Cauchy's integral theorem: Information from Answers.com

In mathematics, the Cauchy integral 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.

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

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

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 one can deduce that 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 : 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

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

The Cauchy integral theorem is valid in slightly stronger forms than given above. Let e.g. 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 $γ$ be a loop in $\textstyle\overline{U}$ which is uniform limit of a sequence $γ k$ of rectifiable loops in U with bounded length. Then, applying the Cauchy theorem to the $γ k$ , and passing to the limit one has

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

See e.g. [KK], Theorem 2.3 for a more general result.

The Cauchy integral theorem leads to Cauchy's integral formula and the residue theorem.

1 Proof
2 See also
3 References
4 External links

Proof

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 $f = u + i v$ must satisfy the Cauchy–Riemann equations in the region bounded by $γ$ .

We can break the integrand $f$ , as well as the differential $d z$ into their real and imaginary components:

$f=u+iv$

$dz=dx+idy$

In this case we have

$\oint_\gamma f(z)\,dz = \oint_\gamma (u+iv)(dx+idy) = \oint_\gamma (udx-vdy) +i\oint_\gamma (vdx+udy)$

By Green's theorem, we may then replace the integrals around the closed contour $γ$ with an area integral throughout the domain $D$ that is enclosed by $γ$ as follows:

$\oint_\gamma (udx-vdy) = \oint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )dxdy$

$\oint_\gamma (vdx+udy) = \oint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )dxdy$

However, being the real and imaginary parts of a function analytic 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

$\oint_D \left( -\frac{\partial v}{\partial x} -\frac{\partial u}{\partial y} \right )dxdy =\oint_D \left( \frac{\partial u}{\partial y} -\frac{\partial u}{\partial y} \right )dxdy =0$

$\oint_D \left( \frac{\partial u}{\partial x}-\frac{\partial v}{\partial y} \right )dxdy=\oint_D \left( \frac{\partial u}{\partial x}-\frac{\partial u}{\partial x} \right )dxdy=0$

This gives the desired result

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

References

[KK] Kunihiko Kodaira (2007). Complex Analysis. Cambridge Stud. Adv. Math., 107. CUP. ISBN 978-0-521-80937-5.

External links

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Cauchy's integral theorem

Contents

Proof

See also

References

External links