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Math Help - Complex Variables

  1. #1
    Senior Member vincisonfire's Avatar
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    Complex Variables

    Let  f : D \rightarrow \mathbb C be holomorphic, let  z_0 \in D and let  r > 0  be such that the disk  |w-z_0| \leq r \in D then  f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta .
    I started from the Cauchy integral formula and got  f'(z) = \frac{1}{2\pi r} \oint\limits_0^{2\pi} f(z+re^{i\theta})e^{-i\theta}d\theta .
    I feel like something is missing in the question.
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  2. #2
    Senior Member vincisonfire's Avatar
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    Ok got it!
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    Quote Originally Posted by vincisonfire View Post
    Ok got it!
    Maybe you could post your solution so that others can enjoy it. You could post it inside spoiler tags.
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    Senior Member vincisonfire's Avatar
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    After the exams. Promised.
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  5. #5
    Senior Member vincisonfire's Avatar
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    We want to show f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta
    Cauchy integral formula is  f'(z) = \frac{1}{2\pi i} \oint\limits_{\gamma} \frac{f(w)}{(w-z)^2}dw
    Let  w= z+r e^{i\theta} and  dw= ire^{i\theta}d\theta.
     f'(z) = \frac{1}{2\pi i} \oint\limits_{\gamma} \frac{f(z+re^{i\theta})}{(z+re^{i\theta}-z)^2} ire^{i\theta}d\theta = \frac{1}{2\pi r} \oint\limits_0^{2\pi} f(z+re^{i\theta})e^{-i\theta}d\theta<br />
.
    Then, since f is holomorphic,  f'(z) = \sum\limits_{n=0}^{\infty} a_nz^n and from that we get that \overline{f'(z)} = f'(\overline{z}).
    Now we integrate around the upper half of the circle from 0 to \pi and around the lower half of the circle from 0 to -\pi. For the second we might as well integrate 0 to \pi but change the sign of  \theta .
     f'(z) = \frac{1}{2\pi r} \oint\limits_0^{\pi} f(z+re^{i\theta})e^{-i\theta}-\oint\limits_0^{\pi}f(z+re^{-i\theta})e^{i\theta}d\theta<br />
.
     f'(z) = \frac{1}{\pi r} \oint\limits_0^{\pi} \frac{f(z+re^{i\theta})e^{-i\theta}-f(z+re^{-i\theta})e^{i\theta}}{2}d\theta<br />
     f'(z) = \frac{1}{\pi r} \oint\limits_0^{\pi} iIm\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta<br />
     f'(z) = \frac{i}{\pi r} \oint\limits_0^{\pi} Im\left(f(z+re^{i\theta})e^{-i\theta}\right)d\theta<br />
    I may have made mistakes though.
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