A friend of mine is taking a course on analytic number theory in the spring and needs to learn basic complex analysis in a couple of weeks. I decided to do a post (self-contained, except for Stokes’ formula) on deducing the Cauchy theorems and their applications from Stokes’ theorem now instead of later–when I’ll talk about several complex variables. It might be objected that Stokes’ theorem is just Green’s theorem for , commonly used in undergraduate treatments, but my goal was to take an expository challenge: write something rigorous on complex variables in as short a space as possible without sacrificing readability. So Stokes’ theorem for manifolds is preferable to Green’s theorem as stated in a vague way about “insides of a curve” (before, say, the Jordan curve theorem is proved) and the traditional proof of Green’s theorem via rectangular decompositions.

So, let’s consider an open set , and a function . We can consider the differential

which is a complex-valued 1-form on . It is also convenient to write the differential using the and -derivatives I talked about earlier, i.e.

The reason these are important is that if , we can choose with

by differentiability, and it is easy to check that . So we can define a function to be **holomorphic** if it satisfies the differential equation

which is equivalent to being able to write

for each and a suitable . In particular, it is equivalent to a difference quotient definition. The **derivative** of a holomorphic function thus satisfies all the usual algebraic rules, under which holomorphic functions are closed.

Recall now that is a multiple of the Laplacian. In particular, it follows that a holomorphic function is harmonic, and thus infinitely differentiable.

The other reason we care about the -derivative is that we can write, for any ,

where naturally enough . It is easy to check that and are dual to each other as complex-valued elements of the cotangent and tangent spaces, respectively. So, as a result, we can compute on (complex-valued) 1-forms:

This will lead to the Cauchy theorem. Note that is a multiple of the usual area element.

**Complex integration **

Cauchy’s theorem is now a simple corollary of Stokes’ theorem. Given a 1-dimensional manifold , we can consider the integral of a complex-valued 1-form

which, given a parametrization, can of course be computed using the change-of-variables formula, and by taking real and imaginary parts. More interestingly, when is the boundary of a compact 2-dimensional submanifold-with-boundary we can compute it using Stokes theorem:

In particular, if is holomorphic on , we find:

Theorem 1 (Cauchy)For the boundary of , we have

Using a similar approach, we can establish the following:

Theorem 2 (Cauchy formula)If is in the interior of and is holomorphic in , then

In the proof, we will actually show something more general. So let us **not** assume holomorphic, only differentiable.

To prove this, let’s consider the manifold-with-boundary for very small, as in the figure.

The boundary is the disjoint union of with the usual orientation and the circle with the opposite orientation. Also, is holomorphic in a neighborhood of . This is the key fact, and we will use this together with a limiting argument on . So

First, we can compute the term by using the parametrization of the circle. Then , so

and as , this approaches . The integrand in the second term becomes times

because the -derivative clearly satisfies the product rule and is holomorphic (by the quotient rule).

So, when we let , using the integrability of , we find:

This is a more general analog of the Cauchy formula, and it clearly implies the usual one when is holomorphic.

**Representability as power series **

Fix a holomorphic function on . In a neighborhood with ; then for any we have

In particular, we have represented as a weighted average of the functions for ranging over the circle . Each function is represented by a power series in , the geometric series. Thus, the weighted average has the same property. We now prove this rigorously; by a translation assume . So, writing out the geometric series:

where is the constant . It is immediate that , so convergence questions don’t arise when , and the interchange of summation and integration was justified. In fact the convergence is uniform on compact subsets (where is bounded by something less than ) by the same argument. We have thus proved that **can be represented as a power series** converging in the whole disk. Differentiation of this power series can be done term-by-term as one can check directly, though the justification is slightly technical.

**Laurent series **

The Taylor series just discussed works for functions in a disk. In an annulus, we can’t necessarily do that, but we end up with terms of negative order. As usual, let be holomorphic. Suppose the closed annulus , where . Then by Cauchy’s formula applied to , we have for ,

Now is holomorphic in so can be represented as a power series as before. is holomorphic in the “disk at infinity” , and we can write for such

The convergence issues can be handled as before because the integrands are and .

In particular, we can represent in the annulus as

where the are uniquely determined constants. The uniqueness is seen because one can choose between , and get

because all but one of the integrals vanish by direct computation. So determines the .

March 27, 2010 at 8:24 am

The next step (without proof), concerning Laurent series.

The expansion of — in the vinicity of — into a Laurent series is

, ,

where the coefficients are

for .

If is an order pole of , then the coefficient is given by

March 27, 2010 at 12:29 pm

Thanks! Of course you’re right- I hadn’t gone into that much detail because it was only a crash course.

March 27, 2010 at 8:33 am

correction

March 27, 2010 at 8:35 am

Please excuse me!

New correction

You can edite my first comment and delete these two.

March 27, 2010 at 8:43 am