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Cauchy's theorem and formula

  1. Feb 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Verify that
    a) ##\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}\bar z}{1-\bar z e^{it}}dt = 0##, if ##f(w)## is analytic for ##|w|<1+\epsilon##, and that
    b) ##\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}}{e^{it}-z}dt = f(z).##
    for ##z = re^{i\theta}## with ##r < 1##.

    Use the above to verify ##f(x) = \frac{1}{2\pi} \int_0^{2\pi} f(e^{it}) P_r(\theta) dt, \; \; z = re^{i\theta}.##

    2. Relevant equations
    Cauchy's theorem:
    Suppose that ##f## is analytic on a domain ##D##. Let ##\gamma## be a piecewise smooth simple closed curve in ##D## whose inside ##\Omega## also lies in ##D##. Then
    ##\int_\gamma f(z)dz = 0.##

    Cauchy's formula:
    Suppose that ##f## is analytic on a domain ##D## and that ##\gamma## is a piecewise smooth, positively oriented simple closed curve in ##D## whose inside ##\Omega## also lies in ##D##. Then
    ##f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(\xi)}{\xi - z}d\xi, \; \; \forall z \in \Omega.##

    Poisson kernel:
    ##P_r(\theta) = \frac{1-r^2}{1-2r\cos (\theta-t)+r^2} = \frac{e^{it}}{e^{it}-z}+\frac{\bar z e^{it}}{1-e^{it}\bar z}##.
    3. The attempt at a solution
    For a) it seems to follow directly from Cauchy theorem since both ##f## and the other part is analytic and the product of two analytic functions is analytic. However in this case the argument, ##|w| = |e^{it}| = 1##, isn't less than one? Perhaps the ##\epsilon## fixes this? But I don't understand the reason for the ##\epsilon##, is that a typo?

    I don't really see the difference from a) here. For the same region if a) is zero then so is b). However If I compute it with Cauchy's formula I indeed get the desired result:

    Substituting ##s = e^{it}## we have ##dt = \frac{ds}{ie^{it}}## and hence we get
    ##\frac{1}{2\pi} \int_0^{2\pi} f(e^{it})\frac{e^{it}}{e^{it}-z}dt = \frac{1}{2\pi i} \int \frac{f(s)}{s-z} = f(z)##
     
  2. jcsd
  3. Feb 15, 2016 #2

    Samy_A

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    To understand the difference between a) and b), consider the following functions defined for ##s \in \mathbb C, |s| \lt 1+\epsilon##, where ##z \in \mathbb C, |z| \lt 1## is fixed:
    ##f_a(s)=\frac{f(s)\bar z}{1-\bar zs}##
    ##f_b(s)=\frac{f(s)}{s-z}##

    Which of these function is holomorphic in an open region that includes ##\{s \in \mathbb C | \ |s|\leq 1 \}##, and which is not? That will hopefully explain why in one case you apply the Cauchy integral theorem, and in the other the Cauchy integral formula.
     
  4. Feb 15, 2016 #3
    Thanks I believe I understand now! Since the ##r < 1## ##f_a## doesn't have a singularity in ##\Omega## while the second one does since ##|s| = r## is allowed.

    Do you have any idea about the ##\epsilon##? Is that pretty much just another way of writing ##|s| \le 1## avoiding the less equal sign?
     
  5. Feb 15, 2016 #4

    Samy_A

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    Exactly.
    No.

    You need that ##1+\epsilon## in order to apply the Cauchy integral theorem.

    From Wikipedia:
    Let ##U## be an open subset of ##\mathbb C## which is simply connected, let ##f : U \to \mathbb C ## be a holomorphic function, and let ##\gamma## be a rectifiable path in ##U## whose start point is equal to its end point. Then
    $$\oint_{\gamma} f(z) dz =0$$

    In your case, it means that the unit circle must lie inside the open subset in which the function is holomorphic. So they take an open subset just a little bit larger, with radius ##1+\epsilon##. That way you can safely apply the Cauchy integral theorem with the unit circle as ##\gamma##.
    Just using ##|s| \le 1## doesn't define an open subset.
     
  6. Feb 15, 2016 #5
    Thanks, nicely explained!
     
  7. Feb 15, 2016 #6

    Samy_A

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    You are welcome. Complex analysis is pure beauty!
    /nerd mode
     
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