Find the integral $$\oint_{|z|=1}\frac{\sin(z)}{z^2}\,dz$$
I can see that there is a pole at $z = 0$. But I don't know how to solve that integral over that way.
2026-04-08 20:58:52.1775681932
Contour Integral of $\oint_{|z|=1}\frac{\sin(z)}{z^2}\,dz$
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We know
$$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \dots$$
In fact, operations on Power/Laurent series end up being exactly what we expect in the naive fashion:
$$\frac{\sin z}{z^2} = \frac{1}{z} - \frac{z}{3!} + \frac{z^3}{5!} - \dots$$
So by the Residue Theorem: $$\oint_{S^2} \frac{\sin z}{z^2} dz = 2\pi i$$