1. let C be the circle |z| = 2 traveled once in the positive sense. Computer the following integrals... a.∫c zez/(2z-3) dz 2. Relevant equations I am confused as to a step in my solution, but i believe a relevant equation is if i am integrating over a circle and the function is analytic at a point, i can remove it by integrating and removing the denominator and replacing it with 2pi*i in the numerator. However i do not know why. This seems to be a product of the cauchy reimann equations used to calculate the integral. 3. The attempt at a solution So i need to get the equation in the form x/z-z0 so i divide everything by 2. I am left with .5*zez/(z-(2/3)) In this form i can see it is analytic at z=3/2 (i believe the word analytic to mean that it goes to infinity) so i remove the denominator, and replace it with a substitution of 2pi*i... 2pi*i*.5*z*ez evaluated at z=3/2 is equal to 3/2pi*i*e3/2 Correct? i Know that sometimes the factor 2pi*i can be negative but i dont understandw hy..