​$$\text{Let } \gamma \text{ be the positively oriented circle in the complex plane given by:} \\\{z \in \mathbb{C} : |z - 1| = \frac{1}{2}\}. \\\text{The line integral:} \int_{\gamma} \frac{z e^{1/z}}{z^2 - 1} \, dz \ is\ ?$$​​

Correct option is A

​$$\text{The given circle is: } \gamma : |z-1| = \frac{1}{2} \\z^2 - 1 = 0 \implies z = \pm 1 \\|1 - 1| = 0 < \frac{1}{2} \quad (\text{inside}) \\|-1 - 1| = 2 > \frac{1}{2} \quad (\text{outside}) \\|0 - 1| = 1 > \frac{1}{2} \quad (\text{outside}) \\\text{Now, evaluate the integral:} \\\int_{\gamma} \frac{z e^{1/z}}{z^2 - 1} \, dz \\\text{Using residue theorem:} \\2\pi i \cdot \text{Res}_{z=1} \left( \frac{z e^{1/z}}{(z-1)(z+1)} \right) \\= 2\pi i \cdot \frac{e^{1/1}}{(1+1)} = (2\pi i) \cdot \frac{e}{2} \\= \pi i e \\\therefore \int_{\gamma} \frac{z e^{1/z}}{z^2 - 1} \, dz = i \pi e$$​​

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