proof of Cauchy integral formula

Let D={z:z-z0<R} be a disk in the complex plane, SD a finite subset, and U an open domain that contains the closed disk D¯. Suppose that

Hence, by a straightforward compactness argument we also have that f(z) is bounded on D¯\S, and hence bounded on D\S.

Let zD\S be given, and set


where S=S{z}. Note that g(ζ) is holomorphic and bounded on D\S. The second assertion is true, because


Therefore, by the Cauchy integral theorem


where C is the counterclockwise circular contour parameterized by

ζ=z0+Reit, 0t2π.


Cf(ζ)ζ-z𝑑ζ=Cf(z)ζ-z𝑑ζ. (1)

𝐋𝐞𝐦𝐦𝐚 If z is such that z1, then

ζ=1dζζ-z={0if z>12πiif z<1

The proof is a fun exercise in elementary integral calculus, an application of the half-angle trigonometric substitutions.

Thanks to the Lemma, the right hand side of (1) evaluates to 2πif(z). Dividing through by 2πi, we obtain


as desired.

Since a circle is a compact set, the defining limit for the derivative


converges uniformly for ζD. Thanks to the uniform convergenceMathworldPlanetmath, the order of the derivative and the integral operations can be interchanged. In this way we obtain the second formula:

Title proof of Cauchy integral formula
Canonical name ProofOfCauchyIntegralFormula
Date of creation 2013-03-22 12:47:23
Last modified on 2013-03-22 12:47:23
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 15
Author rmilson (146)
Entry type Proof
Classification msc 30E20