$\displaystyle \int\frac{sin(\pi.z)}{z^{2}(z-2)}dz$

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