Discussion
The residue of \(f(z) = \dfrac{e^{z} - 1}{z^{2}}\) at \(z = 0\) is :
Answer: C
The residue is 1.
Expand the numerator:
\[ \begin{aligned} \frac{e^{z} - 1}{z^{2}} &= \frac{z + \frac{z^{2}}{2} + \frac{z^{3}}{6} + \ldots}{z^{2}} \\ &= \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \ldots \end{aligned} \]
The residue is the coefficient of \(1/z\), which is 1. The value \(\frac{1}{2}\) is the constant term, not the residue, and 0 or 2 come from expanding the exponential incorrectly.
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