Cauchy Integral Theorem

## Introduction

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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

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### Statement

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

∫ γ f ( z ) d z = 0 {\displaystyle \int _{\gamma }f(z)\,dz=0}

### Proof 1: Antiderivatives of f

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

## Proof 2: Definition of the Antiderivative

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

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

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

### Proof 3: Application of Goursat’s Lemma

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

0 = ∫ ∂ Δ f ( z ) d z = ∫ [ z 0 , z ] f ( ζ ) d ζ − ∫ [ z 0 , w ] f ( ζ ) d ζ + ∫ [ z , w ] f ( ζ ) d ζ = F ( z ) − F ( w ) + ∫ [ z , w ] f ( ζ ) d ζ {\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}}}

### Proof 4: Conclusion Using Goursat's Lemma

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This leads to:

F ( z ) − F ( w ) = ∫ [ w , z ] f ( ζ ) d ζ = ∫ 0 1 f ( w + t ( z − w ) ) ⋅ ( z − w ) d t = ∫ 0 1 f ( w + t ( z − w ) ) d t ⏟ A ( z ) := ⋅ ( z − w ) {\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:

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

### Proof 5: Limit Process

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

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

### Proof 5: Differentiability of  F {\displaystyle F}

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

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

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

### Proof 6: Path Integration

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

∫ γ f ( z ) d z = ∫ a b f ( γ ( t ) ) γ ′ ( t ) d t = ∫ a b F ′ ( γ ( t ) ) γ ′ ( t ) d t = ∫ a b ( F ∘ γ ) ′ ( t ) d t = F ( γ ( b ) ) − F ( γ ( a ) ) = 0. {\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}}}

### Proof 7:

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

| ∫ γ ^ f ( z ) , d z − ∫ γ f ( z ) , d z | < ϵ . {\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  ∫ γ ^ f ( z ) , d z = 0 {\displaystyle \int _{\hat {\gamma }}f(z),dz=0} . Consequently,

| ∫ γ f ( z ) , d z | < ϵ . {\displaystyle \left|\int _{\gamma }f(z),dz\right|<\epsilon .}

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

## For Cycles in Arbitrary Open Sets

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

### Statement

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Let  G ⊆ C {\displaystyle G\subseteq \mathbb {C} } be open, and let  Γ {\displaystyle \Gamma } be a null-homologous cycle in  G {\displaystyle G} . Then, for every holomorphic function  f : G -> C {\displaystyle f\colon G\to \mathbb {C} } , the following holds:

∫ Γ f ( z ) d z = 0 {\displaystyle \int _{\Gamma }f(z)\,dz=0}

### Proof

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Let  w ∈ G ∖ trace ( Γ ) {\displaystyle w\in G\setminus {\text{trace}}(\Gamma )} , and define  g : G -> C {\displaystyle g\colon G\to \mathbb {C} } by

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

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

∫ Γ f ( z ) d z = ∫ Γ g ( z ) z − w d z = 2 π i ⋅ n ( Γ , w ) ⋅ g ( w ) = 0. {\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.}

## See Also

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* Complex Analysis

* Cauchy integral theorem for disks

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