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Complex line integral

  1. May 5, 2012 #1
    I read(and numerically calculated) that the line integral around a circle of radius r centered at 0 for the function conjugate(z)/(z - t) is 0 for all t inside the circle.

    I don't know see how this integration is performed.
     
  2. jcsd
  3. May 5, 2012 #2


    Let's see if I'm still medium lucid at this hour...As the integral is on [itex]\,\,|z|=r\,\,[/itex] , we can put [itex]\,\,z=re^{i\theta}\,,\,0\leq\theta\leq 2\pi\Longrightarrow dz=rie^{i\theta}\,d\theta[/itex] , so that
    [tex]\oint_{|z|=r}\frac{\overline{z}}{z-t}\,dz=r^2i\int_0^{2\pi}\frac{d\theta}{re^{i\theta}-t}[/tex]
    This last integral you can find in integral tables or, funny enough, you can go back to your original variable by means of

    another change of variables and get zero then...

    DonAntonio
     
    Last edited: May 6, 2012
  4. May 5, 2012 #3
    I was able to get where you got to but I don't know how to calculate the final integral.

    Note that it's not nearly as simple as it would appear since

    1/(e^(thetai) - t) = (cos(theta) - sin(theta)i - conjugate(t))/((cos(theta) + real(t))^2 + (sin(theta) + img(t))^2).
     
  5. May 5, 2012 #4


    Look it in Bronshtein-Semendiaev: it appears as [itex]\int\frac{1}{b+ce^{ax}}dx[/itex] , or do what I adviced you: substitute back to z leaving that constant out.

    DonAntonio
     
  6. May 6, 2012 #5
    hmm I am still not seeing the change of variables solution.
     
  7. May 6, 2012 #6


    We had [itex]\,\,\displaystyle{r^2i\int_0^{2\pi}\frac{d\theta}{re^{i\theta}-t}}\,\,[/itex] . Now just substitute back
    [tex]z=re^{i\theta}\Longrightarrow dz=rie^{i\theta}d\theta\Longrightarrow id\theta=\frac{dz}{z}\Longrightarrow r^2i\int_0^{2\pi}\frac{d\theta}{re^{i\theta}-t}=r^2\oint_{|z|=r}\frac{dz}{z(z-t)}=\frac{r^2}{t}\left[\oint_{|z|=r}\frac{dz}{z-t}-\oint_{|z|=r}\frac{dz}{z}\right]\,\,,\,\,t\neq 0[/tex]

    Now, each of both integrals on the right have the same value so

    Of course, if t is zero then the integral is NOT zero.

    DonAntonio
     
  8. May 6, 2012 #7
    Hmm, I'm not seeing how one gets from z = re^(theta*i) to dz = rie^(theta*i)dtheta.
     
  9. May 6, 2012 #8


    Eeer...if you can't differentiate then perhaps this stuff's still a little too advanced for you...and I wonder now how

    come you wrote after my first post that you were able to get where I got...??

    DonAntonio
     
  10. May 6, 2012 #9
    Edit: oh nevermind, I see.

    I was confusing your dz with being the derivative of z, not the notational defintion hahahaha.
     
    Last edited: May 6, 2012
  11. May 7, 2012 #10
    Something is bugging me, why do you say that if t is zero the integral is not zero?

    integral of 1/e^(thetai) with respect to theta = integral of 1/(cos(theta) + sin(theta)i)
    = integral of cos(theta) - sin(theta)i
    = 0.

    If that's a typo it's no problem but I want to make sure I am not missing something obvious.
     
  12. May 7, 2012 #11

    That wasn't a typo but a mistake: I forgot what the function was and thought it was only [itex]\,\,\displaystyle{\frac{1}{z}\,\,,\,\,instead\,\,of\,\,\frac{\overline{z}}{z}}\,\,[/itex] . I'd go


    [itex]\displaystyle{\oint_{|z|=r}\frac{\overline{z}}{z}dz=ri\int_0^{2\pi}e^{-i\theta}d\theta=\left.-re^{-i\theta}\right]_0^{2\pi}=-r(1-1)=0}\,\,[/itex] , but what you did is correct, too.

    DonAntonio
     
    Last edited: May 7, 2012
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