Cauchy Integral Theorem

Suppose I’m thinking of a path between two complex numbers {z_0} and {z_1}, and I have a holomorphic function {f:U\rightarrow{\mathbb C}} that I want you to integrate on the path. The problem is, I’m obnoxious, and I didn’t tell you what the path was. Can you do it?

The short answer is unfortunately no. The slightly longer answer is: it depends on the open set {U}. If {U} has no holes, then you can do it. Sometimes, you can even if {U} does have holes, but that’s not what we’re going to worry about. The claim is that if I can continuously morph one path {\gamma_0} into another path {\gamma_1}, then

\displaystyle \int_{\gamma_0} f(z)dz=\int_{\gamma_1}f(z)dz.

Such a “continuous morphing” is called a homotopy. Maybe you can see, roughly speaking, how a hole could be an obstruction to morphing one path into another. The intuition is, if you tie down two ends of a string, and you stick a big “pole” in the ground. You can’t get the string to go around the other side of the pole (without lifting it off the ground).

Here’s another way to look at it. Suppose I first integrate over the curve {\gamma_0}, and then I integrate over the curve {\gamma_1}, but I do it backwards. The result would be the difference between the two integrals. If {\beta} is the curve given by {\gamma_0} and then {\gamma_1} backwards, then

\displaystyle \int_\beta f(z)dz=\int_{\gamma_0} f(z)dz-\int_{\gamma_1}f(z)dz.

I’d like to show that this integral is zero. One way to do this would be to show that if I integrate over any loop which is homotopic (can be shrunk down) to a point, then the integral must be zero.

Indeed these two things are equivalent, but I’m not going to prove either. To avoid some messy details, and sweep some others under the rug, I’m going to prove a result that is slightly weaker, though I hope strong enough to give you the right idea. The details I’m leaving out aren’t complex analysis. They’re topology, and in my opinion, boring. I’ll prove this weaker result, and then use the stronger one. If you’re interested, Ahlfors has a proof of the stronger result.

This weaker result is about loops which enclose a simply-connected region. That is, it’s the same result as I want to prove, but only for loops which are the boundary of some region.

Theorem 1 Let {R} be a simply-connected region contained in an open set {U\subseteq{\mathbb C}}, and {\gamma:[0,1]\rightarrow\partial R} is a parameterization of the boundary, then for {f\in{\mathcal H}(U)},

\displaystyle \int_\gamma f(z)dz=0.

Proof: Let {f=u+iv}, and let {z=x+iy}. Then we have the relation on 1-forms {dz=dx+idy}. As before, we’ll be using the notation {u_x} as shorthand for {\frac{\partial u}{\partial x}}. Now

\displaystyle \int_\gamma fdz=\int_\gamma(udx-vdy)+i\int_\gamma(vdx+udy).

From Green’s theorem (which we won’t prove because it is real analysis), we know that

\displaystyle \int_\gamma (udx-vdy)=\iint_R(-v_x-u_y)dxdy,

and that

\displaystyle \int_\gamma(vdx+udy)=\iint_R(u_x-v_y)dxdy.

Of course, the Cauchy-Riemann equations tell us that {-v_x-u_y=0} and {u_x-v_y=0}, so

\displaystyle \int_\gamma f(z)dz=0,

as desired. \Box

A discussion with some friends revealed that this is sort of a hack. I just swept the difficult part of the theoerm into Green’s theorem and then didn’t both to prove it. I know. But I like it anyway, and the reason I like it is because I had never seen the connection between the Cauchy-Riemann equations and the result of Green’s theorem. In some sense, one might say that complex analysis is the study of a very special class of functions that behave in remarkably nice ways with respect to Green’s theorem.

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2 Responses to Cauchy Integral Theorem

  1. Pingback: Cauchy integral formula « Andy Soffer

  2. Pingback: Meromorphic functions and residues « Andy Soffer

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