# 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 $f(z)$ is holomorphic in a simply connected domain Ω, then for any simply closed contour $C$ in Ω, that contour integral is zero.

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

### Fundamental theorem for complex line integrals

If f(z) is a holomorphic function on an open region U, and $\gamma$  is a curve in U from $z_{0}$  to $z_{1}$  then,

$\int _{\gamma }f'(z)\,dz=f(z_{1})-f(z_{0}).$

Also, when f(z) has a single-valued antiderivative in an open region U, then the path integral ${\textstyle \int _{\gamma }f'(z)\,dz}$  is path independent for all paths in U.

#### Formulation on Simply Connected Regions

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

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

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

#### General Formulation

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

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

(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 $\gamma$  does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:

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

which traces out the unit circle. Here the following integral:
$\int _{\gamma }{\frac {1}{z}}\,dz=2\pi i\neq 0,$

is nonzero. The Cauchy integral theorem does not apply here since $f(z)=1/z$  is not defined at $z=0$ . Intuitively, $\gamma$  surrounds a "hole" in the domain of $f$ , so $\gamma$  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 $f'(z)$  exists everywhere in $U$ . 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 $U$  be simply connected means that $U$  has no "holes" or, in homotopy terms, that the fundamental group of $U$  is trivial; for instance, every open disk $U_{z_{0}}=\{z:\left|z-z_{0}\right| , for $z_{0}\in \mathbb {C}$ , qualifies. The condition is crucial; consider

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

which traces out the unit circle, and then the path integral
$\oint _{\gamma }{\frac {1}{z}}\,dz=\int _{0}^{2\pi }{\frac {1}{e^{it}}}(ie^{it}\,dt)=\int _{0}^{2\pi }i\,dt=2\pi i$

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

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 $U$  be a simply connected open subset of $\mathbb {C}$ , let $f:U\to \mathbb {C}$  be a holomorphic function, and let $\gamma$  be a piecewise continuously differentiable path in $U$  with start point $a$  and end point $b$ . If $F$  is a complex antiderivative of $f$ , then

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

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

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 $f=u+iv$  must satisfy the Cauchy–Riemann equations in the region bounded by $\gamma$ , 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 $f$ , as well as the differential $dz$  into their real and imaginary components:

$f=u+iv$

$dz=dx+i\,dy$

In this case we have

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

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

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

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

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

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

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

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

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

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

This gives the desired result

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