- #1

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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

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- Thread starter KNUCK7ES
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- #1

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

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

i think the answer is cos(pi)*2πi

- #2

Dick

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- #3

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

- #4

Dick

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

That's better. But the answer still depends on the closed contour. I'd feel better if you described what it was.

- #5

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(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

- #6

Dick

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(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

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?

- #7

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- #8

Dick

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Yes! Now continue with the other parts and pay attention to which poles are inside of the contour.

- #9

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kk thanks alot

- #10

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- #11

Dick

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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).

- #12

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- #13

Dick

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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.

- #14

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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 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

- #15

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- #16

Dick

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How did you get from z/((9-z^2)*(2z-i)) to 1/(z^2-z+2)??

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