Cauchy's Integral Formula - Proof Sketch

Proof Sketch

By using the Cauchy integral theorem, one can show that the integral over C (or the closed rectifiable curve) is equal to the same integral taken over an arbitrarily small circle around a. Since f(z) is continuous, we can choose a circle small enough on which f(z) is arbitrarily close to f(a). On the other hand, the integral

over any circle C centered at a. This can be calculated directly via a parametrization (integration by substitution) where 0 ≤ t ≤ 2π and ε is the radius of the circle.

Letting ε → 0 gives the desired estimate

\begin{align} \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)}{z-a} \,dz - f(a) \right | &= \left | \frac{1}{2 \pi i} \oint_C \frac{f(z)-f(a)}{z-a} \,dz \right |\\ &\leq \frac{1}{2 \pi} \int_0^{2\pi} \frac{ |f(z(t)) - f(a)| } {\varepsilon} \,\varepsilon\,dt\\ &\leq \max_{|z-a|=\varepsilon}|f(z) - f(a)| \xrightarrow{} 0. \end{align}

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