The lecture before spring break

I let myself go a little bit since I was overjoyed with the prospect of a week's freedom. I covered a great deal so I though it might be useful to record the result. Today we leave real calculus behind. What will occur is quite novel.

The one integral you'll need ...
We know that |z|=1(1/z)dz=2Pi i. This is a vital fact.

And almost all the others are 0, anyway!
Well, here is a version of Cauchy's Theorem which relies totally on Green's Theorem for its verification.
Cauchy's Theorem Suppose f(z) is analytic on and inside a simple closed curve C. Then Cf(z) dz=0.
Proof Suppose that f(z)=u(x,y)+iv(x,y) where u and v are real-valued functions (the real and imaginary parts of f). Then f(z) dz=[u(x,y)+iv(x,y)][dx+idy]=[u(x,y)dx-v(x,y)dy]+i[v(x,y)dx+u(x,y)dy]. The statement of Green's Theorem we will use is that Cpdx+qdy=Rqx-pydA where R is the region bounded by C, dA means area (dxdy, if you would rather) and both p and q are continuously differentiable everywhere in C and R (even one point with a problem might make the equation invalid!).
Let's look at C[u(x,y)dx-v(x,y)dy] with p=u and q=-v. Then qx-py is -vx-uy. But one of the Cauchy-Riemann equations is uy=-vx, so the integrand in the double integral over R is 0.
Let's look at C[v(x,y)dx+u(x,y)dy] with p=v and q=u. Then qx-py is ux-vy. The other Cauchy-Riemann equation is ux=vy, so the integrand in this use of Green's Theorem is 0 also.
Both the real and imaginary parts are 0, so the integral is 0. We're done.

A note about Goursat
The key hypothesis in Cauchy's Theorem is that u and v are continuously differentiable and satisfy the Cauchy-Riemann equations in the appropriate region. But Goursat, late in the 19th century, showed that that if f´(z) exists, then the conclusion of Cauchy's Theorem follows. The continuity of the derivative is unnecessary. The proof is in the text, and is quite clever. I won't give it here, because to my knowledge the techniques of the proof are rarely used in most of complex analysis.

Deforming curves, not changing integrals
Here is a more complicated corollary of Cauchy's Theorem. It will allow us to compute many more integrals by comparing them with one another.
Corollary Suppose that f(z) is analytic in a region R, and that C1 and C2 are simple closed curves in R. Suppose also that the curves can be deformed, one into the other, all inside the region R. Then C1f(z) dz=C2f(z) dz.
About the picture The picture attempts to illustrate the situation described in the corollary. The region R is a connected open set. The green arrows show how the curve C2 is deformed into C1 through part of R (but not through any of the holes!). If we picked a point w2 on C2, the deformation would move it along a curve to a point w1 on C1: let's call that curve B.
Proof Look at this curve, which is a sum of curves: start at w2 and move along C2 until we get to w2 again. Then move on B until we arrive at w1. Now move backwards on C1, returning to w1. Finish by moving backwards on B. The curve described is C2+B-C1-B. We apply Cauchy's Theorem to that curve, because f(z) is analytic on and inside the curve. The integral is 0, but that means 0=C2+B-C1-Bf(z) dz=C2f(z) dz+Bf(z) dz-C1f(z) dz-Bf(z) dz, so (since the B integrals cancel) we get C1f(z) dz=C2f(z) dz.
Comment Here is some (negative) discussion about this "proof". What do I mean precisely by "deformation"? I haven't described this. Well, the notion can be described precisely, but I won't do it in this course, and any situation I use this result in this course will be one where a picture verifying the proof can be explicitly drawn. Another objection can be applying Cauchy's Theorem to the curve C2+B-C1-B. This really isn't a simple closed curve. Again, this is true, but it is a limit of simple closed curves (think of twin B's slightly separated, with accompanying twin w2's and w1's). Each of those curves would be simple closed curves, and each would have integral equal to 0. The limiting curve with limiting integral would then also have integral equal to 0. The idea is what I'm showing here, and I am willing to admit that I've given up some precision!

A strange choice of function
What I'm describing is a somewhat complicated setup. I don't think I'll make it better by declaring that the whole procedure is one which is used repeatedly in many areas of applied mathematics, engineering, and physics. I will try, though, to describe the simplest incarnation of the situation. So here are the ingredients:

Now I want to consider the function g(z)=[f(z)-f(p)]/[z-p]. This is sort of bizarre, but stay with me. This function is certainly analytic everywhere that f is analytic, except perhaps at p. Because of the division by z-p, we can't be sure about the formula's behavior at p. So everywhere except p (and inside R) g(z) is surely analytic.
Now let's consider a very small circle centered at p, all inside the curve C, as shown in the picture. The curve C can be deformed to the small circle through points in R which are not p. But g is analytic at those points, so the integral doesn't change. That is, Cg(z) dz=small circleg(z) dz.
Now we'll compute small circleg(z) dz. Let's suppose the circle has radius r. Of course we know that the modulus of the integral is bounded by ML, where L is 2Pi r and M is the maximum of |g(z)| on the circle. But the formula for g(z) is not random. It is the difference quotient defining the derivative of f(z) at p. That means limz-->pg(z)=f´(p). More specifically, the limit statement itself means this: given e>0, there is d>0 so that when 0<|z-p|<d then |g(z)-f(p)|<e. That is, g(z)=f(p)+error where |error|<e. So for small r (any r less than d) we know that |g(z)|<|f(p)|+e which we can take as M. The integral of small circleg(z) dz therefore has modulus less than (2Pi r)(f(p)|+e). Let me be more precise by stating this carefully: small circleg(z) dz doesn't depend on the radius of the small circle, and, given any e>0, when the radius r is small, the value of the integral's modulus is at most (2Pi r)(f(p)|+e). But that estimate can be made as small as you like by taking r very small. The only way the estimate can be correct is if the value of the integral itself is 0.
Therefore small circleg(z) dz=0. But let's write in the formula for g, so this becomes small circle[f(z)-f(p)]/[z-p]dz. Break up the fraction in the integrand, so we get small circle[f(z)]/[z-p]-[f(p)]/[z-p]dz and this is further small circle[f(z)]/[z-p]dz-small circle[f(p)]/[z-p]dz. The second integral is small circle[f(p)]/[z-p]dz=f(p)small circle1/[z-p]dz because f(p) is a multiplicative constant with respect to integration by z. And, finally, small circle1/[z-p]dz is 2Pi i, the only integral we'll ever need. So small circleg(z) dz=0 becomes small circle[f(z)]/[z-p]dz-2Pi f(p)=0. But the integral of f(z)/[z-p] over C is the same as the integral of f(z)/[z-p] over the small circle. We've just verified a major result of the subject.

