Discussion
The value of \(\displaystyle\int_{|z| = 1} \frac{\sin z}{z}\, dz\) is :
Answer: B
The integral is 0.
At \(z = 0\) the integrand has a removable singularity, because
\[ \begin{aligned} \frac{\sin z}{z} &= 1 - \frac{z^{2}}{3!} \\ &\quad + \frac{z^{4}}{5!} - \ldots \end{aligned} \]
has no negative powers. Defining it as 1 at 0 makes it an entire function, and by Cauchy's theorem the integral of an entire function around the closed curve \(|z| = 1\) is 0. Equivalently, the residue at 0 is the coefficient of \(1/z\), which is 0.
\(2\pi i\) is the answer for \(\cos z/z\) or \(e^{z}/z\), where the numerator is 1 at the origin; \(-2\pi i\) is the same with the orientation reversed, and \(2\pi\) drops the factor \(i\).
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