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Complex Countour Integral. not sure how to solve.

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    So, as far as I understand, there is not much difference between integrals in complex variables and those pesky line integrals from vector calculus. I'm not particullary good with line integrals, so it follows that I'm not quite understanding the complex variables ones.
    Anyway, here is the problem
    [itex]\oint (z^*)^2 dz [/itex] around the circunferences [itex] |z| = 1 [/itex] and [itex]|z-1|= 1[/itex] ([itex] z^*[/itex] is the complex conjugate.)


    2. Relevant equations

    [itex] z=x+iy [/itex]
    [itex] f(z) = u(x,y) +iv(x,y) [/itex]
    [itex] \int_c f(z)dz = \int_c (u dx -v dy) + i\int_c (v dx + u dy) [/itex] (1)

    3. The attempt at a solution

    First of all I use the Cauchy-Rienmann equations to see if the function is analytic inside the circuferences, so I could use Cauchy's theorem. Turned out, the function is analytic only at [itex]z=0[/itex]
    [itex] f(z) = (x-iy)^2 = (x^2 - y^2) + i(-2xy) = u + iv [/itex]
    [itex] \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \iff 2x = -2x \iff x = 0 [/itex]
    [itex] \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y} \iff -2y = 2y \iff y = 0 [/itex]
    so the function is analytic only on [itex] z=0 [/itex] and I have to integrate the hard way.

    So I use (1) and now I have: [itex] \oint f(z)dz = \int_c (x^2 - y^2) dx +\int_c (2xy)dy + i\int_c (-2xy)dx +i\int_c (x^2 - y^2)dy [/itex]

    so here is my problem: how do I integrate this?

    what I tried is the following (and it might or migh not be correct);

    using polar coordinates:
    [itex] x =r\cos\theta[/itex]
    [itex] y= r\sin\theta [/itex]
    [itex]dx =-r\sin\theta d\theta [/itex]
    [itex]dy =r\cos\theta d\theta [/itex]


    the first integral [itex] \oint_c (x^2 - y^2)dx [/itex] becomes [itex] \int_0^{2\pi}( \cos^2 \theta - \sin^2\theta)(-sin\theta)d\theta = \int_0^{2\pi} (-\cos^2\theta sin\theta +sin^3\theta) d\theta [/itex] (remember that r = 1)

    the idea is to do the same for the other integrals, integrate and then add.
    I won't do it now because, like I said, I'm not sure this is the correct approach.

    Can you you tell me if this would work or not?
     
    Last edited: May 30, 2014
  2. jcsd
  3. May 30, 2014 #2

    BiGyElLoWhAt

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    Gold Member

    if your problem is the path integral of (z*)^2, and z = x + iy, then (z*)^2 is not x^2 -y^2,
    z*z = x^y -y^2, is that what you're looking for? I don't have extensive experience with complex integrals, but they really aren't too terribly different from real integrals.
    z* = x-iy, and (z*)^2 = x^2 -2ixy -y^2
     
  4. May 30, 2014 #3

    CAF123

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    Gold Member

    Analyticity is a term reserved for the differentiability of a function in some open disk. The function is differentiabie only at z=0, and is therefore nowhere analytic.
    Yes, it works. It is the same method that is familiar from computing line integrals in multivariable calculus. Alternatively, you should be able to use the Residue theorem, a generalisation of Cauchy's theorem.
     
  5. May 30, 2014 #4
    You're right. I was a bit in doubt about this to tell you the truth.
    But either way, the point is, I can't use Cauchy's theorem.

    Ah, yes, I've heard about. Although I havent studied it yet so...
    Guess I'll be revisiting this problem later
     
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