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 counterclockwise.
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 selfintersections and which is oriented counterclockwise. 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 farreaching 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
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