.
\(\mathrm{cosec}\,w = 1/\sin w\) has poles where \(\sin w = 0\), at \(w = k\pi\). With \(w = 1/(z - 1)\), the poles of \(f\) are at
\[ z_{k} = 1 + \frac{1}{k\pi},\quad k = \pm 1, \pm 2, \ldots \]
These accumulate at \(z = 1\). Every punctured disc around 1 contains poles, so \(f\) is not analytic on any punctured neighbourhood of 1, and the singularity there is not isolated.
Removable singularities, poles and essential singularities are all isolated by definition, so options A, B and C do not apply.
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