Homework Help: Complex contour integration problem.

1. The problem statement, all variables and given/known data

i have the closed ∫ (cos(z))/(3z-π)) dz

2. Relevant equations

i presume this works closed ∫ f(z)/(z-w) dz = f(w)2πi

3. The attempt at a solution

i think the answer is cos(pi)*2πi

 

i can't believe i didn't factor. now i get 1/3*pi*i

 

1. evaluate each integral around the counter-clockwise circle lzl =1 repeat for lzl=2.

(a) ∮ cos(z)/(3z-pi) dz, (b) ∮ z /((9-z^2)*(2z-i)) dz

thats all it says. i think that the pole may be not be significant

 

so that would mean that the integrand is analytic in the interior and on the contour, so it would equal 0?

 

kk thanks alot

 

so being aware of the poles, in the case of (b) we have poles at z=-3,3,i/2. but the only one that matters is the i/2 because it is the only one inside the contour line. my only problem is now trying to get ∮ z/((9-z^2)(2z-i)) dz into the form f(w) ∮ 1/(z-w) dz

 

i did both, and the partial fractions is a little more ugly. but where i became stuck, is that i began to wonder if i had to do it twice the other one being where f(z) = z/(2z-i)

 

i guess my confusion is whether we can set f(z) = z/(9-z^2) OR f(z)= z/(2z-i). but i'd much rather just do z/(2z-i). so i suppose my question is just inquisitive. but i could see it expanded out to something like ∮ 1/(z^2-z+2) dz for lzl=1. in this case it isn't so easy to find an f(x) to set up the nice form of 2*pi*i f(w) from f(w) ∮ 1/(z-w) dz

 

actually upon further thinking of my question, it would have to have bounds ∮ 1/(z^2-z+2) dz for lzl=1 would be 0 but if i were to extend the radius out to lzl =2 then it gets more interesting