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Evaluate the contour integral: $$\oint \limits_{\left | z \right |=3}\frac{\sin \left ( \pi z^2 \right )+\cos\left ( \pi z^2 \right )}{z^2-3z+2}\,dz$$

Let \[f(z) = \frac{\sin(\pi z^2) + \cos(\pi z^2)}{z^2 - 3z+2}.\] Then \(f\) has exactly two simple poles inside the countour of integration, at \(z=1\) and at \(z=2\). So the integral \(I\) is equal to \(2\pi i (\mathrm{Res}(f;z=1) + \mathrm{Res}(f;z=2)).\)

We have \[ \mathrm{Res}(f;z=1) = \lim_{z\to 1} (z-1)f(z) = \lim_{z\to 1} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{z-2} = 1 \] and \[ \mathrm{Res}(f;z=2) = \lim_{z\to 2} (z-2)f(z) = \lim_{z\to 1} \frac{\sin(\pi z^2) + \cos(\pi z^2)}{z-1} = 1. \]

Therefore we finally get \(I = 4\pi i.\)