Cauchy Integral Theorem

Statement

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:

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

Proof 1: Antiderivatives of f

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.

Proof 3: Application of Goursat’s Lemma

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:

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

Proof 4: Conclusion Using Goursat's Lemma

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:

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

Proof 5: Limit Process

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

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

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

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.

Proof 6: Path Integration

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

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

Proof 7:

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.

Statement

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

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

Proof

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:

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

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