Complex Analysis/Cauchy Integral Theorem

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.

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

${\displaystyle \int _{\gamma }f(z)\,dz=0}$

First, we observe that ${\displaystyle f}$ has a antiderivative in ${\displaystyle G}$. Fix a point ${\displaystyle z_{0}\in G}$. For any point ${\displaystyle z\in G}$, let ${\displaystyle [z_{0},z]}$ denote the straight-line segment connecting ${\displaystyle z_{0}}$ and ${\displaystyle z}$ as path.

Define ${\displaystyle F\colon G\to \mathbb {C} }$ by:

${\displaystyle F(z):=\int _{[z_{0},z]}f(\zeta ),d\zeta }$.

Due to the convexity of ${\displaystyle G}$, the triangle ${\displaystyle D}$ with vertices ${\displaystyle z_{0},z,w}$ lies entirely within ${\displaystyle G}$ for ${\displaystyle z,w\in G}$.

By Goursat's Lemma for the boundary ${\displaystyle \partial \Delta }$ of a triangle ${\displaystyle \Delta }$ with vertices ${\displaystyle z_{0},z,w\in \mathbb {C} }$, we have:

${\displaystyle {\begin{array}{rl}0&=\int _{\partial \Delta }f(z)\,dz\\&=\int _{[z_{0},z]}f(\zeta )\,d\zeta -\int _{[z_{0},w]}f(\zeta )\,d\zeta +\int _{[z,w]}f(\zeta )\,d\zeta \\&=F(z)-F(w)+\int _{[z,w]}f(\zeta )\,d\zeta \end{array}}}$

This leads to:

${\displaystyle {\begin{array}{rl}F(z)-F(w)&=\int _{[w,z]}f(\zeta )\,d\zeta \\&=\int _{0}^{1}f{\bigl (}w+t(z-w){\bigr )}\cdot (z-w)\,dt\\&=\underbrace {\int _{0}^{1}f{\bigl (}w+t(z-w){\bigr )}\,dt} _{A(z):=}\cdot (z-w)\end{array}}}$

Thus, we have:

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

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

${\displaystyle A(w)=\lim _{z\to w}A(z)=\lim _{z\to w}{\frac {F(z)-F(w)}{(z-w)}}=F'(w).}$

Therefore, ${\displaystyle A\colon G\to \mathbb {C} }$ is continuous, and ${\displaystyle F}$ is differentiable in ${\displaystyle w\in G}$, with:

${\displaystyle F'(w)=A(w)=f(w).}$

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

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

${\displaystyle {\begin{array}{rl}\int _{\gamma }f(z)\,dz&=\int _{a}^{b}f(\gamma (t))\gamma '(t)\,dt\\&=\int _{a}^{b}F'(\gamma (t))\gamma '(t)\,dt\\&=\int _{a}^{b}(F\circ \gamma )'(t)\,dt\\&=F(\gamma (b))-F(\gamma (a))=0.\end{array}}}$

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

${\displaystyle \left|\int _{\hat {\gamma }}f(z),dz-\int _{\gamma }f(z),dz\right|<\epsilon .}$

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

${\displaystyle \left|\int _{\gamma }f(z),dz\right|<\epsilon .}$

As ${\displaystyle \epsilon >0}$ was arbitrary, the claim follows.

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 ${\displaystyle f(z)={\frac {1}{z}}}$ and ${\displaystyle \gamma (t):=e^{it}}$ in a domain ${\displaystyle G=\mathbb {C} \setminus \{0\}}$ . Even though ${\displaystyle f}$ is holomorphic in ${\displaystyle G}$, the integral is not zero but ${\displaystyle 2\pi i}$ (see nullhomologous cycle).

Let ${\displaystyle G\subseteq \mathbb {C} }$ be open, and let ${\displaystyle \Gamma }$ be a nullhomologous cycle in ${\displaystyle G}$. Then, for every holomorphic function ${\displaystyle f\colon G\to \mathbb {C} }$, the following holds:

${\displaystyle \int _{\Gamma }f(z)\,dz=0}$

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

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

Then, ${\displaystyle g}$ is holomorphic, and by the global integral formula, we have:

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

• Complex Analysis

• Cauchy integral theorem for disks

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• Source: Integralsatz von Cauchy - URL: https://de.wikiversity.org/wiki/Integralsatz_von_Cauchy
• Date: 12/18/2024