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Homework Statement
i have the closed ∫ (cos(z))/(3z-π)) dz
Homework Equations
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
That's better. But the answer still depends on the closed contour. I'd feel better if you described what it was.i can't believe i didn't factor. now i get 1/3*pi*i
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
Yes! Now continue with the other parts and pay attention to which poles are inside of the contour.so that would mean that the integrand is analytic in the interior and on the contour, so it would equal 0?
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