**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 _{C}f(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 _{C}pdx+qdy=_{R}q_{x}-p_{y}dA 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
q_{x}-p_{y} is -v_{x}-u_{y}. But one
of the Cauchy-Riemann equations is u_{y}=-v_{x}, 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
q_{x}-p_{y} is u_{x}-v_{y}. The other
Cauchy-Riemann equation is u_{x}=v_{y}, 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 19^{th}
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
C_{1} and C_{2} are simple closed curves in R. Suppose
also that the curves can be deformed, one into the other, all
*inside* the region R. Then _{C1}f(z) dz=_{C2}f(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
C_{2} is *deformed* into C_{1} through part of R
(but *not* through any of the holes!). If we picked a point
w_{2} on C_{2}, the deformation would move it along a curve to a point w_{1} on
C_{1}: let's call that curve B.

**Proof** Look at this curve, which is a sum of curves: start at
w_{2} and move along C_{2} until we get to
w_{2} again. Then move on B until we arrive at
w_{1}. Now move *backwards* on C_{1}, returning
to w_{1}. Finish by moving backwards on B. The curve described
is C_{2}+B-C_{1}-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-B}f(z) dz=_{C2}f(z) dz+_{B}f(z) dz-_{C1}f(z) dz-_{B}f(z) dz, so (since the B integrals cancel) we get
_{C1}f(z) dz=_{C2}f(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 C_{2}+B-C_{1}-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 w_{2}'s and w_{1}'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:

- A connected open set R and a function f(z) which is analytic in R.
- A simple closed curve C whose inside is contained entirely in R.
- A
*random*(!) point p inside C.

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

Now we'll compute

Therefore

**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=0}^{}p^{n}/z^{n+1}= _{n=0}^{N}p^{n}/z^{n+1}+_{n=N+1}^{}p^{n}/z^{n+1}=_{n=0}^{N}p^{n}/z^{n+1}+[p^{N+1}/z^{N+2}]/[1-{p/z}].

Here the `error`, the difference between
the partial sum and the whole series, is
[p^{N+1}/z^{N+2}]/[1-{p/z}]. The modulus of this is at
most (for any z on the circle)
[|p|^{N+1}/R^{N+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:

_{|z|=R}[f(z)]/[z-p]dz=_{|z|=R}[f(z)](_{n=0}^{N}p^{n}/z^{n+1}+`error`)dz

The finite sum is really
_{n=0}^{N}_{|z|=R}[f(z)]
p^{n}/z^{n+1}dz= _{n=0}^{N}(_{|z|=R}[f(z)]/z^{n+1}dz)p^{n}.

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}/R^{N+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=0}^{}a_{n}p^{n} where
a_{n}=[1/{2Pi i}]_{|z|=R}[f(z)]/z^{n+1}dz.

**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 z_{0}. Then there are
complex numbers {a_{n}}_{n=0}^{} so that the series _{n=0}^{}a_{n}(z-z_{0})^{n} converges at
least for all z in that disc centered at z_{0}. Additionally,
the coefficients a_{n} are given by the formula
a_{n}=[1/{2Pi i}]_{|z-z0|=R}[f(w)]/(w-z_{0})^{n+1}dw
where R is any positive number less than the radius of the disc.

**Consequences**

**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 x^{2}/2 for x>=0 and by
-x^{2}/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=0}^{}a_{n}(z-z_{0})^{n} and
f(z)=_{n=0}^{}b_{n}(z-z_{0})^{n}, both
converging near z_{0}, then each a_{n} is equal to the
corresponding b_{n}. In fact, since we can differentiate power
series, the power series must be a Taylor series:
a_{n}=(1/n!)f^{(n)}(z_{0}). But we can compare
the two descriptions of the coefficients a_{n} 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
greenfie@math.rutgers.edu and last modified 2/21/2008.
**