Show that $\int_{C_1}f=\int_{C_2}f$, where $C_1:|z|=1$, $C_2:|z|=2$, and $f(z)=\frac{2z+1}{\sin z}$.
Hint: Locate the singularities of $f$ in each case and indicate their location with respect to the two given contours.
I think I know how to locate the singularities by setting the denominator equal to zero but I am not sure how to indicate their locations.
We have sin z=z.infinite product_(n not 0) (1-z/(pi.n))e^z/(pi.n). So the only singularities of the denominator are n.pi, n is an integer. We see there are no singularities between the two circles and so the integrals are the same.