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

g(ζ)=f(ζ)-f(z)ζ-z,ζD\S,

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

g(ζ)f(z),asζz.

Therefore, by the Cauchy integral theorem

Cg(ζ)𝑑ζ=0,

where C is the counterclockwise circular contour parameterized by

ζ=z0+Reit, 0t2π.

Hence,

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

f(z)=12πiCf(ζ)ζ-z𝑑ζ,zD,

as desired.

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

ddzf(ζ)ζ-z=f(ζ)(ζ-z)2,zD

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:

f(z)=12πiddzCf(ζ)ζ-z𝑑ζ=12πiCf(ζ)(ζ-z)2𝑑ζ,zD.
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