The Cauchy Integral Formula
Suppose R is a connected open set, C is a simple closed curve whose inside is contained entirely in R, p is a point inside C, and f is analytic in R. Then f(p)={1/[2Pi i]}C[f(z)]/[z-p]dz.
Comment This is a fundamental and remarkable result. Differentiability, that is, complex differentiability, has the consequence that knowing the values of f on C determines the values of f inside C. We need to sort of "average" the values properly with weights determined by 1/[z-p] but everything is determined by the boundary values. This has no counterpart in real calculus. Certainly a differentiable function's value inside an interval is not determined by what happens at the endpoints of the interval.

Expanding the Cauchy kernel
The Cauchy kernel is the function 1/[z-p]. It can be manipulated algebraically in a variety of ways to give properties of analytic functions. I'll do a basic example of this manipulation here. Later we will do others.
I will suppose f(z) is analytic in some big disc centered at 0. For my simple closed curve I will take a circle of radius R centered at 0. The radius will be smaller than the radius of the "big disc". Also p will be some point inside the circle of radius R. Then I know f(p)={1/[2Pi i]}|z|=R[f(z)]/[z-p]dz. Consider 1/[z-p]. Since z is on |z|=R and p is inside the circle, |p|<R=|z|. |p/z| is less than 1. So do this:
1/[z-p]=[1/z]/[1-{p/z}]: this is supposed to make you think of the formula for the sum of a geometric series, a/[1-r], since we have arranged for r to be p/z, a complex number of modulus less than 1. We could write the whole infinite series, or just stop after a few terms. Let me show you.
[1/z]/[1-{p/z}]=n=0pn/zn+1= n=0Npn/zn+1+n=N+1pn/zn+1=n=0Npn/zn+1+[pN+1/zN+2]/[1-{p/z}].
Here the error, the difference between the partial sum and the whole series, is [pN+1/zN+2]/[1-{p/z}]. The modulus of this is at most (for any z on the circle) [|p|N+1/RN+2]/[1-{|p|/R}] and since |p|/R<1, this will go to 0 as N increases.
Now let's stuff the preceding stuff into the integral:
The finite sum is really n=0N|z|=R[f(z)] pn/zn+1dz= n=0N(|z|=R[f(z)]/zn+1dz)pn.
The other term can be estimated by ML: the length is 2Pi R while the M is the maximum of |f| on the circle |z|=R multiplied by [|p|N+1/RN+2]/[1-{|p|/R}]. Since |p|/R<1, this will go to 0 as N increases. But put everything together. We have just shown that a certain infinite series converges, and since we know the Cauchy integral formula, we also know the sum of the infinite series.

f(p)=n=0anpn where an=[1/{2Pi i}]|z|=R[f(z)]/zn+1dz.

Complex differentiability implies power series
Here is a general statement of the preceding result.
Theorem Suppose f(z) is analytic in a connected open set which includes some disc centered at a point z0. Then there are complex numbers {an}n=0 so that the series n=0an(z-z0)n converges at least for all z in that disc centered at z0. Additionally, the coefficients an are given by the formula an=[1/{2Pi i}]|z-z0|=R[f(w)]/(w-z0)n+1dw where R is any positive number less than the radius of the disc.

First, since power series can be differentiated inside their radius of convergence, we see that a complex differentiable function has a derivative which is itself a complex differentiable function! This is totally unlike real calculus. For example, we know that |x| can't be differentiated at 0. It has an antiderivative, let's call it h(x), given piecewise by x2/2 for x>=0 and by -x2/2 for z<0. The function h(x) has a derivative (just |x|) and it does not have a second derivative. In the complex "realm", functions always have more derivatives! There aren't any problems like |x|. One derivative implies having infinitely many derivatives!

Second, we saw when we studied power series that a function can have only one power series representation centered at a point. That is, if f(z)=n=0an(z-z0)n and f(z)=n=0bn(z-z0)n, both converging near z0, then each an is equal to the corresponding bn. In fact, since we can differentiate power series, the power series must be a Taylor series: an=(1/n!)f(n)(z0). But we can compare the two descriptions of the coefficients an we have to deduce the following result.

The Cauchy Integral Formula for Derivatives
Suppose R is a connected open set, C is a simple closed curve whose inside is contained entirely in R, p is a point inside C, and f is analytic in R. Then for any positive integer n, f is n times differentiable, and f(n)(p)={n!/[2Pi i]}C[f(z)]/[(z-p)n+1]dz.

I think I proved this only for a circle which has p inside it, but any C described in the theorem can be deformed to such a circle, so the result is true also.

Maintained by and last modified 2/21/2008.