Intuition - Cauchy integral theorem

[tiago23]

So folks, I'm learning complex analysis right now and I've come across one thing that simply fails to enter my mind: the Cauchy Integral Theorem, or the Cauchy-Goursat Theorem. It says that, if a function is analytic in a certain (simply connected) domain, then the contour integral over a simple closed path must be zero. My question is:Why? Sure, I've seen the proof, but I don't get the intuition, I can't visualize it. Can someone help me out with the intuition?

 

[fresh_42]

I was always satisfied by the thought, that it is simply a generalization of fundamental theorem of calculus: $\int_a^b f(x)\,dx=F(a)-F(b)$ or a bit sloppy: What you gather all the way up is lost again on the way down. I'm sure there are other explanations, e.g. considering differential forms and chain complexes, however, whether they are more intuitive depends on how one is "wired" (e.g. https://www.physicsforums.com/threads/why-the-terms-exterior-closed-exact.871875/#post-5474443)

 

[lavinia]

Writing out the integrand $f(z)dz$ along a curve $z(t)=x(t) + iy(t)$ one gets $(ux'-vy')dt + i(uy'+vx')dt$ so the integral $∫_{z(t)}f(z)dz$ is the sum of two integrals:

$∫ux'-vy'dt$ and $i∫(uy'+vx')dt$.

The Cauchy Integral Theorem says that both of these integrals are zero.

Consider the vector field $V=(u,-v)$.

The first integral $∫ux'-vy'dt$ is the work done by $V$ along the curve $z(t)$. This integral is zero if $V$ is conservative which in a simply connected domain is the same as saying that its curl is zero. That is: $0= ∇×V = ∂v/∂x-∂u/∂y$. But this is just the second Cauchy-Riemann equation.

The second integral $i∫(uy'+vx')dt$ equals zero if the divergence of $V$ is zero. That is: $∂u/∂x+∂(-v)/∂y=0$ and this is just the first Cauchy Riemann equation.

So $V$ may be thought of as an irrotational divergence free flow.

 

[mathwonk]

i agree with Lavinia that the intuition is the same as for green's theorem or the equivalent (plane) divergence theoem.

of course it all depends on your definition of "analytic". I presume the definition you are using is that the function has one derivative, which i call "holomorphic". with my definition of "analytic", which requires the function to have a local power series expansion everywhere, the theorem is more elementary, and closer to the one variable fundamental theorem of calculus. I.e. then it reduces to the fact that the function then has locally an antiderivative, hence the integral along an arc equals the difference of the values of the antiderivative at the end points. in particular if the endpoints are equal, the integral is zero. (this uses simple connectivity to decompose the loop of integration into small loops whose interiors lie in the domain.)

of course that's why the stronger theorem, that the integral is zero is also true just with assuming one derivative, is really useful. I.e. it implies the amazing result that assuming one derivative actually gives you not just infinitely many derivatives but also a local power series representation, i.e. gives you "analyticity" in my sense. If the theorem is truly to be considered intuitive, one might try to extend the intuition to understanding why assuming only one derivative actually gives you a power series.

of course there are other amazing subtleties in goursat's proof, which does not even assume the derivative is continuous, which is needed for the green's theorem argument.

 

[Delta2]

If you are familiar with vector calculus and conservative vector fields, i could tell this: Holomorphic(equivalently analytic) functions are to closed contour integrals, what conservative vector fields are to closed line(curve) integrals.
But which proof did you cover, the one that uses Green's theorem from vector calculus or the other one that doesn't use theorems from vector calculus and is essentially the proof provided by Goursat?