Cauchy's integral theorem

Theorem in complex analysis
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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 (and Édouard Goursat), is an important statement about line integrals for holomorphic functions in the complex plane. Essentially, it says that if f ( z ) {\displaystyle f(z)} is holomorphic in a simply connected domain Ω, then for any simply closed contour C {\displaystyle C} in Ω, that contour integral is zero.

∫ C f ( z ) d z = 0. {\displaystyle \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 γ {\displaystyle \gamma } is a curve in U from z 0 {\displaystyle z_{0}} to z 1 {\displaystyle z_{1}} then,

∫ γ f ′ ( z ) d z = f ( z 1 ) − f ( z 0 ) . {\displaystyle \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 ∫ γ f ′ ( z ) d z {\textstyle \int _{\gamma }f'(z)\,dz} is path independent for all paths in U.

Formulation on simply connected regions

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

∫ γ f ( z ) d z = 0. {\displaystyle \int _{\gamma }f(z)\,dz=0.}
(The condition that U {\displaystyle U} be simply connected means that U {\displaystyle U} has no "holes", or in other words, that the fundamental group of U {\displaystyle U} is trivial.)

General formulation

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

∫ γ f ( z ) d z = 0. {\displaystyle \int _{\gamma }f(z)\,dz=0.}
(Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within U {\displaystyle U} ) 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 γ {\displaystyle \gamma } does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve:

γ ( t ) = e i t t ∈ [ 0 , 2 π ] , {\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right],}
which traces out the unit circle. Here the following integral:
∫ γ 1 z d z = 2 π i ≠ 0 , {\displaystyle \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 {\displaystyle f(z)=1/z} is not defined at z = 0 {\displaystyle z=0} . Intuitively, γ {\displaystyle \gamma } surrounds a "hole" in the domain of f {\displaystyle f} , so γ {\displaystyle \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 ) {\displaystyle f'(z)} exists everywhere in U {\displaystyle 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 {\displaystyle U} be simply connected means that U {\displaystyle U} has no "holes" or, in homotopy terms, that the fundamental group of U {\displaystyle U} is trivial; for instance, every open disk U z 0 = { z : | z − z 0 | < r } {\displaystyle U_{z_{0}}=\{z:\left|z-z_{0}\right|<r\}} , for z 0 ∈ C {\displaystyle z_{0}\in \mathbb {C} } , qualifies. The condition is crucial; consider

γ ( t ) = e i t t ∈ [ 0 , 2 π ] {\displaystyle \gamma (t)=e^{it}\quad t\in \left[0,2\pi \right]}
which traces out the unit circle, and then the path integral
∮ γ 1 z d z = ∫ 0 2 π 1 e i t ( i e i t d t ) = ∫ 0 2 π i d t = 2 π i {\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 f ( z ) = 1 / z {\displaystyle f(z)=1/z} is not defined (and is certainly not holomorphic) at z = 0 {\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 {\displaystyle U} be a simply connected open subset of C {\displaystyle \mathbb {C} } , let f : U → C {\displaystyle f:U\to \mathbb {C} } be a holomorphic function, and let γ {\displaystyle \gamma } be a piecewise continuously differentiable path in U {\displaystyle U} with start point a {\displaystyle a} and end point b {\displaystyle b} . If F {\displaystyle F} is a complex antiderivative of f {\displaystyle f} , then

∫ γ f ( z ) d z = F ( b ) − F ( a ) . {\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 {\displaystyle U} , a simply connected open subset of C {\displaystyle \mathbb {C} } , we can weaken the assumptions to f {\displaystyle f} being holomorphic on U {\displaystyle U} and continuous on U ¯ {\textstyle {\overline {U}}} and γ {\displaystyle \gamma } a rectifiable simple loop in U ¯ {\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 proven as a direct consequence of Green's theorem and the fact that the real and imaginary parts of f = u + i v {\displaystyle f=u+iv} must satisfy the Cauchy–Riemann equations in the region bounded by γ {\displaystyle \gamma } , 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 f {\displaystyle f} , as well as the differential d z {\displaystyle dz} into their real and imaginary components:

f = u + i v {\displaystyle f=u+iv}
d z = d x + i d y {\displaystyle dz=dx+i\,dy}

In this case we have

∮ γ f ( z ) d z = ∮ γ ( u + i v ) ( d x + i d y ) = ∮ γ ( u d x − v d y ) + i ∮ γ ( v d x + u d y ) {\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 γ {\displaystyle \gamma } with an area integral throughout the domain D {\displaystyle D} that is enclosed by γ {\displaystyle \gamma } as follows:

∮ γ ( u d x − v d y ) = ∬ D ( − ∂ v ∂ x − ∂ u ∂ y ) d x d y {\displaystyle \oint _{\gamma }(u\,dx-v\,dy)=\iint _{D}\left(-{\frac {\partial v}{\partial x}}-{\frac {\partial u}{\partial y}}\right)\,dx\,dy}
∮ γ ( v d x + u d y ) = ∬ D ( ∂ u ∂ x − ∂ v ∂ y ) d x d y {\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 D {\displaystyle D} , u {\displaystyle u} and v {\displaystyle v} must satisfy the Cauchy–Riemann equations there:

∂ u ∂ x = ∂ v ∂ y {\displaystyle {\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}}}
∂ u ∂ y = − ∂ v ∂ x {\displaystyle {\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}}

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

∬ D ( − ∂ v ∂ x − ∂ u ∂ y ) d x d y = ∬ D ( ∂ u ∂ y − ∂ u ∂ y ) d x d y = 0 {\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}
∬ D ( ∂ u ∂ x − ∂ v ∂ y ) d x d y = ∬ D ( ∂ u ∂ x − ∂ u ∂ x ) d x d y = 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

∮ γ f ( z ) d z = 0 {\displaystyle \oint _{\gamma }f(z)\,dz=0}

See also

References

  1. ^ Walsh, J. L. (1933-05-01). "The Cauchy-Goursat Theorem for Rectifiable Jordan Curves". Proceedings of the National Academy of Sciences. 19 (5): 540–541. doi:10.1073/pnas.19.5.540. ISSN 0027-8424. PMC 1086062. PMID 16587781.

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