$$\int_{\text{c}}\frac{\sin \pi z^2 + \cos \pi z^2}{(z+1)(z+2)}$$
Where $\text{C}$ is the circle $|z| =3$
I'm a little confused about how to do this. Should this be done the normal way ? How do I use the concept of what C is and the concepts of complex analysis ?
First, use partial fraction expansion to write
$$\frac{1}{(z+1)(z+2)}=\frac{1}{z+1}-\frac{1}{z+2}$$
Then, applying Cauchy's Integral Formula, we obtain
$$\begin{align} \oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{(z+1)(z+2)}\,dz&=\oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{z+1}\,dz-\oint_{|z|=3}\frac{\sin(\pi z^2)+\cos(\pi z^2)}{z+2}\,dz\\\\ &=2\pi i\left(\sin(\pi)+\cos(\pi)-\sin(4\pi)-\cos(4\pi)\right)\\\\ & =-4\pi i \end{align}$$