# 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* : *U* → **C** is a holomorphic function, and the disk *D* = { *z* : |*z* - *z*_{0}| ≤ *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*:

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

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.

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*) d*z*

over this small circle. It turns out that the value of this integral is independent of the circle's radius: it is equal to 2π*i*.