Cauchy's Integral Formula

In mathematics, Cauchy's integral formula, named after Augustin 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:
f(a) = {1 \over 2\pi i} \oint_C {f(z) \over z-a}\, dz
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
f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz.
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.

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. We may do it by choosing the parametrization (variable substitution)
z = a + \epsilon\cdot\exp{(i\cdot t)}
where 0 \le t \le 2\cdot\pi and \epsilon \rightarrow 0 . It turns out that the value of this integral is independent of the circle's radius: it is equal to 2πi.

Example usage

Consider the function
f(z)={z^2 \over z^2+2z+2}
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:
f(z)={z^2 \over (z-z_1)(z-z_2)}, \mbox{where}\ z_1=-1+i, z_2=-1-i.
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:
f(z)={\left({z^2 \over z-z_2}\right) \over z-z_1}
and now
{1 \over 2 \pi i}\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}.
Doing likewise for the other contour:
f(z)={\left({z^2 \over z-z_1}\right) \over z-z_2}
{1 \over 2 \pi i}\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}.
The integral around the original contour C then is the sum of these two integrals:
\oint_C {z^2 \over z^2+2z+2}\,dz = {1 \over 2 \pi i}\oint_{C_1} {\left({z^2 \over z-z_2}\right) \over z-z_1}\,dz + {1 \over 2 \pi i}\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.\;\!

 

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