Cauchy's integral formula

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In mathematics, Cauchy's integral formula, named after Augustin Louis Cauchy, 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.

Suppose U is an open subset of the complex plane C, and f : U → C 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:

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

Some call this identity Cauchy's differentiation formula. A proof of this last identity is a by-product of the proof that holomorphic functions are analytic.

One may replace 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.

[edit] 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

over this small circle. We may perform the integral by changing the parametrization (integration by substitution). Let

<math> z = a + \varepsilon e^{it} </math>

where <math> 0 \le t \le 2\pi </math> and <math> \varepsilon \rightarrow 0 </math>. It turns out that the value of this integral is independent of the circle's radius: it is equal to 2πi.

[edit] Example usage

Image:ComplexResiduesExample.png Consider the function

and the contour described by |z| = 2, call it C.

To find out the integral of f(z) around the contour, we need to know the singularities of f(z). Observe that we can rewrite f as follows:

<math>f(z)={z^2 \over (z-z_1)(z-z_2)}, \mbox{where}\ z_1=-1+i, z_2=-1-i.</math>

Clearly the poles become evident, their moduli are less than 2 and thus lie inside the contour and are subject to consideration by the formula. By the Cauchy-Goursat theorem, we can express the integral around the contour as the sum of the integral around z₁ and z₂ where the contour is a small circle around each pole. Call these contours C₁ around z₁ and C₂ around z₂.

Now, around C₁, f is analytic (since the contour does not contain the other singularity), and this allows us to write f in the form we require, viz:

and now

<math>\oint_{C_1} {\left({z^2 \over z-z_2}\right) \over z-z_1}\,dz=2\pi i{z_1^2 \over z_1-z_2}.</math>

Doing likewise for the other contour:

<math>\oint_{C_2} {\left({z^2 \over z-z_1}\right) \over z-z_2}\,dz=2\pi i{z_2^2 \over z_2-z_1}.</math>

The integral around the original contour C then is the sum of these two integrals:

<math>\begin{align}\oint_C {z^2 \over z^2+2z+2}\,dz &{}= \oint_{C_1} {\left({z^2 \over z-z_2}\right) \over z-z_1}\,dz + \oint_{C_2} {\left({z^2 \over z-z_1}\right) \over z-z_2}\,dz \\ \\

&{}= 2\pi i\left({z_1^2 \over z_1-z_2}+{z_2^2 \over z_2-z_1}\right) \\ \\ &{}=2\pi i(-2) \\ \\ &{}=-4\pi i.\end{align}</math>