Discussion
The function \(f(z) = \dfrac{\sin z}{1 - e^{z}}\) has :
Answer: A
The function has a removable singularity at z = 0.
Both numerator and denominator vanish at 0, each to first order:
\[ \begin{aligned} \sin z &= z - \tfrac{z^{3}}{6} + \ldots \\ 1 - e^{z} &= -z - \tfrac{z^{2}}{2} - \ldots \end{aligned} \]
so \(f(z) = \frac{1 - z^{2}/6 + \ldots}{-1 - z/2 - \ldots} \to -1\) as \(z \to 0\). A finite limit means the singularity is removable; defining \(f(0) = -1\) makes f analytic there.
A pole would need the denominator to vanish to higher order than the numerator, and an essential singularity would need infinitely many negative powers. The poles of this function are at \(z = 2\pi i k\), \(k \neq 0\), not at the origin.
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