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Cauchy's Integral Formula

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data

    integral of (z^2/(4-z^2)) with respect to z, over |z+1|=2

    2. Relevant equations
    Cauchy's Formula(I'm attempting to do it in the more fancy and easily readable sense, if it's not readable then go here.. http://en.wikipedia.org/wiki/Cauchy's_integral_formula )
    [tex]f^k(z)=\frac{k!}{2i\pi}\int_{C}\frac{f(\zeta)d\zet a}{(\zeta-z)^{k+1}}[/tex]

    3. The attempt at a solution

    So, the first thing I did was try and get it in the form of the formula. I did a partial fraction expansion and have integral of(1/(z-2) + 1/(z+2)). You can split these apart if I recall correctly, however, once I get there I have absolutely no idea what to do. I have some examples and things that I've looked at, but I can't seem to really put it together.

    I don't understand how to apply the formula is what I mean.
     
  2. jcsd
  3. Dec 12, 2007 #2

    Gib Z

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    Homework Helper

    [tex]\int_{|z+1|=2} \frac{z^2}{4-z^2} dz = - \int_{|z+1|=2} \frac{-z^2 + 4 -4}{4-z^2} dz [/tex]

    So basically your only problem is to evaluate the integral only the same region, but now with integrand [tex]\frac{1}{4-z^2}[/tex].

    Partial fractions should give a form [tex]\frac{a}{2+z} + \frac{b}{2-z}[/tex], which might have turned out to be your evaluations if b is negative. Sadly, it isn't. Check that.
     
  4. Dec 12, 2007 #3

    mjsd

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    These integrals are best handled by the residue theory. But I can show you how to use Cauchy Integral formula to do this.
    Firstly understand the domain bounded by the curve |Z+1|=2, it is a circle of radius 2, centre at (-1,0)
    Secordly, find out all the poles of your function [tex]g(z)=\frac{z^2}{4-z^2}[/tex]
    well this language of "finding the poles" may sound a bit strange, but it all relates back to the language used in residue theory. Anyway, without confusing you too much, all it really means here is to work out all the [tex]a\text{'s}[/tex] in the Cauchy formula:

    [tex]f(a) = \frac{1}{2\pi i}\int_{\partial D}\frac{f(z)}{a-z}dz,\quad a\in D[/tex]
    where [tex]\partial D[/tex] is the boundary of domain [tex]D[/tex].

    Here, [tex]g(z) = \frac{z^2}{4-z^2} = \frac{z^2}{(2-z)(2+z)}[/tex], the poles are -2 and 2. If you look at it carefully, in this case, you do not need to employ any tricks such as partial fractions or "excising any holes to your domain D". This is because only -2 is inside your domain [tex]|z+1|\leq 2[/tex]. This observation means that now you only need to evaluate an integral that looks like:

    [tex]\int_{|z+1|=2}\frac{h(z)}{(2+z)}dz,\quad -2\in \{z:|z+1|\leq 2\};\quad
    \text{ with }\;\; h(z) = \frac{z^2}{2-z}[/tex]

    Note that h(z) is analytic in the domain given so you can now apply Cauchy Integral Formula to complete the job.
    .
     
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