# Cauchy's integral theorem

In mathematics, the Cauchy integral theorem (also known as the Cauchy–Goursat theorem) in complex analysis, named after Augustin-Louis Cauchy (and Édouard Goursat), is an important statement about line integrals for holomorphic functions in the complex plane. Essentially, it says that if ${\displaystyle f(z)}$ is analytic in a simply connected domain Ω, then for any simply closed contour ${\displaystyle C}$ in Ω, that contour integral is zero.

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

## Statement

Formulation on Simply Connected Regions

Let ${\displaystyle U\subseteq \mathbb {C} }$ be a simply connected open set, and let ${\displaystyle f:U\to \mathbb {C} }$ be a holomorphic function. Let ${\displaystyle \gamma :[a,b]\to U}$ be a smooth closed curve. Then:

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

(The condition that ${\displaystyle U}$ be simply connected means that ${\displaystyle U}$ has no "holes", or in other words, that the fundamental group of ${\displaystyle U}$ is trivial.)

General Formulation

Let ${\displaystyle U\subseteq \mathbb {C} }$ be an open set, and let ${\displaystyle f:U\to \mathbb {C} }$ be a holomorphic function. Let ${\displaystyle \gamma :[a,b]\to U}$ be a smooth closed curve. If ${\displaystyle \gamma }$ is homotopic to a constant curve, then:

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

(Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy from the curve to the constant curve. Intuitively, this means that one can shrink the curve into a point without exiting the space.) The first version is a special case of this because on a simply connected set, every closed curve is homotopic to a constant curve.

Main Example

In both cases, it is important to remember that the curve ${\displaystyle \gamma }$ not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:

${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right],}$

which traces out the unit circle. Here the following integral:

${\displaystyle \int _{\gamma }{\frac {1}{z}}\,dz=2\pi i\neq 0,}$

is nonzero. The Cauchy integral theorem does not apply here since ${\displaystyle f(z)=1/z}$ is not defined at ${\displaystyle z=0}$. Intuitively, ${\displaystyle \gamma }$ surrounds a "hole" in the domain of ${\displaystyle f}$, so ${\displaystyle \gamma }$ cannot be shrunk to a point without exiting the space. Thus, the theorem does not apply.

## Discussion

As Édouard Goursat showed, Cauchy's integral theorem can be proven assuming only that the complex derivative ${\displaystyle f'(z)}$ exists everywhere in ${\displaystyle U}$. This is significant because one can then prove Cauchy's integral formula for these functions, and from that deduce these functions are infinitely differentiable.

The condition that ${\displaystyle U}$ be simply connected means that ${\displaystyle U}$ has no "holes" or, in homotopy terms, that the fundamental group of ${\displaystyle U}$ is trivial; for instance, every open disk ${\displaystyle U_{z_{0}}=\{z:\left|z-z_{0}\right|, for ${\displaystyle z_{0}\in \mathbb {C} }$, qualifies. The condition is crucial; consider

${\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]}$

which traces out the unit circle, and then the path integral

${\displaystyle \oint _{\gamma }{\frac {1}{z}}\,dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it}\,dt)=\int _{0}^{2\pi }i\,dt=2\pi i}$

is nonzero; the Cauchy integral theorem does not apply here since ${\displaystyle f(z)=1/z}$ is not defined (and is certainly not holomorphic) at ${\displaystyle z=0}$.

One important consequence of the theorem is that path integrals of holomorphic functions on simply connected domains can be computed in a manner familiar from the fundamental theorem of calculus: let ${\displaystyle U}$ be a simply connected open subset of ${\displaystyle \mathbb {C} }$, let ${\displaystyle f:U\to \mathbb {C} }$ be a holomorphic function, and let ${\displaystyle \gamma }$ be a piecewise continuously differentiable path in ${\displaystyle U}$ with start point ${\displaystyle a}$ and end point ${\displaystyle b}$. If ${\displaystyle F}$ is a complex antiderivative of ${\displaystyle f}$, then

${\displaystyle \int _{\gamma }f(z)\,dz=F(b)-F(a).}$

The Cauchy integral theorem is valid with a weaker hypothesis than given above, e.g. given ${\displaystyle U}$, a simply connected open subset of ${\displaystyle \mathbb {C} }$, we can weaken the assumptions to ${\displaystyle f}$ being holomorphic on ${\displaystyle U}$ and continuous on ${\textstyle {\overline {U}}}$ and ${\displaystyle \gamma }$ a rectifiable simple loop in ${\textstyle {\overline {U}}}$.[1]

The Cauchy integral theorem leads to Cauchy's integral formula and the residue theorem.

## Proof

If one assumes that the partial derivatives of a holomorphic function are continuous, the Cauchy integral theorem can be proved as a direct consequence of Green's theorem and the fact that the real and imaginary parts of ${\displaystyle f=u+iv}$ must satisfy the Cauchy–Riemann equations in the region bounded by ${\displaystyle \gamma }$, and moreover in the open neighborhood U of this region. Cauchy provided this proof, but it was later proved by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.

We can break the integrand ${\displaystyle f}$, as well as the differential ${\displaystyle dz}$ into their real and imaginary components:

${\displaystyle f=u+iv}$
${\displaystyle dz=dx+i\,dy}$

In this case we have

${\displaystyle \oint _{\gamma }f(z)\,dz=\oint _{\gamma }(u+iv)(dx+i\,dy)=\oint _{\gamma }(u\,dx-v\,dy)+i\oint _{\gamma }(v\,dx+u\,dy)}$

By Green's theorem, we may then replace the integrals around the closed contour ${\displaystyle \gamma }$ with an area integral throughout the domain ${\displaystyle D}$ that is enclosed by ${\displaystyle \gamma }$ as follows:

${\displaystyle \oint _{\gamma }(u\,dx-v\,dy)=\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy}$
${\displaystyle \oint _{\gamma }(v\,dx+u\,dy)=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right)\,dx\,dy}$

But as the real and imaginary parts of a function holomorphic in the domain ${\displaystyle D}$, ${\displaystyle u}$ and ${\displaystyle v}$ must satisfy the Cauchy–Riemann equations there:

${\displaystyle {\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}}$
${\displaystyle {\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}}$

We therefore find that both integrands (and hence their integrals) are zero

${\displaystyle \iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy=\iint _{D}\left({\frac {\partial u}{\partial y}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy=0}$
${\displaystyle \iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial v}{\partial y}}\right)\,dx\,dy=\iint _{D}\left({\frac {\partial u}{\partial x}}-{\frac {\partial u}{\partial x}}\right)\,dx\,dy=0}$

This gives the desired result

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