nLab Cauchy's integral formula

Statement

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

$$f(\zeta) = \frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z - \zeta} \,\mathrm{d}z.$$

Hence the contour integral picks out the enclosed residues.

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

$$f^{(n)}(\zeta) = \frac{n!}{2\pi i} \oint_C \frac{f(z)}{(z - \zeta )^{n+1}} d z \,.$$

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 $\epsilon$ be a nilpotent. Let $S_\epsilon$ denote the circle of radius $\epsilon$ centered at $\zeta$. By the holomorphicity of $f$, the differential 1-form $f(z)/(z-\zeta) \,\mathrm{d}z$ is closed in the region bounded by $C$ and $S_\epsilon$. By the Stokes theorem,

$$\frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z - \zeta} \,\mathrm{d}z = \frac{1}{2\pi\mathrm{i}}\oint_{S_\epsilon} \frac{f(z)}{z - \zeta} \,\mathrm{d}z.$$

Parametrize $S_\epsilon$ by $(t\mapsto \zeta + \epsilon \,\exp(2\pi\mathrm{i}t))$ to transform the above integral to

$$\frac{1}{2\pi\mathrm{i}}\int_0^1 \frac{f(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t))}{\epsilon \,\exp(2\pi\mathrm{i}t))} \,\mathrm{d}(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t)) = \int_0^1 f(\zeta + \epsilon\, \exp(2\pi\mathrm{i}t)) \, \mathrm{d}t$$

By the infinitesimal Taylor formula and the holomorphicity of $f$,

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

Hence the above integral is equal to

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

Related concepts

• Cauchy's integral theorem

• Cauchy's theorems

• Goursat theorem

References

• Wikipedia, Cauchy’s integral formula