Cauchy's Theorem for Multiply Connected Domains

The general form of Cauchy's theorem extends to domains with holes, relating the integral over the outer boundary to integrals over inner boundary components. This is essential for residue theory.

General Cauchy Theorem

Theorem4.10Cauchy's theorem (general form)

Let $D$ be a domain bounded by a finite number of piecewise smooth simple closed curves. Let $\Gamma = \gamma_0 - \gamma_1 - \cdots - \gamma_n$ where $\gamma_0$ is the outer boundary (positively oriented) and $\gamma_1, \ldots, \gamma_n$ are the inner boundaries (negatively oriented). If $f$ is holomorphic on $D$ and continuous on $\overline{D}$ , then

$\oint_\Gamma f(z)\,dz = 0.$

That is, $\oint_{\gamma_0} f\,dz = \sum_{k=1}^n \oint_{\gamma_k} f\,dz$ where all contours are positively oriented.

RemarkHomological version

The most general version uses homology: if $f$ is holomorphic on a domain $D$ and $\gamma$ is a cycle (formal sum of closed curves) that is homologous to zero in $D$ (i.e., $n(\gamma, z) = 0$ for all $z \notin D$ ), then $\oint_\gamma f\,dz = 0$ and $f(z) n(\gamma, z) = \frac{1}{2\pi i}\oint_\gamma \frac{f(w)}{w-z}\,dw$ for $z \notin \gamma$ .

Proof by Cross-Cut Method

Proof

Step 1. Connect each inner boundary $\gamma_k$ to the outer boundary $\gamma_0$ by line segments (cross-cuts) that do not intersect each other. This decomposes the multiply connected region into simply connected sub-regions.

Step 2. Apply the Cauchy-Goursat theorem to each simply connected sub-region. The integral over the boundary of each sub-region vanishes.

Step 3. Sum over all sub-regions. The integrals over the cross-cuts cancel (each cross-cut is traversed twice in opposite directions). The remaining integrals are precisely $\oint_\Gamma f\,dz = 0$ .

More precisely, each cross-cut from $\gamma_0$ to $\gamma_k$ appears in two adjacent sub-regions with opposite orientations, so their contributions cancel in the sum. $\blacksquare$

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Applications

ExampleIntegral over an annular region

Let $f(z) = \frac{1}{z(z-2)}$ and consider the annulus $1 < |z| < 3$ . The function $f$ is holomorphic on this annulus. With $\gamma_0: |z| = 3$ (positive) and $\gamma_1: |z| = 1$ (positive):

$\oint_{|z|=3} f\,dz = \oint_{|z|=1} f\,dz + \oint_{|z-2|=\varepsilon} f\,dz.$

Computing: $\oint_{|z|=1} \frac{dz}{z(z-2)} = \oint_{|z|=1} \frac{-1/2}{z}\,dz = -\pi i$ (only the pole at $z=0$ is inside).

Also $\oint_{|z-2|=\varepsilon} \frac{dz}{z(z-2)} = 2\pi i \cdot \frac{1}{2} = \pi i$ .

Therefore $\oint_{|z|=3} f\,dz = -\pi i + \pi i = 0$ , consistent with the fact that the residues at $z=0$ and $z=2$ are $-1/2$ and $1/2$ respectively.

ExampleKeyhole contour

To evaluate integrals involving branch cuts, one uses a keyhole contour: a large circle $|z| = R$ minus a small circle $|z| = \varepsilon$ connected by line segments along the branch cut. By the general Cauchy theorem, the integral over the full keyhole contour vanishes (if $f$ is holomorphic in the slit region), relating the contributions from the circles and the branch-cut segments.

RemarkFoundation for residue theory

The general Cauchy theorem directly leads to the residue theorem: if $f$ has isolated singularities $z_1, \ldots, z_n$ inside $\gamma_0$ , replace $\gamma_k$ by small circles around each $z_k$ to obtain

$\oint_{\gamma_0} f\,dz = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k).$