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Cauchy's integral formula

Cauchy's integral formula is a central statement in complex analysis. It expresses the fact that a holomorphic function defined on a disk is completely determined by its values on the boundary of the disk. It can also be used to formulate integral formulas for all derivatives of a holomorphic function.

Supppose U is an open subset of the complex plane C, and f : UC is a holomorphic function, and the disk D = { z : |z - z0| ≤ r} is completely contained in U. Let C be the circle forming the boundary of D. Then we have for every a in the interior of D:

<math>f(a) = {1 \over 2\pi i} \int_C {f(z) \over (z-a)} dz </math>

where the integral is to be taken counter-clockwise.

The proof of this statement uses the Cauchy integral theorem and, just like that theorem, only needs that f is complex differentiable. One can then deduce from the formula that f must actually be infinitely often continuously differentiable, with

<math>f^{(n)}(a) = {n! \over 2\pi i} \int_C {f(z) \over (z-a)^{n+1}} dz</math>

One may replaces the circle C with any closed rectifiable curve in U which doesn't have any self-intersections and which is oriented counter-clockwise. The formulas remain valid for any point a from the region enclosed by this path. Moreover, just as in the case of the Cauchy integral theorem, it is sufficient to require that f be holomorphic in the open region enclosed by the path and continuous on that region's closure.

These formulas can be used to prove the residue theorem, which is a far-reaching generalization.

Sketch of the proof of Cauchy's integral formula

By using the Cauchy integral theorem, one can show that the integral over C (or the closed rectifiable curve) is equal to the same integral taken over a tiny circle around a. Since f(z) is continuous, we can choose a circle small enough on which f(z) is almost constant and equal to f(a). We then need to evaluate the integral

∫ 1/(z-a) dz
over this small circle. It turns out that the value of this integral is independent of the circle's radius: it is equal to 2πi.

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