Cauchy Integral Theorem

Sources

Here are some discussions of the Cauchy Integral Theorem:

The Theorem

And here is how the above sources describe the theorem.

Ash and Novinger

3.3 Cauchy’s Theorem

This section is devoted to a discussion of the global (or homology) version of Cauchy’s theorem. The elementary proof to be presented below is due to John Dixon, and appeared in Proc. Amer. Math. Soc. 29 (1971), pp. 625-626, but the theorem as stated is originally due to E.Artin.

3.3.1 Cauchy’s Theorem

Let $\gamma$ be [a] closed path in $\Omega$ such that $n(\gamma, z) = 0$ for all $z\in \mathbb{C}\backslash \Omega$. [Then,]

(i) For all analytic functions $f$ on $\Omega, \int_\gamma f(w) dw = 0;$

(ii) If $z\in\Omega\backslash \gamma*,$ then

$n(\gamma, z) f(z) = \frac{1}{2\pi i} \int_\gamma \frac{f(w)}{w-z}dw$.

The authors add

3.3.4 Remarks

Part (i) of (3.3.1) is usually referred to as Cauchy’s theorem, and part (ii) as Cauchy’s integral formula. In the above proof we derived (i) from (ii); see Problem 1 for the reverse implication.

Also, there is a converse to part (i): If $\gamma$ is a closed path in $\Omega$ such that $\int_\gamma f (w) dw = 0$ for every $f$ analytic on $\Omega$, then $n(\gamma, z ) = 0$ for every $z \notin \Omega.$ To prove this, take $f (w) = 1/(w - z )$ and apply (3.2.3).

—where

3.2.3 Theorem

Let $\gamma$ be a closed path, and $z_0$ a point not belonging to $\gamma\ast$. Then

$n(\gamma, z_0) = \frac{1}{2\pi i}\int_\gamma \frac{1}{z-z_0}dz.$

Chen

5.1. A Restricted Case

Cauchy’s integral theorem states that in a simply connected domain, the integral of an analytic function over a closed contour is zero. The proof of this general result is rather involved. Here we first study a special case of the theorem in order to develop the basic properties of analytic functions.

THEOREM 5A. Suppose that a function f is analytic in a domain D. Suppose further that the closed triangular region T lies in D, and that C denotes the boundary of T in the positive (anticlockwise) direction.
Then

$\int_C f(z)dz=0$.

We shall give two proofs of this result, usually known as Cauchy’s integral theorem for a triangular path. The first of these proofs, given next, is based on an additional assumption that the derivative f’(z) is continuous in D.

Mathews

Theorem 6.5 (Cauchy-Goursat Theorem). Let f(z) be analytic in a simply connected domain D. If C is a simple closed contour that lies in D, then

$\int_C f(z)dz=0.$

Proof.

Proof of Theorem 6.5 is in the book.
Complex Analysis for Mathematics and Engineering

Planetmath.org

In the latter part of the 19th century E. Goursat found a proof of the integral theorem that merely required that $f'(z)$ exist. Continuity of the derivative, as well as the existence of all higher derivatives, then follows as a consequence of the Cauchy integral formula. Not only is Goursat's version a sharper result, but it is also more elementary and self-contained, in that sense that it is does not require Green's theorem. Goursat's argument makes use of rectangular contour (many authors use triangles though), but the extension to an arbitrary simply-connected domain is relatively straight-forward.

Theorem 3 (Goursat) Let $U$ be an open domain containing a rectangle

$R = \{ x+iy\in\mathbb{C}: a\leq x\leq b\,, c\leq y\leq d\}.$

If the complex derivative of a function $f:U\rightarrow \mathbb{C}$ exists at all points of $U$, then the contour integral of $f$ around the boundary of $R$ vanishes; in symbols

$\oint_{\partial R} f(z)\,dz = 0.$

Advisory

Please note that I am not a mathematician and so the presentation of proofs that I make may be deeply flawed. I'm using this writing process to figure out what I'm reading. Please consult more authoritative sources as well.

Feel free to contact me by leaving a comment or sending me a private message.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License