How to evaluate this, $$\oint_{c} \frac{\sin\pi z^2+\cos\pi z^2}{(z-1)(z-2)}dz$$ where $C$ is the circle, $|z|=3$
I tried below things
I believe 1 and 2 are simple poles here and the equation can be reduced to $\frac{1}{z-1}+\frac{1}{z-2}$
How to continue from here? I am a beginner in contour integration, Any hints approach would be good.
Let $\;C_1\,,\,\,C_2\;$ be little circles around $\;1,2\;$ resp. By Cauchy's Theorem:
$$\oint_{C_1}\frac{\sin\pi z^2+\cos\pi z^2}{(z-1)(z-2)}dz=\oint_{C_1}\frac{\frac{\sin\pi z^2+\cos\pi z^2}{z-2}}{z-1}dz=\left.2\pi i\frac{\sin\pi z^2+\cos\pi z^2}{z-2}\right|_{z=1}=$$
$$=-2\pi i \left(0-1\right)=2\pi i $$
Do now something similar with $\;C_2\;$ and check that your integral is the sum of these two integrals.