Discussion
Let \(\Gamma\) be the circle given by \(z = 4e^{i\theta}\), where \(\theta\) varies from 0 to \(2\pi\). Then
\[ \oint_{\Gamma} \frac{e^{z}}{z^{2} - 2z}\, dz = \]
Answer: C
The integral equals \(\pi i(e^{2} - 1)\).
The integrand \(\frac{e^{z}}{z(z-2)}\) has simple poles at \(z = 0\) and \(z = 2\), and both lie inside the circle \(|z| = 4\). By the residue theorem the integral is \(2\pi i\) times the sum of the residues:
\[ \begin{aligned} \mathrm{Res}_{z=0} &= \frac{e^{0}}{0 - 2} = -\frac{1}{2} \\ \mathrm{Res}_{z=2} &= \frac{e^{2}}{2} \\ \oint_{\Gamma} &= 2\pi i \cdot \frac{e^{2} - 1}{2} = \pi i(e^{2} - 1) \end{aligned} \]
\(2\pi i(e^{2} - 1)\) forgets the factor \(\frac{1}{2}\) in each residue. The options with \(1 - e^{2}\) have the sign reversed, which is what you get by taking the residue at 0 as \(+\frac{1}{2}\) and at 2 as \(-\frac{e^{2}}{2}\).
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