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Complex Integral

  1. Dec 20, 2005 #1
    I'm studying for a math exam for Complex Variable Analysis. We had a bunch of problems, and we keep having to let an integral just equal zero, and we're not sure why, but it always works.
    What we do is this...
    we're integrated over a closed contour. and we're supposed to chose a point [tex]z_0[/tex] which is inside the countour and the we use the equation:[tex] \int_C \frac {\frac {1}{f(z)}}{z-z_0} = 2\pi i f(z_0) [/tex] or something like that. there's also a formula using the derivatives of f. so actually, i don't quite understand any of this... cause i'm not sure why we use the above equation when we need to get the integral of f(z) over a countour C... but anyway, we use that, and thats always the answer. We've been using partial fractions in order to get it into that form i guess. oh yeah, we use partial fractions to change f(z) into something that looks like the above, and then we apply that formula to each fraction and sum them... right.
    anyway, sometimes we get it so that [tex]z_0[/tex] isn't in the interior of C and then the integral equals zero. I've looked through the book, and i don't quite understand why, but its always the case. So i was hoping someone could explain it to me.
    Last edited: Dec 20, 2005
  2. jcsd
  3. Dec 20, 2005 #2


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  4. Dec 20, 2005 #3
    well, first, we didn't go over the laurent series in class at all, we didn't have time. And i understood the first link fine. what i didn't understand was how when z_0 was in the countour, we got a non-zero answer, but when z_0 was outside we didn't. i understand that when z_0 is outside, it equals zero based on the first liknk you provided. but i don't really understand the second link very well, which explains about z_0 on the inside. my professor wasn't ever particularly good at explaining the concepts, neither is the book really. its more about calculating things than understanding them.

    but, if i look at that second link, and believe the first line, then i understand the second line alright. i don't really understand why cauchy-reimann says the first and last terms vanish in line 3. and then i'm lost when it calls a_-1 the complex residue? never heard of it? yeah, i get way lost after that.
  5. Dec 20, 2005 #4

    Tom Mattson

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    Assuming that [itex]\frac{1}{f(z)}[/itex] has no poles, your integrand only has one simple pole at [itex]z=z_0[/itex]. If your contour excludes that pole then the integral is immediately seen to vanish by the residue theorem. If there are no poles inside your contour then the sum of the residues of the poles must be zero.
  6. Dec 20, 2005 #5


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    (Disclaimer: I'm not an expert in complex analysis, so temper my remarks with this knowledge. :smile:)

    Probably the most important fact about analytic functions is that if [itex]f(z)[/itex] is analytic on a closed set [itex]C[/itex], then:

    \oint_{\delta C} f(z) \, dz = 0

    The whole subject begins with this pretty fact! (remember my disclaimer)

    Another wonderful fact is that for any closed contour [itex]\gamma[/itex] that doesn't pass through the origin, and for [itex]n \in \mathbb{Z}[/itex]:

    \oint_{\gamma} z^n = \left\{ \begin{array}{ll}
    0 \quad & n \neq -1 \\
    2 \pi i & n = -1

    Armed with the knowledge that analytic functions can generally be written as a Laurent series, this pretty theorem tells us that when the function is not analytic inside our contour, the only part that matters is the part that looks like [itex]z^{-1}[/itex]. This leads to the residue theorem that Tide mentioned.

    This knowledge naturally suggests an approach to evaluating contour integrals: figure out the points [itex]z_k[/itex] where the integrand isn't analytic, and extract the residue (the part that looks like [itex](z - z_k)^{-1}[/itex]) at each of them.

    Now, think about the method you're using in terms of what I've suggested...
  7. Dec 20, 2005 #6
    alright, this makes some sense to me, i had to look up what a pole was though... they aren't mentioned in the book.. weird? maybe we went over them in class, but lately i play the newpaper's sudoku... anyway.

    ok, i looked it up, we haven't gotten to residues and poles yet. so i don't understand any more. so i don't know what we're doing at all... When i was reading the mathworld page on the Cauchy Integral Formula, it said right on it that by the Cauchy Integral theorem, that the integral over a closed countour that doesn't contain a pole is zero. but i don't see why, i went and checked the theorem page, and it just says the integral over a closed contour is zero, and says nothing about poles. i feel like i'm just missing something??
  8. Dec 21, 2005 #7


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    It's ok that you don't know residues yet (assuming it hasn't been covered in your class yet while you were sudoku'ing).

    You know what an analytic function is right? f(z) analytic on a region D, means f has no poles in D. If you integrate this analytic f over any closed contour contained in D, then you get zero. This is Cauchy's Integral Theorem.The proof of this very important theorem is sometimes glossed over in a "Complex Varibales" course as it's somewhat complicated, so it might be sufficient for your course to take it "as fact". The mathworld link Tide gave proves it via Green's Theorem if you are interested.

    The mathworld page on Cauchy's integral formula appears a little lazy, they are assuming that f is analytic (hence it has no poles). By the way, you've mis-stated this theorem in your first post.

    Cauchy's integral theorem is really key though. One thing it lets you do is 'bend' your contours without affecting the integral provided your function is analytic where the 'bending' occurs. You also get path independance of an integral for analytic functions, that is no matter what contours you pick going from two points a and b, integrating your analytic function over these paths always yields the same result.

    A little more detail on what I mean by 'bending' a contour. Think of your contours as rubber bands. You can stretch or compress them anyway you like without affecting the integral, provided your rubber band never moves out of the region where your integrand in analytic. My own imagery is an actual physical pole sticking out of the complex plane where the 'pole' of an analytic function is, and there's no way my rubber band contour can get around this pole (it's confined to the complex plane). If I want to pass my rubber band through a pole, I need the residue theorem, which comes latter in the theory. For now, think of them as obstructions that you can breach only in certain circumstances (where you're able to apply the Integral Formula).

    Cauchy's integral theorem makes proving cauchy's integral formula a snap, essentially the diagram at the top of mathworlds "Cauchy's Integral Formula" page. On the [tex]\gamma_0[/tex] contour and it's interior f(z)/(z-z_0) is analytic and hence the integral over it is 0. By taking the [tex]\gamma_1[/tex] contour to be a circle centered at z_0 of small enough radius, you can determine the integral of f(z)/(z-z_0) over this contour (note it's NOT analytic here, specifically it 'blows up' at z=z_0). Did they at least prove the Integral Formula in class?
    Last edited: Dec 21, 2005
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