Cauchy's Theorems.

Let $C:[a,b] \longrightarrow \mathbb{C}$ be a path of integration. It is smooth at a point $t_0 \in [a,b]$ if $C$ is derivable at $t_0$ and if its first derivative is continuous at $t_0$ . The path is smooth if it is smooth at every point; it is smooth by parts if it is not smooth at only a finite number of points.

Example 5.3.1   The arc $AB$ of the parabola whose equation is $y=x^2$ is smooth.

Figure 2: Paths.

A loop is an integrating path whose origin and endpoints are identical. If it is smooth and does not intersect itself at another point, we will call it a Jordan curve. If we travel exactly once along the loop, we will call it a simple loop. The integral of a function $f$ along a simple loop $\mathcal{C}$ will be denoted as follows:

$$\oint_{\mathcal{C}} f(z) \; dz.$$

Note the little circle on the integration symbol.

A Jordan curve determines in the plane three disjoint regions: the curve itself, the interior (= a bounded region) and the exterior (= an unbounded region).

Theorem 5.3.2 (Cauchy-Goursat)   Let $f$ be a function, analytic on an open simply connected subset $U$ of $\mathbb{C}$ . If $C$ is a Jordan curve in $U$ , then $\int_C f(z) \; dz =0$ .

Note that the converse is not true: if $f$ is not analytic on the interior of $C$ , the integral $\oint_C f(z) \; dz$ can either vanish or not. For example, compute the following integral:

$$I=\oint_{\vert z\vert=1} \frac {1}{z^2} \; dz.$$

Using the parametrization $z=\cos t + i \; \sin t, \; t\in [0,2 \pi ]$ , we can show that $I=0$ . The integral vanishes, despite the fact that the function fails to be analytic at 0, which is an interior point of the unit circle (defined here by the equation $\vert z\vert=1$ ).

Example 5.3.3  

1. $\int_C \sin z \; dz =0$ , for every Jordan curve $C$ in the Cauchy-Argand plane.
2. Let $f(z)=\frac {e^z}{z^2-4}$ and let $C$ be the unit circle. As $f$ is analytic on the closed unit disk, we have: $\int_C \frac {e^z}{z^2-4} \; dz =0$ .

Corollary 5.3.4   Let $f$ be a function defined and analytic on a connected domain $\mathcal{D}$ . Let $\mathcal{C}_1$ and $\mathcal{C}_2$ be two paths with the same origin $A$ and the same endpoint $B$ such that both path contain only interior points of $\mathcal{D}$ (see Figure 3). Then $\int_{\mathcal{C}_1} f(z) \; dz = \int_{\mathcal{C}_2} f(z) \; dz$ .

Figure 3: Two paths

Theorem 5.3.5   Consider two Jordan curves $C_1$ and $C_2$ such that all the points of $C_1$ are interior to $C_2$ (Figure 4). Let $f(z)$ be a function, analytic on $C_1$ , on $C_2$ and at every point of the ``annulus'' bounded by $C_1$ and $C_2$ . Then:

$$\int_{C_1} f(z) \; dz = \int_{C_2} f(z) \; dz$$

Figure 4: Loop within loop.