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Calculate closed curve integral

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Calculate ∫C [itex]\frac{dz}{z(z+a)}[/itex] where C is the unit disk Δ(1) with counterclockwise orientation where a is a complex number with |a| < 1.

    2. Relevant equations

    L [itex]\frac{dz}{(z-p)}[/itex]= 2πiν(L,p)

    3. The attempt at a solution

    Using partial fraction decomposition,

    C [itex]\frac{dz}{z(z+a)}[/itex] = ∫C[itex]\frac{-1/a}{z} + \frac{1/a}{(z+a)}[/itex] = (-1/a)2πi + (1/a)2πi = 0
     
  2. jcsd
  3. Mar 24, 2015 #2

    Dick

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    Well, I think you an overall negative sign wrong in your partial fraction decomposition, but that doesn't matter since the answer is zero anyway. Yes, that's a fine way to do it. Assuming ##\nu## is winding number?
     
  4. Mar 24, 2015 #3
    Yes, it's the winding number. I forgot to mention that. Generally, an integral wouldn't vanish because it contains the origin, correct? This is just a special case that happens to vanish?
     
  5. Mar 24, 2015 #4

    Dick

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    The integral of ##1/z^2## or ##z## vanishes when you integrate around the origin. So that's not a valid general rule. In this case you have two poles inside the contour and they cancel. Isn't that what the calculation is telling you?
     
  6. Mar 24, 2015 #5
    The complex inversion's analyticity is destroyed by the origin so that it doesn't vanish, and the two poles are at z=0 and z=a. Apparently, that's what it's telling me, yes. Heh.
     
  7. Mar 25, 2015 #6

    SteamKing

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    Isn't the second pole at z = -a?
     
  8. Mar 25, 2015 #7

    HallsofIvy

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    Yes, that was probably a typo. The poles are at 0 and -a.
     
  9. Mar 25, 2015 #8
    Yes, of course, I apologize for my oversight. Your previous comment about an incorrect negative sign makes sense how. Heh. Thanks again for the help.
     
  10. Mar 27, 2015 #9
    Lemma: Every curve C(φ: [0,1]→D) is homotopic to a piecewise linear curve in D.

    Therefore, it is homotopic to the straight line curve connecting (1,1) and (2,3). By the Independence Theorem and Cauchy Integral Theorem, the integrals are equal. I can parametrize the line connecting the points by z(t) = 1 + i + t(1+2i) and insert this into f(z) to get the monstrosity f(z(t)). There has to be a better way of finding the integral.
     
  11. Mar 27, 2015 #10
    Theorem: Let f be a continuous function on D and let F a primitive function of f. Let C = C(φ:[T0,T1→ℂ) be a curve with finite length with the initial point z0 and the terminal point z1. Then ∫ f(z)dz = Fz1 - Fz0.

    So I don't need to go through the rigmarole of finding f(z(t)) and df(z(t))/dt, etc. I can simply integrate with respect to z.
     
    Last edited: Mar 27, 2015
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