Discussion
The value of \(\oint z\exp\left(\frac{1}{z}\right)dz\), where the closed contour is a unit circle traversed anticlockwise is :
Answer: A
The value of the integral is \(\pi i\).
The only singularity inside the unit circle is the essential singularity at z = 0, so expand in a Laurent series:
\[ \begin{aligned} z\,e^{1/z} &= z\left(1 + \frac{1}{z} + \frac{1}{2!z^{2}} + \cdots\right) \\ &= z + 1 + \frac{1}{2z} + \cdots \end{aligned} \]
The residue is the coefficient of \(1/z\), which is 1/2, and the integral is \(2\pi i \times \frac{1}{2} = \pi i\).
\(2\pi i\) is what you get by taking the residue as 1, the coefficient of \(1/z\) in \(e^{1/z}\) alone, forgetting the extra factor z. Zero would follow only if the integrand were analytic inside the circle, and \(\pi\) loses the factor i.
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