Integrating a complex function with Cauchy formula We have I =$\oint_{C}^{} \frac{(z-1)\sin(z)}{z^2 - 2z - 3}$, C is a circle for which $|z-2| = 2$. 
I wrote 
$I =  \oint_{C}^{} \frac{(z-1)\sin(z)}{4(z-3)} - \oint_{C}^{} \frac{(z-1)\sin(z)}{4(z+1)}$
By using the Cauchy formula
$I =  \frac{1}{4} 2\pi i (3-1)\sin(3) - \frac{1}{4} 2\pi i (-1-1)\sin(-1)$. 
Is the second term zero because we have been integrating within the circle C that doesn't include -1?
The result would then be
$I =  \pi i \sin(3)$


Is the second term zero because we have been integrating within the circle $C$ that doesn't include $-1$?

Yes. The second integrand is holomorphic in the entire region enclosed by $C$, hence by Cauchy's integral theorem, the second integral is $0$.
