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I Question about Contour Integration

  1. Nov 3, 2016 #1
    This question deals specifically with complex analysis.

    Let C be the unit circle in the complex plane (|z| = 1). If you calculate the contour integral of (1/z)dz over C using Cauchy's Integral Formula, you get 2*pi*i. If you calculate it using the path z(t)=e^(it), t in [0,2pi], you also receive 2*pi*i.

    However, if I do the second method, but make the substitution u = e^(it), the limits of integration (specifically 0 and 2pi) both become 1. Hence, after substitution, the integral is 0. Why is this the case? I imagine that u-substitutions work differently with complex integrals, but I'm trying to understand what is wrong with making this substitution.
     
  2. jcsd
  3. Nov 3, 2016 #2

    andrewkirk

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    If you use the second method you end up with the integral ##\int_0^{2\pi}i\,dt## which gives ##2\pi i##. If you substitute ##u=e^{it}## in there you'll just get back to the original integral ##\int_C\frac{dz}z##, since ##u=z##. It's not clear why you believe that gives an integral of zero. If you show your working we may be able to show where it goes wrong.
     
  4. Nov 3, 2016 #3
    I assume you mean the following situation:

    ##\displaystyle\biggr. \oint \frac{1}{z}dz= log(z)\biggr |_1^1=(0+2\pi i)-0=2\pi i##

    And that is because the antiderivative is a multi-valued function. So in Complex Analysis, it is not always the case that the integral is zero when the limits "appear" to be the same. One must consider the (analytically-continuous) path along a multivalued antiderivative (if antiderivatives are used) which in general does not return to the same spot over the antiderivative when the path is closed. Now, if you plot the real or imaginary component of the log(z) function and plot an analytically-continuous path along it starting from z=1, wrap around the origin ##2\pi## and go back to 1, you'll see that you end up at a point on the function equal to ##0+2\pi i## or if you just plot the imaginary part of log(z), you'll see that you end up at the point ##2\pi##. So that if you understand this, you can explain why, if the winding number is 2, then we have:

    ##\displaystyle\biggr. \oint \frac{1}{z}dz= log(z)\biggr |_1^1=4\pi i##

    But this is tough to understand I know when seeing for the first time. Better for now to just use the substitution ##z(t)=e^{it}## which is of course an analytically continuous path, then the integral becomes

    ##\displaystyle \oint \frac{1}{z}dz=\int_0^{2\pi} i dt=2\pi i##
     
  5. Nov 3, 2016 #4
    Thank you very much! I understand what the issue is now. logz is multivalued. So, though the limits of integration are the same, each results in two different values of the antiderivative. So, the result 2*pi*I is still recovered. :)
     
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