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Determining whether the Cauchy-Goursat theorem applies

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##C## be the negatively oriented contour ##2e^{-i \pi t}## ##0 \le t \le 1## For which of
    the following functions does the Cauchy-Goursat theorem apply so that the integral of the
    function along ##C## is zero?

    2. Relevant equations


    3. The attempt at a solution

    The first function is ##f(z) = e^{z^7}##, where ##z^7 = e^{7 \log z}##.

    Obviously, the contour over which we wish to integrate ##f(z)## is the unit circle. For ##f(z)## to even be integrable, it must first be a function, which can be accomplished by removing a branch so that ##\log z## becomes a function. By removing a branch, we are actually removing a ray which makes some angle ##\theta## with the positive real-axis that begins at the origin and extends away from it. For instance, I could remove the ray which corresponds to the negative real-axis, all those points of the form ##(a,0)##, where ##a \in \mathbb{R} \setminus \{0\}##; consequently, there can be no points whose angle (argument) is ##\theta = \pi##.

    Here is why I do not believe one can apply the Cauchy-Goursat theorem: there are singularities interior to the contour (infinitely many, actually), and one singularity on the contour--namely, ##(-1,0)##.

    Despite this, I believe we can integrate the function over the contour because there is only one singularity point. This would be analogous to an improper integral because we are integrating up to a singularity point, right?

    My question is, how would I set up the integral? When ##t = 1/2##, we get ##C(1/2) = (-1,0)##. So, would I do an improper integral on the variable ##t##?

    Also, would the integral be the same independent of which branch I choose to remove?
     
  2. jcsd
  3. Dec 7, 2014 #2

    Ray Vickson

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    Why would you define ##z^7## as ##e^{7 \log z}##? Did YOU do this, or did the person setting the problem force you to use that weird definition?
     
  4. Dec 7, 2014 #3
  5. Dec 7, 2014 #4

    pasmith

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    [itex]z^n[/itex] is single-valued for integer [itex]n[/itex]: [itex]\exp(n(\mathrm{Log}(z) + 2im\pi)) = \exp(n \mathrm{Log}(z))[/itex] for any [itex]m \in \mathbb{Z}[/itex], since [itex]\exp(2imn\pi) = 1[/itex].
     
  6. Dec 7, 2014 #5

    Ray Vickson

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    But, BY DEFINITION, ##z^7 = z \cdot z \cdot z \cdot z \cdot z \cdot z \cdot z##. You only need ##z^a = e^{a \log z}## for non-integer values of ##a##.
    But, BY DEFINITION, ##z^7 = z \cdot z \cdot z \cdot z \cdot z \cdot z \cdot z##.

    You only need ##z^a = e^{a \log z}## for non-integer values of ##a##. While the book does not specifically say so, I think the intention is that definition 183 applies to non-integer values of the power---at least, that is how it is explained in all complex variable books I have ever seen before. One hint that this might really be the author's intention is the fact that in various places much before page 28 he writes things like ##z^2##, ##z^3##, etc, as though he expects the reader to understand what these mean without having to reach page 28.
     
  7. Dec 8, 2014 #6
    Ah, of course. I should have been able to easily deduce that from the definition. So, then ##f(z) = e^{z^7}## would be analytic in and on the contour, thereby allowing us to apply the Cauchy-Goursat theorem.

    Okay, suppose we didn't have such a function, an analytic one, but one for which we had to remove a branch. Would all I said in my first post be valid? For instance, ##f(z) = \log z##, where ##\log z = \ln|z| + i \arg z## would be one such function, right?
     
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