# Cauchy's integral theorem

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In mathematics, the Cauchy integral theorem (also known as the Cauchy–Goursat theorem) in complex analysis, named after Augustin-Louis Cauchy, 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 of theorem

The theorem is usually formulated for closed paths as follows: let U be an open subset of C which is simply connected, let f : UC be a holomorphic function, and let $\!\,\gamma$ be a rectifiable path in U whose start point is equal to its end point. Then

$\oint_\gamma f(z)\,dz = 0.$

A precise (homology) version can be stated using winding numbers. The winding number of a closed curve around a point a not on the curve is the integral of f(z)/[2 iπ], where f(z) = 1/(z − a) around the curve. It is an integer. Briefly, the path integral along a Jordan curve of a function holomorphic in the interior of the curve, is zero. Instead of a single closed path we can consider a linear combination of closed paths, where the scalars are integers. Such a combination is called a closed chain, and one defines an integral along the chain as a linear combination of integrals over individual paths. A closed chain is called a cycle in a region if it is homologous to zero in the region; that is, the winding number, expressed by the integral of 1/(z − a) over the closed chain, is zero for each point 'a' not in the region. This means that the closed chain does not wind around points outside the region. Then Cauchy's theorem can be stated as the integral of a function holomorphic in an open set taken around any cycle in the open set is zero. 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 was shown by Goursat, 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 in fact 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 $U=\{ z: |z-z_{0}| < r\}$ 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

$\oint_\gamma \frac{1}{z}\,dz = \int_0^{2\pi} { ie^{it} \over e^{it} }\,dt= \int_0^{2\pi}i\,dt = 2\pi i$

is non-zero; the Cauchy integral theorem does not apply here since $f(z) = 1/z$ is not defined (and 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 real 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

$\int_\gamma f(z)\,dz=F(b)-F(a).$

The Cauchy integral theorem is valid in slightly stronger forms than given above. e.g. Let U be a simply connected open subset of C and f a function which is holomorphic on U and continuous on $\textstyle\overline{U}$. Let $\gamma$ be a loop in $\textstyle\overline{U}$ which is uniform limit of a sequence $\gamma_k$ of rectifiable loops in U with bounded length. Then, applying the Cauchy theorem to the $\gamma_k$, and passing to the limit one has

$\oint_\gamma f(z)\,dz=0.$

See e.g. (Kodaira 2007, Theorem 2.3) for a more general result.

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:

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

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:

$\oint_\gamma (u\,dx-v\,dy) = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )\,dx\,dy$
$\oint_\gamma (v\,dx+u\,dy) = \iint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )\,dx\,dy$

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

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

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

$\oint_\gamma f(z)\,dz =0$