Cauchy's integral theorem explained

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)

is holomorphic in a simply connected domain Ω, then for any simply closed contour

C

in Ω, that contour integral is zero.

\int_C f(z)\,dz = 0.

Statement

Fundamental theorem for complex line integrals

If is a holomorphic function on an open region, and

\gamma

is a curve in from

z0

to

z1

then,\int_f'(z) \, dz = f(z_1)-f(z_0).

Also, when has a single-valued antiderivative in an open region, then the path integral \int_f'(z) \, dz is path independent for all paths in .

Formulation on simply connected regions

Let

U\subseteq\Complex

be a simply connected open set, and let

f:U\to\Complex

be a holomorphic function. Let

\gamma:[a,b]\toU

be a smooth closed curve. Then:\int_\gamma f(z)\,dz = 0. (The condition that

U

be simply connected means that

U

has no "holes", or in other words, that the fundamental group of

U

is trivial.)

General formulation

Let

U\subseteq\Complex

be an open set, and let

f:U\to\Complex

be a holomorphic function. Let

\gamma:[a,b]\toU

be a smooth closed curve. If

\gamma

is homotopic to a constant curve, then:\int_\gamma f(z)\,dz = 0. (Recall that a curve is homotopic to a constant curve if there exists a smooth homotopy (within

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

\gamma

does not surround any "holes" in the domain, or else the theorem does not apply. A famous example is the following curve: \gamma(t) = e^ \quad t \in \left[0, 2\pi\right],which traces out the unit circle. Here the following integral:\int_ \frac\,dz = 2\pi i \neq 0, is nonzero. The Cauchy integral theorem does not apply here since

f(z)=1/z

is not defined at

z=0

. Intuitively,

\gamma

surrounds a "hole" in the domain of

f

, so

\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)

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
U
z0

=\{z:\left|z-z0\right|<r\}

, for

z0\in\Complex

, qualifies. The condition is crucial; consider\gamma(t) = e^ \quad t \in \left[0, 2\pi\right]which traces out the unit circle, and then the path integral\oint_\gamma \frac\,dz = \int_0^ \frac(ie^ \,dt) = \int_0^i\,dt = 2\pi i is nonzero; the Cauchy integral theorem does not apply here since

f(z)=1/z

is not defined (and is 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 calculus: let

U

be a simply connected open subset of

\Complex

, let

f:U\to\Complex

be a holomorphic function, and let

\gamma

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 with a weaker hypothesis than given above, e.g. given

U

, a simply connected open subset of

\Complex

, we can weaken the assumptions to

f

being holomorphic on

U

and continuous on \overline and

\gamma

a rectifiable simple loop in \overline.[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+iv

must satisfy the Cauchy–Riemann equations in the region bounded by and moreover in the open neighborhood 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 as well as the differential

dz

into their real and imaginary components:

f=u+iv 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 -\frac \right) \,dx\,dy \oint_\gamma (v\,dx+u\,dy) = \iint_D \left(\frac -\frac \right) \,dx\,dy

But as the real and imaginary parts of a function holomorphic in the domain

u

and

v

must satisfy the Cauchy–Riemann equations there:\frac = \frac \frac = -\frac

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

\iint_D \left(-\frac -\frac \right)\,dx\,dy = \iint_D \left(\frac - \frac \right) \, dx \, dy =0\iint_D \left(\frac-\frac \right)\,dx\,dy = \iint_D \left(\frac - \frac \right) \, dx \, dy = 0

This gives the desired result\oint_\gamma f(z)\,dz = 0

See also

External links

Notes and 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. 10.1073/pnas.19.5.540. 16587781. 1086062. 0027-8424. free.