i have the closed ∫ (cos(z))/(3z-π)) dz
i presume this works closed ∫ f(z)/(z-w) dz = f(w)2πi
The Attempt at a Solution
i think the answer is cos(pi)*2πi
I think it's very significant. The pole in (a) is at z=pi/3. That's outside of the counter-clockwise circle |z|=1. What's the answer in that case?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
Split the expression 1/((9-z^2)(2z-i)) up using partial fractions. Remember that trick? Or even simpler, just put f(z)=z/(9-z^2) and make the expression f(z)/(2z-i).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'm not sure where the stuckness comes in. f(z)=z/(9-z^2) is analytic inside of both |z|=1 and |z|=2, isn't it? Just do f(z)/(2z-i). Just apply your theorem to that.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) dzI'm not sure where the stuckness comes in. f(z)=z/(9-z^2) is analytic inside of both |z|=1 and |z|=2, isn't it? Just do f(z)/(2z-i). Just apply your theorem to that.
How did you get from z/((9-z^2)*(2z-i)) to 1/(z^2-z+2)??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