Solution: Let us write \[ I = \oint \limits_{|z| = 3} \frac{\sin \pi z^2 + \cos \pi z^2}{(z-1)(z-2)} \mathrm{d} z \] Note that \[ \frac{1}{(z-1)(z-2)} = \frac{1}{z-2} - \frac{1}{z-1}. \] Thus, \begin{align*} I & = \oint\limits_{|z|=3} \frac{\sin \pi z^2 + \cos \pi z^2}{z - 2}\mathrm{d} z - \oint \limits_{|z| = 3} \frac{\sin \pi z^2 + \cos \pi z^2}{z - 1}\mathrm{d} z \\ & = \oint\limits_{|z|=3} \frac{f(z)}{z - 2}\mathrm{d} z - \oint \limits_{|z| = 3} \frac{f(z)}{z - 1}\mathrm{d} z,~ f(z) = \sin \pi z^2 + \cos \pi z^2 \\ & = 2\pi \iota f(2) - 2\pi \iota f(1) \\ & = 2\pi \iota \left( 0 + 1 \right) - 2\pi \iota \left( 0 + (-1) \right) \\ & = 4\pi \iota. \end{align*} Here we have used the Cauchy integral theorem. Since $z=2$ and $z=1$ both lie inside the contour $\vert z \vert =3$ and $f(z)$ is analytic inside the same contour. Therefore, $I = 4\pi \iota $.