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Cauchy's integral formula

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Statement

For UU \subset \mathbb{C} an open subset of the complex plane \mathbb{C} and for CC a Jordan curve in UU, a holomorphic function ff of UU sends a point ζ\zeta enclosed by CC to the contour integral

f(ζ)=12πi Cf(z)zζdz.. f(\zeta) = \frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z - \zeta} \,\mathrm{d}z. \,.

Hence the contour integral picks out the enclosed residues.

More generally, this implies, by Taylor series expansion of ff, that for nn \in \mathbb{N} the nnth complex derivative f (n)f^{(n)} is

f (n)(ζ)=n!2πi Cf(z)(zζ) n+1dz. f^{(n)}(\zeta) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - \zeta )^{n+1}} d z \,.

This is also known as Cauchy’s differentiation formula.

Proof in synthetic differential geometry

Here is a proof written in terms of synthetic infinitesimals as in synthetic differential geometry:

Let ϵ\epsilon be a nilpotent. Let S ϵS_\epsilon denote the circle of radius ϵ\epsilon centered at ζ\zeta. By the holomorphicity of ff, the differential 1-form f(z)/(zζ)dzf(z)/(z-\zeta) \,\mathrm{d}z is closed in the region bounded by CC and S ϵS_\epsilon. By the Stokes theorem,

12πi Cf(z)zζdz=12πi S ϵf(z)zζdz.\frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z - \zeta} \,\mathrm{d}z = \frac{1}{2\pi\mathrm{i}}\oint_{S_\epsilon} \frac{f(z)}{z - \zeta} \,\mathrm{d}z.

Parametrize S ϵS_\epsilon by (tζ+ϵexp(2πit))(t\mapsto \zeta + \epsilon \,\exp(2\pi\mathrm{i}t)) to transform the above integral to

12πi 0 1f(ζ+ϵexp(2πit))ϵexp(2πit))d(ζ+ϵexp(2πit))= 0 1f(ζ+ϵexp(2πit))dt\frac{1}{2\pi\mathrm{i}}\int_0^1 \frac{f(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t))}{\epsilon \,\exp(2\pi\mathrm{i}t))} \,\mathrm{d}(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t)) = \int_0^1 f(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t)) \, \mathrm{d}t

By the infinitesimal Taylor formula and the holomorphicity of ff,

f(ζ+ϵexp(2πit))=f(ζ)+ϵexp(2πit)f(ζ)z.f(\zeta + \epsilon\exp(2\pi\mathrm{i}t)) = f(\zeta) + \epsilon\,\exp(2\pi\mathrm{i}t) \frac{\partial f(\zeta)}{\partial z}.

Hence the above integral is equal to

0 1[f(ζ)+ϵexp(2πit)f(ζ)z]dt=[f(ζ)+ϵf(ζ)z 0 1exp(2πit)dt]=f(ζ).\int_0^1 \left[f(\zeta) + \epsilon\,\exp(2\pi\mathrm{i}t) \frac{\partial f(\zeta)}{\partial z}\right] \, \mathrm{d}t = \left[f(\zeta) + \epsilon\frac{\partial f(\zeta)}{\partial z} \int_0^1 \exp(2\pi\mathrm{i}t) \, \mathrm{d}t\right] = f(\zeta).

References

Revised on September 6, 2017 11:11:54 by Urs Schreiber (77.56.177.247)