Question
Find all the singularities of the function
Answer :
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The singularities of the function ( f(z) = \exp!\left(\dfrac{z}{\sin z}\right) ) arise from the points where the denominator (\sin z) vanishes. The zeros of (\sin z) occur at (z = n\pi), where (n \in \mathbb{Z}). First consider (z = 0). Using the Taylor expansion (\sin z = z - \dfrac{z^{3}}{6} + \cdots), we obtain [ \frac{z}{\sin z} = \frac{z}{z\left(1 - \frac{z^{2}}{6} + \cdots\right)} = \frac{1}{1 - \frac{z^{2}}{6} + \cdots} = 1 + \frac{z^{2}}{6}
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