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    Cauchy integral formula on a disc

    T-6DEUH

    The Cauchy kernel reconstructs every derivative at the center from boundary values.Suggest a slogan

    Let \(\Omega\subseteq\CC\) be open and let \(f\) be holomorphic on \(\Omega\). Let \(z_0\in\Omega\) and \(R>0\) satisfy \(\overline{D_R(z_0)}\subseteq\Omega\), and let \(\gamma=\bd D_R(z_0)\) be the positively oriented circle. Then \[ f(z_0)=\frac{1}{2\pi i}\int_\gamma\frac{f(\xi)}{\xi-z_0}\dxi, \] and for every \(n\ge0\), \[ f^{(n)}(z_0)=\frac{n!}{2\pi i}\int_\gamma\frac{f(\xi)}{(\xi-z_0)^{n+1}}\dxi. \]

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