## Sources

Here are some discussions of the Cauchy Integral Theorem:

- The General Cauchy Theorem, chapter three of an on-line book on complex variables by Robert B. Ash and W.P. Novinger

- Cauchy's Integral Theorem, chapter five of Introduction to Complex Analysis by W.W.L. Chen

## The Theorem

And here is how the above sources describe the theorem.

### Ash and Novinger

3.3 Cauchy’s TheoremThis 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 TheoremLet $\gamma$ 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’stheorem, and part (ii) as Cauchy’sintegral 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 CaseCauchy’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 inD.

### Mathews

Theorem 6.5 (Cauchy-Goursat Theorem).Letf(z)be analytic in a simply connected domainD. IfCis a simple closed contour that lies inD, 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.*

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