This was the the problem i was give and it asks to integrate around the contours center a=-2, a=0, a=2 and radius 1.

i) a=-2 gives and answer of 0 because it is holomorphic everywhere

ii) a=0 gives and answer of \(\displaystyle \pi i\)

iii) a=2 gives and answer of 0 as \(\displaystyle \int \frac{f(z)}{z-2} dz\), when \(\displaystyle f(z)=\frac{sin(\pi.z)}{z^{2}}\), = \(\displaystyle 2\pi.i \frac{sin(\pi.2)}{2^{2}}=0\)

could some please check my working for me as am a bit confused about \(\displaystyle sin(\pi.z)\) around the contours.

thanks bobisback