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 obtain integral formulas for all derivatives of a holomorphic function. The analytic significance of Cauchy's formula is that it shows that in complex analysis "differentiation is equivalent to integration": thus complex differentation, like integration, behaves well under uniform limits, a result which is not true in real analysis.

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Suppose U is an open subset of the complex plane C, f : UC is a holomorphic function and the closed disk D = { z : | zz0| ≤ r} is completely contained in U. Let C be the circle forming the boundary of D. Then 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 contour integral is taken counter-clockwise.

The proof of this statement uses the Cauchy integral theorem and similarly only requires f to be complex differentiable. The formula implies that f is actually infinitely differentiable, with

f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz.

This formula is sometimes referred to as Cauchy's differentiation formula; it is also a consequence of the fact that holomorphic functions are analytic.

The circle C can be replaced by any closed rectifiable curve in U which has winding number one about a. Moreover, as for the Cauchy integral theorem, it is sufficient to require that f be holomorphic in the open region enclosed by the path and continuous on its closure.

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 an arbitrarily small circle around a. Since f(z) is continuous, we can choose a circle small enough on which f(z) is arbitrarily close to f(a). On the other hand, the integral

\oint_C { {1 \over z-a} \,dz}

over any circle C centered at a is 2πi. (This can be calculated directly via parametrization (integration by substitution)

 z = a + \varepsilon e^{it}

where 0 ≤ t ≤ 2π and ε is the radius of the circle.) Letting ε → 0 gives the desired estimate

\left | \frac{1}{2 \pi i} \oint_C { {f(z) \over z-a} \,dz}  - f(a) \right |

\leq \frac{1}{2 \pi i} \oint_C \frac{ |f(z) - f(a)| } {z-a} \,dz \rightarrow 0.

Surface of the function g(z) = z2 / (z2 + 2z + 2) and its singularities, with the contours described in the text.
Surface of the function g(z) = z2 / (z2 + 2z + 2) and its singularities, with the contours described in the text.

Consider the function

g(z)={z^2 \over z^2+2z+2}

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

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

g(z)={z^2 \over (z-z_1)(z-z_2)}

where z1 = − 1 + i, z2 = − 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)={z^2 \over z-z_2}

and now

\oint_C {g(z)} = \oint_C {f(z) \over z-a}\, dz=2\pi i*f(a)


\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)={z^2 \over z-z_1},


\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:

\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}

f^{(n)}(a) = {n! \over 2\pi i} \oint_C {f(z) \over (z-a)^{n+1}}\, dz.

Because one may deduce from the formula that f must be infinitely often continuously differentiable, the Cauchy integral theorem has broad implications. It is used to prove the residue theorem, which is a far-reaching generalization that removes the requirement that the function be analytic in the enclosed region. The Cauchy integral theorem has no counterpart in real analysis because for a real valued function possession of a first derivative by a function will not guarantee the existence of higher order derivatives. In contrast to this, the proof of the Cauchy integral formula for nth derivatives shows that analytic functions posses derivatives of all orders.[1]

It is known from Morera's theorem that the uniform limit of holomorphic functions is holomorphic. This can also be deduced from Cauchy's integral formula: indeed the formula also holds in the limit and the integrand, and hence the integral, can be expanded as a power series. In addition the Cauchy formulas for the higher order derivatives show that all these derivatives also converge uniformly.

  1. ^ Transform Calculus: with an Introduction to Complex Variables by E. J. Scott

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