Cauchy's integral theorem

Short description: Theorem in complex analysis

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 [math]\displaystyle{ f(z) }[/math] is holomorphic in a simply connected domain Ω, then for any simply closed contour [math]\displaystyle{ C }[/math] in Ω, that contour integral is zero.

[math]\displaystyle{ \int_C f(z)\,dz = 0. }[/math]

Statement

Fundamental theorem for complex line integrals

If f(z) is a holomorphic function on an open region U, and [math]\displaystyle{ \gamma }[/math] is a curve in U from [math]\displaystyle{ z_0 }[/math] to [math]\displaystyle{ z_1 }[/math] then, [math]\displaystyle{ \int_{\gamma}f'(z) \, dz = f(z_1)-f(z_0). }[/math]

Also, when f(z) has a single-valued antiderivative in an open region U, then the path integral [math]\displaystyle{ \int_{\gamma}f'(z) \, dz }[/math] is path independent for all paths in U.

Formulation on simply connected regions

Let [math]\displaystyle{ U \subseteq \Complex }[/math] be a simply connected open set, and let [math]\displaystyle{ f: U \to \Complex }[/math] be a holomorphic function. Let [math]\displaystyle{ \gamma: [a,b] \to U }[/math] be a smooth closed curve. Then: [math]\displaystyle{ \int_\gamma f(z)\,dz = 0. }[/math] (The condition that [math]\displaystyle{ U }[/math] be simply connected means that [math]\displaystyle{ U }[/math] has no "holes", or in other words, that the fundamental group of [math]\displaystyle{ U }[/math] is trivial.)

General formulation

Let [math]\displaystyle{ U \subseteq \Complex }[/math] be an open set, and let [math]\displaystyle{ f: U \to \Complex }[/math] be a holomorphic function. Let [math]\displaystyle{ \gamma: [a,b] \to U }[/math] be a smooth closed curve. If [math]\displaystyle{ \gamma }[/math] is homotopic to a constant curve, then: [math]\displaystyle{ \int_\gamma f(z)\,dz = 0. }[/math] (Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within [math]\displaystyle{ U }[/math]) 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 [math]\displaystyle{ \gamma }[/math] does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: [math]\displaystyle{ \gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right] , }[/math] which traces out the unit circle. Here the following integral: [math]\displaystyle{ \int_{\gamma} \frac{1}{z}\,dz = 2\pi i \neq 0 , }[/math] is nonzero. The Cauchy integral theorem does not apply here since [math]\displaystyle{ f(z) = 1/z }[/math] is not defined at [math]\displaystyle{ z = 0 }[/math]. Intuitively, [math]\displaystyle{ \gamma }[/math] surrounds a "hole" in the domain of [math]\displaystyle{ f }[/math], so [math]\displaystyle{ \gamma }[/math] 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 [math]\displaystyle{ f'(z) }[/math] exists everywhere in [math]\displaystyle{ U }[/math]. 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 [math]\displaystyle{ U }[/math] be simply connected means that [math]\displaystyle{ U }[/math] has no "holes" or, in homotopy terms, that the fundamental group of [math]\displaystyle{ U }[/math] is trivial; for instance, every open disk [math]\displaystyle{ U_{z_0} = \{ z : \left|z-z_{0}\right| \lt r\} }[/math], for [math]\displaystyle{ z_0 \in \Complex }[/math], qualifies. The condition is crucial; consider [math]\displaystyle{ \gamma(t) = e^{it} \quad t \in \left[0, 2\pi\right] }[/math] which traces out the unit circle, and then the path integral [math]\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 }[/math] is nonzero; the Cauchy integral theorem does not apply here since [math]\displaystyle{ f(z) = 1/z }[/math] is not defined (and is certainly not holomorphic) at [math]\displaystyle{ z = 0 }[/math].

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 [math]\displaystyle{ U }[/math] be a simply connected open subset of [math]\displaystyle{ \Complex }[/math], let [math]\displaystyle{ f: U \to \Complex }[/math] be a holomorphic function, and let [math]\displaystyle{ \gamma }[/math] be a piecewise continuously differentiable path in [math]\displaystyle{ U }[/math] with start point [math]\displaystyle{ a }[/math] and end point [math]\displaystyle{ b }[/math]. If [math]\displaystyle{ F }[/math] is a complex antiderivative of [math]\displaystyle{ f }[/math], then [math]\displaystyle{ \int_\gamma f(z)\,dz=F(b)-F(a). }[/math]

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

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 proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of [math]\displaystyle{ f=u+iv }[/math] must satisfy the Cauchy–Riemann equations in the region bounded by [math]\displaystyle{ \gamma }[/math], and moreover in the open neighborhood U of this region. Cauchy provided this proof, but it was later proven by Goursat without requiring techniques from vector calculus, or the continuity of partial derivatives.

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

[math]\displaystyle{ f=u+iv }[/math] [math]\displaystyle{ dz=dx+i\,dy }[/math]

In this case we have [math]\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) }[/math]

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

[math]\displaystyle{ \oint_\gamma (u\,dx-v\,dy) = \iint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right) \,dx\,dy }[/math] [math]\displaystyle{ \oint_\gamma (v\,dx+u\,dy) = \iint_D \left( \frac{\partial u}{\partial x} -\frac{\partial v}{\partial y} \right) \,dx\,dy }[/math]

But as the real and imaginary parts of a function holomorphic in the domain [math]\displaystyle{ D }[/math], [math]\displaystyle{ u }[/math] and [math]\displaystyle{ v }[/math] must satisfy the Cauchy–Riemann equations there: [math]\displaystyle{ \frac{ \partial u }{ \partial x } = \frac{ \partial v }{ \partial y } }[/math] [math]\displaystyle{ \frac{ \partial u }{ \partial y } = -\frac{ \partial v }{ \partial x } }[/math]

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

[math]\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 }[/math] [math]\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 }[/math]

This gives the desired result [math]\displaystyle{ \oint_\gamma f(z)\,dz = 0 }[/math]

See also

• Morera's theorem
• Methods of contour integration
• Star domain

References

• Kodaira, Kunihiko (2007), Complex Analysis, Cambridge Stud. Adv. Math., 107, Cambridge University Press, 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 

External links

• Hazewinkel, Michiel, ed. (2001), "Cauchy integral theorem", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/c020900 
• Weisstein, Eric W.. "Cauchy Integral Theorem". http://mathworld.wolfram.com/CauchyIntegralTheorem.html. 
• Jeremy Orloff, 18.04 Complex Variables with Applications Spring 2018 Massachusetts Institute of Technology: MIT OpenCourseWare Creative Commons.