Cauchy Integral Theorem

Introduction

The Cauchy integral theorem is one of the central results of Complex Analysis. It exists in various versions, and in this article, we aim to present a basic one for convex regions and a relatively general one for nullhomologous cycles.

For Convex Regions

Statement

Let $G\subseteq ℂ$ be a convex region, and let $\gamma$ be a closed rectifiable curve Trace of Curve in $G$. Then, for every holomorphic function $f:G\to ℂ$, the following holds:

$\underset{\gamma }{\int }f(z)dz=0$

Proof 1: Antiderivatives of f

First, we observe that $f$ has a antiderivative in $G$. Fix a point $z_0\in G$. For any point $z\in G$, let $[z_0,z]$ denote the straight-line segment connecting $z_0$ and $z$ as path.

Proof 2: Definition of the Antiderivative

Define $F:G\to ℂ$ by:

$F(z):=\underset{[z_0,z]}{\int }f(\zeta ),d\zeta$.

Due to the convexity of $G$, the triangle $D$ with vertices $z_0,z,w$ lies entirely within $G$ for $z,w\in G$.

Proof 3: Application of Goursat’s Lemma

By Goursat's Lemma for the boundary $\partial \mathrm{\Delta }$ of a triangle $\mathrm{\Delta }$ with vertices $z_0,z,w\in ℂ$, we have:

$\begin{array}{cc}0& =\underset{\partial \mathrm{\Delta }}{\int }f(z)dz\\ & =\underset{[z_0,z]}{\int }f(\zeta )d\zeta -\underset{[z_0,w]}{\int }f(\zeta )d\zeta +\underset{[z,w]}{\int }f(\zeta )d\zeta \\ & =F(z)-F(w)+\underset{[z,w]}{\int }f(\zeta )d\zeta \end{array}$

Proof 4: Conclusion Using Goursat's Lemma

This leads to:

$\begin{array}{cc}F(z)-F(w)& =\underset{[w,z]}{\int }f(\zeta )d\zeta \\ & =\underset{0}{\overset{1}{\int }}f(w+t(z-w))\cdot (z-w)dt\\ & =\underset{A(z):=}{\underbrace{\underset{0}{\overset{1}{\int }}f(w+t(z-w))dt}}\cdot (z-w)\end{array}$

Thus, we have:

$A(z)=\frac{F(z)-F(w)}{(z-w)}$

Proof 5: Limit Process

Since $A$ is continuous in $w$, taking the limit as $z\to w$ gives:

$A(w)=\underset{z\to w}{\lim }A(z)=\underset{z\to w}{\lim }\frac{F(z)-F(w)}{(z-w)}=F^{\prime }(w).$

Proof 5: Differentiability of $F$

Therefore, $A:G\to ℂ$ is continuous, and $F$ is differentiable in $w\in G$, with:

$F^{\prime }(w)=A(w)=f(w).$

Since $w\in G$ was arbitrary, we conclude $F^{\prime }=f$, proving that $f$ has a antiderivative.

Proof 6: Path Integration

Now, let $\gamma :[a,b]\to G$ be a piecewise continuously differentiable, closed curve. Then:

$\begin{array}{cc}\underset{\gamma }{\int }f(z)dz& =\underset{a}{\overset{b}{\int }}f(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\underset{a}{\overset{b}{\int }}{F}^{\prime }(\gamma (t)){\gamma }^{\prime }(t)dt\\ & =\underset{a}{\overset{b}{\int }}(F\circ \gamma {)}^{\prime }(t)dt\\ & =F(\gamma (b))-F(\gamma (a))=0.\end{array}$

Proof 7:

Let $\gamma :[a,b]\to G$ be an arbitrary integration path in $G$, and let $ϵ>0$. As shown here, we choose a polygonal path $\hat{\gamma }:[a,b]\to ℂ$ such that $\hat{\gamma }(a)=\gamma (a)$, $\hat{\gamma }(b)=\gamma (b)$, and

$|\underset{\hat{\gamma }}{\int }f(z),dz-\underset{\gamma }{\int }f(z),dz|<ϵ.$

Since polygonal paths are piecewise continuously differentiable, the above result implies $\underset{\hat{\gamma }}{\int }f(z),dz=0$. Consequently,

$|\underset{\gamma }{\int }f(z),dz|<ϵ.$

As $ϵ>0$ was arbitrary, the claim follows.

For Cycles in Arbitrary Open Sets

In arbitrary open sets, one must ensure that cycles do not enclose singularities or poles in the complement of the domain. Enclosing such singularities may contribute a non-zero value to the integral (e.g., the function $f(z)=\frac{1}{z}$ and $\gamma (t):=e^{it}$ in a domain $G=ℂ\setminus \{0\}$ . Even though $f$ is holomorphic in $G$, the integral is not zero but $2\pi i$ (see null-homologous Chain).

Statement

Let $G\subseteq ℂ$ be open, and let $\mathrm{\Gamma }$ be a null-homologous cycle in $G$. Then, for every holomorphic function $f:G\to ℂ$, the following holds:

$\underset{\mathrm{\Gamma }}{\int }f(z)dz=0$

Proof

Let $w\in G\setminus \text{trace}(\mathrm{\Gamma })$, and define $g:G\to ℂ$ by

$g(z):=(z-w)\cdot f(z).$

Then, $g$ is holomorphic, and by the global integral formula, we have:

$\underset{\mathrm{\Gamma }}{\int }f(z)dz=\underset{\mathrm{\Gamma }}{\int }\frac{g(z)}{z-w}dz=2\pi i\cdot n(\mathrm{\Gamma },w)\cdot g(w)=0.$

See Also

• Complex Analysis

• Cauchy integral theorem for disks

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