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Math Help - complex integral

  1. #1
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    complex integral

    can somebody explain to me why ∫(1/((z-2)^3)dz=0(integral is bounded by |z-1|=2)?????? z is complex number consder
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  2. #2
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    Re: complex integral

    You say the integral is "bounded by |z- 1|= 2" but this is a path integral. Do you mean that |z- 1|= 2 is the path?

    If so, the only point at which 1/(z- 2)^3 is not analytic is at z= 2 which is inside the curve. That means that the integral around any simple closed path that also encloses z= 2 will be the same. In particular, the integral around, say |z- 2|= 1, the circle of radius 1 with center at 2, must have the same integral. But on the path, we can use the parameterization z= 2+ e^{it} with t going from 0 to 2\pi. Now, z-2= e^{it}, dz= ie^{it}dt so the integral becomes \int_0^{2\pi} (e^{it})^{-3}(ie^{it}dt)= i\int_0^{2\pi} e^{-2it}dt
    Can you see that that is 0?

    (More generally, if f(z) has a "pole" at z= z_0 then the integral around z_0 will be \frac{1}{2\pi i} times the residue of f at z_0. However, the "residue" of function f at z= z_0 is the coefficient of the (z- z_0)^{-1} in the Laurent series for f around z_0. Here the entire Laurent series for f around the z= 2 is that single term \frac{1}{(z- 2)^3}= (z- 2)^{-3}. There is NO (z- 2)^{-1} term. Its coefficient is 0 so the residue is 0 so the integral is 0.)
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  3. #3
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    Re: complex integral

    OK,If i used the way to solve out the case ∫(1/(z-2))dz the path is |z|=1,then let z=exp(it) dz=iexp(it)dt so we get ∫(1/(z-2))dz = ∫(iexp(it)/(exp(it)-2) dt=ln|exp(it)-2|=0 (the integral from 0 to 2pi) is my argument right? i am not sure the step from ∫(iexp(it)/(exp(it)-2) dt to ln|exp(it)-2|here
    Last edited by cummings123321; November 30th 2012 at 11:21 AM.
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