# 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 two different paths connect the same two points, and a function is holomorphic everywhere in between the two paths, then the two path integrals of the function will be the same.

## Statement

Let ${\displaystyle U}$ be an open subset of ${\displaystyle \mathbb {C} }$ which is simply connected, let ${\displaystyle f:U\to \mathbb {C} }$ be a holomorphic function, and let ${\displaystyle \!\,\gamma }$ be a rectifiable path in ${\displaystyle U}$ whose start point is equal to its end point. Then

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

An alternate version can be stated as follows. Let ${\displaystyle U}$ be an open subset of ${\displaystyle \mathbb {C} }$ which is connected, and let ${\displaystyle f:U\to \mathbb {C} }$ be a holomorphic function. Then for all cycles ${\displaystyle \gamma }$ homologous to zero in ${\displaystyle U}$:

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

We define cycles and homologous to zero below.

• A cycle is a linear combination of closed paths, where the scalars are integers. The integral over the cycle is defined as a linear combination of integrals over individual paths.
• The winding number of a closed curve ${\displaystyle \gamma }$ around a point ${\displaystyle a}$ (not on the curve) is defined as: ${\displaystyle {\dfrac {1}{2\pi i}}\int _{\gamma }{\dfrac {dz}{z-a}}}$. Intuitively, it is the number of times the curve ${\displaystyle \gamma }$ wraps around the point ${\displaystyle a}$.
• A cycle is homologous to zero in ${\displaystyle U}$ if the winding number of ${\displaystyle \gamma }$ is zero for each point ${\displaystyle a}$ not in ${\displaystyle U}$.

An example is furnished by the ring-shaped region. This version is crucial for rigorous derivation of Laurent series and Cauchy's residue formula without involving any physical notions such as cross cuts or deformations. The version enables the extension of Cauchy's theorem to multiply-connected regions analytically.

## 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 ${\displaystyle U_{z_{0}}=\{z:|z-z_{0}|, 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 U be a simply connected open subset of C, let f : UC be a holomorphic function, and let γ 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

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

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

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

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:

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

However, being 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:

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

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

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