Cauchy's integral theorem

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.

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 ${\displaystyle U_{z_{0}}=\{z:|z-z_{0}|<r\}}$, 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 : U → C 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.

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:

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

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

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:

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

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

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

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

• Morera's theorem
• Methods of contour integration

• Kodaira, Kunihiko (2007), Complex Analysis, Cambridge Stud. Adv. Math., 107, CUP, ISBN 978-0-521-80937-5
• Ahlfors, Lars (2000), Complex Analysis, McGraw-Hill series in Mathematics, McGraw-Hill, ISBN 0-07-000657-1
• Lang, Serge (2003), Complex Analysis, Springer Verlag GTM, Springer Verlag
• Rudin, Walter (2000), Real and Complex Analysis, McGraw-Hill series in mathematics, McGraw-Hill

• "Cauchy integral theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Weisstein, Eric W. "Cauchy Integral Theorem". MathWorld.