Laurent series

I must confess, this post will be hand-waivy. My goal in these posts has always been to understand mathematics, and this mostly coincides with careful proofs and decent intuitive explanations. The contents of this post I think are best explained intuitively, leaving the proofs to you, the reader.

Rest assured, you can prove everything in this post. As you will see, this post generalizes much of what we have done thus far. Rather than reprove everything in greater generality, I invite you to generate the new proofs. The proofs we have seen thus far can be altered slightly to get the results presented in this post.

Let’s start with a motivating example. Consider the function {f(z)=1/z}. It’s holomorphic on {{\mathbb C}-\{0\}}, so for any {z_0} other than zero, we can expand it as a power series centered at {z_0}. The expansion is valid in the open ball of radius {|z_0|}; that is, right up to, but not including the origin. Well, that’s fine, but a bit annoying for two reasons. For one, we have to pick some {z_0}, and different choices of z_0 gives us different power series coefficients. Further, no choice is valid everywhere. In fact, we need infinitely many of these choices to cover all of {{\mathbb C}-\{0\}}.

So here’s an idea. What if we allow negative terms in a power series expansion? Then, in this “super-expansion,” we could just write {1/z} as {z^{-1}}.

Definition 1 A Laurent series centered at {z_0} is a series of the form:

\displaystyle \cdots+a_{-2}(z-z_0)^{-2}+a_{-1}(z-z_0)^{-1}+\cdots+a_0+a_1(z-z_0)+a_2(z-z_0)^2+\cdots

As for the radius of convergence of such an expansion, now there are two radii to consider. How far out can we go and have it be valid, and how close to {z_0} can we get and have it be valid. The first has the same formula. The outer radius of convergence is still given by

\displaystyle R=\left(\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}\right)^{-1}.

The inner radius of convergence, {r}, has a formula which is not terribly surprising:

\displaystyle r=\displaystyle\limsup_{n\rightarrow\infty}\sqrt[n]{|a_{-n}|}.

Recognize that power series are just special cases of Laurent series, and the inner radius of convergence just evaluates to zero (since all the terms are zero). As for the proofs of these facts, they contain no new ideas, so I’ll omit them.

You might not expect the Cauchy integral formula to still hold, but it does. The proof is pretty much the same:

Theorem 2 (Cauchy integral formula) Let {f(z)=\sum_{-\infty}^\infty a_n(z-z_0)^n}. Let {\gamma} denote a counter-clockwise path around {z_0} on which we can integrate {f}. Then

\displaystyle a_n=\displaystyle\frac{1}{2\pi i}\int_\gamma\frac{f(z)}{(z-z_0)^{n+1}}dz.

Yes, this is true, even for {n<0}. Quite remarkable.

Advertisements

One Response to Laurent series

  1. Pingback: Meromorphic functions and residues « Andy Soffer

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s