Discussion
If \(\gamma = \{z \in \mathbf{C} / |z - \frac{\pi}{2}| = 1\}\), then the value of \(\displaystyle\int_{\gamma} \frac{\sin z \, dz}{\left(z - \frac{\pi}{2}\right) z}\) is :
Answer: C
The integral equals \(4i\).
The integrand \(\frac{\sin z}{(z - \pi/2)z}\) has singularities at \(\pi/2\) and 0. The contour is the circle of radius 1 about \(\pi/2 \approx 1.571\), which contains \(\pi/2\) but not 0 (its distance from \(\pi/2\) is 1.571, more than 1).
The residue at the simple pole \(\pi/2\) is \(\frac{\sin(\pi/2)}{\pi/2} = \frac{2}{\pi}\), so by the residue theorem
\[ \int_{\gamma} = 2\pi i \cdot \frac{2}{\pi} = 4i \]
The value 0 would need no singularity inside; \(2\pi i\) treats the residue as 1, forgetting the factor \(\frac{1}{z}\) evaluated at \(\pi/2\); and \(-2\pi i\) adds a sign error as well.
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