Discussion
\[ \int_{|z| = 1} z^{4} \sin\frac{1}{z}\, dz = \]
Answer: B
The integral equals \(\frac{\pi i}{60}\).
The only singularity inside \(|z| = 1\) is the essential singularity at 0, so the integral is \(2\pi i\) times the coefficient of \(1/z\) in the Laurent series:
\[ \begin{aligned} z^{4}\sin\frac{1}{z} &= z^{4}\left(\frac{1}{z} - \frac{1}{3!\,z^{3}} + \frac{1}{5!\,z^{5}} - \cdots\right) \\ &= z^{3} - \frac{z}{6} + \frac{1}{120 z} - \cdots \end{aligned} \]
\[ \begin{aligned} \oint z^{4}\sin\frac{1}{z}\, dz &= 2\pi i \cdot \frac{1}{120} \\ &= \frac{\pi i}{60} \end{aligned} \]
The option \(\pi i/30\) doubles the residue (using 60 instead of 120), and \(60\pi i\) and \(12\pi i\) put the factorial in the numerator.
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