Discussion
The value of the integral \(\int \frac{\sin z}{z}\,dz\) over the contour C, C given by |z| = 1 and traversed counter-clockwise, is :
Answer: C
The integral is 0.
\(\frac{\sin z}{z}\) looks singular at \(z = 0\), but its Taylor series is \(1 - \frac{z^{2}}{3!} + \frac{z^{4}}{5!} - \cdots\), so the singularity is removable and the function is analytic inside and on \(|z| = 1\). The residue at 0 is the coefficient of \(\frac{1}{z}\), which is zero.
By Cauchy's theorem, the integral of an analytic function round a closed contour is 0. \(2\pi i\) would be right for \(\frac{\cos z}{z}\) or \(\frac{e^{z}}{z}\), whose residue at 0 is 1. The options with \(\sqrt{3}\) do not arise from any residue here.
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