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5.3: Proof of Cauchy's integral formula

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    6497
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    Useful theorem

    Before proving the theorem we’ll need a theorem that will be useful in its own right.

    Theorem \(\PageIndex{1}\): A second extension of Cauchy's theorem

    Suppose that \(A\) is a simply connected region containing the point \(z_0\). Suppose \(g\) is a function which is

    1. Analytic on \(A\) - {\(z_0\)}
    2. Continuous on \(A\). (In particular, \(g\) does not below up at \(z_0\).)

    Then

    \[\int_{C} g(z)\ dz = 0 \nonumber \]

    for all closed curves \(C\) in \(A\).

    Proof

    The extended version of Cauchy’s theorem in the Topic 3 notes tells us that

    \[\int_C g(z)\ dz = \int_{C_r} g(z)\ dz, \nonumber \]

    where \(C_r\) is a circle of radius \(r\) around \(z_0\).

    屏幕快照 2020-09-08 下午4.42.33.png

    Since \(g(z)\) is continuous we know that \(|g(z)|\) is bounded inside \(C_r\). Say, \(|g(z)| < M\). The corollary to the triangle inequality says that

    \[|\int_{C_r} g(z)\ dz| \le M \text{ (length of } C_r) = M2 \pi r. \nonumber \]

    Since \(r\) can be as small as we want, this implies that

    \[\int_{C_r} g(z) \ dz = 0. \nonumber \]

    Note

    Using this, we can show that \(g(z)\) is, in fact, analytic at \(z_0\). The proof will be the same as in our proof of Cauchy’s theorem that \(g(z)\) has an antiderivative.

    Proof of Cauchy’s integral formula

    We reiterate Cauchy’s integral formula from Equation 5.2.1: \(f(z_0) = \dfrac{1}{2\pi i} \int_C \dfrac{f(z)}{z - z_0} \ dz\).

    屏幕快照 2020-09-08 下午4.49.07.png

    \(Proof\). (of Cauchy’s integral formula) We use a trick that is useful enough to be worth remembering. Let

    \[g(z) = \dfrac{f(z) - f(z_0)}{z - z_0}. \nonumber \]

    Since \(f(z)\) is analytic on \(A\), we know that \(g(z)\) is analytic on \(A - \{z_0\}\). Since the derivative of \(f\) exists at \(z_0\), we know that

    \[\lim_{z \to z_0} g(z) = f'(z_0). \nonumber \]

    That is, if we define \(g(z_0) = f'(z_0)\) then \(g\) is continuous at \(z_0\). From the extension of Cauchy's theorem just above, we have

    \[\int_{C} g(z)\ dz = 0, \text{ i.e. } \int_C \dfrac{f(z) - f(z_0)}{z - z_0} \ dz = 0. \nonumber \]

    Thus

    \[\int_{C} \dfrac{f(z)}{z - z_0}\ dz = \int_C \dfrac{f(z_0)}{z - z_0}\ dz = f(z_0) \int_C \dfrac{1}{z - z_0} \ dz = 2\pi i f(z_0). \nonumber \]

    The last equality follows from our, by now, well known integral of \(1/(z - z_0)\) on a loop around \(z_0\).


    This page titled 5.3: Proof of Cauchy's integral formula is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.