proof of Cauchy integral formula
Let D={z∈ℂ:∥z-z0∥<R} be a disk in the complex plane, S⊂D a finite subset, and U⊂ℂ an open domain that contains the closed disk ˉD. Suppose that
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f:U\S→ℂ is holomorphic, and that
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f(z) is bounded near all z∈D\S.
Hence, by a straightforward compactness argument we also have that f(z) is bounded on ˉD\S, and hence bounded on D\S.
Let z∈D\S be given, and set
g(ζ)=f(ζ)-f(z)ζ-z,ζ∈D\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),asζ→z. |
Therefore, by the Cauchy integral theorem
∮Cg(ζ)𝑑ζ=0, |
where C is the counterclockwise circular contour parameterized by
ζ=z0+Reit, 0≤t≤2π. |
Hence,
∮Cf(ζ)ζ-z𝑑ζ=∮Cf(z)ζ-z𝑑ζ. | (1) |
𝐋𝐞𝐦𝐦𝐚 If z∈ℂ is such that ∥z∥≠1, 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)=12πi∮Cf(ζ)ζ-z𝑑ζ,z∈D, |
as desired.
Since a circle is a compact set, the defining limit for the derivative
ddzf(ζ)ζ-z=f(ζ)(ζ-z)2,z∈D |
converges uniformly for ζ∈∂D. Thanks to the uniform convergence, the order of the derivative and the integral operations can be interchanged. In this way we obtain the second formula:
f′(z)=12πiddz∮Cf(ζ)ζ-z𝑑ζ=12πi∮Cf(ζ)(ζ-z)2𝑑ζ,z∈D. |
Title | proof of Cauchy integral formula |
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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 |