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Proof Using Cauchy's Formula

  1. Apr 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Let D be a simply connected region in C (complex domain) and let C be a simple closed curve contained in D. Let f(z) be analytic in D. Suppose that z0 is a point which is not enclosed by C. What is 1/(2πi)∫f(z)/(z-z0)dz?


    2. Relevant equations
    Cauchy's formula: f(z0) = 1/(2πi)∫f(z)/(z-z0)dz


    3. The attempt at a solution
    I have a gut feeling that since z0 is not in enclosed by C, it is also not part of D. Since f(z) is analytic in D, this somehow means that f(z) = 0 outside of D so f(z0) = 0. I know I'm missing a lot of connections and mathematical reasoning, but this is a guess I have because I don't actually know how to show it mathematically. Help appreciated!
     
  2. jcsd
  3. Apr 26, 2014 #2

    haruspex

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    What f does outside of D is unknown, so irrelevant.
    C is contained in D; z0 is in D but outside C. Are you perhaps confused by the use of C at the start of the question as a reference to the complex plane? Everywhere else it is referring to a closed curve inside D.
    For Cauchy's integral formula to apply, what would be the relationship between z0 and the path of the integral?
     
  4. Apr 26, 2014 #3

    HallsofIvy

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    The whole point of this question is that the integral of an function, analytic inside a given closed curve, around that closed curve, is 0. Did you not know that?

    The proof is fairly simple: any analytic function, pretty much by definition, can be written as a Taylor's series. If you integrate term by term, the integral of [itex](z- a)^n[/itex] about a closed curve containing a, is equivalent to integration around a circle with center at a and that is 0:
    Let [tex]z= a+ Re^{i\theta}[/tex] and [tex]\int_C (z- a)^n dz= \int_0^{2\pi} (Re^{i\theta})(iRe^{i\theta}d\theta)= iR^2\int_0^{2\pi} e^{2i\theta}d\theta= \left[(R^2/2)e^{2i\theta}\right]_0^{2\pi}= 0[/tex]
     
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