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.
\begin{figure}\mbox{ \subfigure[smooth]{\epsfig{file=smootharc1.eps,height=3cm}}...
...mooth at one
point]{\epsfig{file=nonsmootharc1.eps,height=3cm}} }\end{figure}

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:

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

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:

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

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
\mbox{\epsfig{file=TwoPaths.eps, height=3cm}}\end{figure}

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:

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

Figure 4: Loop within loop.

Noah Dana-Picard 2007-12-24