Cauchy-Goursat Theorem

The Cauchy-Goursat theorem eliminates the need to assume continuity of $f'$ in Cauchy's original theorem. This refinement, due to Goursat, shows that holomorphicity alone suffices to guarantee vanishing of contour integrals.

Statement of the Theorem

Theorem3.9Cauchy-Goursat theorem

Let $f$ be holomorphic on a simply connected domain $D$ . Then for every closed contour $\gamma$ in $D$ :

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

More precisely, if $f$ is holomorphic inside and on a simple closed contour $\gamma$ , then $\oint_\gamma f(z)\,dz = 0$ .

RemarkHistorical significance

Cauchy's original proof (1825) assumed that $f'$ is continuous. Goursat (1900) showed this assumption is unnecessary: the mere existence of $f'(z)$ at each point suffices. This is a remarkable strengthening, as in real analysis, a differentiable function need not have a continuous derivative.

Proof Strategy

Proof

We prove the theorem for a triangle $T$ ; the general case follows by triangulation.

Step 1. Subdivide $T$ into four congruent triangles $T_1, T_2, T_3, T_4$ by connecting edge midpoints. Then

$\oint_{\partial T} f\,dz = \sum_{j=1}^4 \oint_{\partial T_j} f\,dz$

since integrals over shared interior edges cancel (traversed in opposite directions).

Step 2. Choose $T^{(1)}$ among $\{T_j\}$ with $\left|\oint_{\partial T^{(1)}} f\,dz\right| \geq \frac{1}{4}\left|\oint_{\partial T} f\,dz\right|$ .

Iterate to obtain a nested sequence $T \supset T^{(1)} \supset T^{(2)} \supset \cdots$ with $\text{diam}(T^{(n)}) = 2^{-n}\text{diam}(T)$ , $L(\partial T^{(n)}) = 2^{-n}L(\partial T)$ , and

$\left|\oint_{\partial T^{(n)}} f\,dz\right| \geq 4^{-n}\left|\oint_{\partial T} f\,dz\right|.$

Step 3. By the nested intervals property, $\bigcap_n T^{(n)} = \{z_0\}$ for some $z_0 \in D$ . Since $f$ is differentiable at $z_0$ :

$f(z) = f(z_0) + f'(z_0)(z - z_0) + \varepsilon(z)(z - z_0)$

where $\varepsilon(z) \to 0$ as $z \to z_0$ .

Step 4. The integral of $f(z_0) + f'(z_0)(z-z_0)$ over any closed contour vanishes (it has an antiderivative). Therefore:

$\left|\oint_{\partial T^{(n)}} f\,dz\right| = \left|\oint_{\partial T^{(n)}} \varepsilon(z)(z-z_0)\,dz\right| \leq \sup_{z \in T^{(n)}} |\varepsilon(z)| \cdot \text{diam}(T^{(n)}) \cdot L(\partial T^{(n)}).$

This gives $4^{-n}\left|\oint_{\partial T} f\,dz\right| \leq \sup|\varepsilon| \cdot 4^{-n}\text{diam}(T) \cdot L(\partial T)$ .

Since $\sup|\varepsilon| \to 0$ , we conclude $\oint_{\partial T} f\,dz = 0$ . $\blacksquare$

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Extensions and Generalizations

Theorem3.10Cauchy's theorem for multiply connected domains

If $f$ is holomorphic on a domain bounded by a positively oriented outer contour $\gamma_0$ and negatively oriented inner contours $\gamma_1, \ldots, \gamma_n$ , then

$\oint_{\gamma_0} f\,dz + \oint_{\gamma_1} f\,dz + \cdots + \oint_{\gamma_n} f\,dz = 0.$

Equivalently, $\oint_{\gamma_0} f\,dz = \sum_{k=1}^n \oint_{\gamma_k^+} f\,dz$ where $\gamma_k^+$ denotes positive orientation.

ExampleApplication to annular regions

Let $f(z) = 1/(z^2 + 1)$ and consider the annular region between $|z| = 2$ and $|z - i| = 1/2$ . Since the only singularity inside $|z| = 2$ in the upper half-plane is $z = i$ :

$\oint_{|z|=2} \frac{dz}{z^2+1} = \oint_{|z-i|=1/2} \frac{dz}{z^2+1} = \oint_{|z-i|=1/2} \frac{1/(z+i)}{z-i}\,dz = 2\pi i \cdot \frac{1}{2i} = \pi.$

RemarkMorera's theorem — the converse

Morera's theorem provides a converse: if $f$ is continuous on a domain $D$ and $\oint_\gamma f\,dz = 0$ for every closed triangle (or closed contour) in $D$ , then $f$ is holomorphic on $D$ . This is remarkably useful for proving holomorphicity of functions defined by integrals.