We’ve established that if has an isolated singularity at and that has a pole of order at then where the coefficients are as follows: and where is some circle taken once in the standard direction that encloses but no other singular point of .

So here is what we will prove: if there exits where is analytic on the set then has a Laurent series expansion about valid for all in this set. Now IF is an singular point and our set is a punctured disk () then the singularity is essential, if and only if the principal part (the part of the Laurent series with the negative powers of has an infinite number of terms, and we have NOT proved that such a series exists in this case. We will remedy this now.

In this diagram: the (possible) singularity is at , (this construction does not require that be a singular point) the large is a circle between and another singularity (that is why this works for isolated singularities), is some point enclosed by where we want to evaluate and small is a circle between and . The green circle about is one between and and note that is analytic inside and just outside of this circle, so Cauchy’s Theorem applies.

Let denote the small circle enclosing taken half way around in the standard direction and note that . But by the theory of cuts (integrate along the in the standard direction, along one cut line, half way around in the opposite direction, once around in the opposite direction, half around in the opposite direction, and back along the cut line to we see that as together, these paths bound a simply connected region where is analytic (and the integrals along the cut lines cancel out).

So we get We switch the sign on the second integral:

We now make the following observations: in the first integral (along ) we note that and in the second integral (along )

Now look at the first integral..in particular the fraction

because we are in the region where the geometric series converges absolutely.

So our integral becomes . In previous work we’ve shown that we can interchange integration and summation in this case so, we end up with : which yields the regular part of the Laurent series.

We know turn to what will become the principal part: the second integral.

. Let’s focus on the fraction

Now we write and recall that, on , so we can expand this into an infinite series (bounded by a convergent geometric series)

So now going back to the integral

and, once again, because the series is bounded by a convergent geometric series (details similar to those we used in developing the power series) we can interchange summation and integration to obtain

. It is traditional to shift the index to write it as:

So, adding the two series together we have:

Now divide both sides by to obtain the desired result.

**One caveat** To define a Laurent series, one needs to know: the center about which one is expanding about and the annulus or disk of convergence. The derivation that we used above does not require to be a singularity or for there to be only 1 singularity inside of ; we just require that be analytic on some open region containing the annulus with boundary circles

**Actually obtaining the Laurent series**

I did NOT show (but it is true) that Laurent series are unique (we did that for power series), but the general principle still applies. Once you get a Laurent series for a function about a singularity (often valid on some punctured disk about a singularity), you have them all. But if you looked at the proof, what was required is that be analytic in the region bounded by and (an open annulus where the smaller boundary circle encloses the singularity in question).

So our Laurent series is valid on an punctured disk or on an annulus..with the later being true if, say, there were two singularities.

Now as far as obtaining: we do that by “hook or crook” and almost never by actually calculating those dreadful integrals.

1) has an isolated essential singularity at and its Laurent series can be obtained by substituting for in the usual Taylor series:

2) Now let’s look at something like Here there are two isolated poles of order 1: and . So the series, we get, will depend on where we expand from AND where we want the expansion to be valid.

Two of the “obvious” expansion points would be and and we’d expect a radius of validity to be 2 for each of these. We could also expand about, say, zero and expect a power series expansion to have radius 1. Now if we want a series to be, say, centered about 0 and be valid for a radius that is GREATER than 1, we can look at the infinite annulus whose inner boundary circle encloses both singularities.

So let us start:

1) about : here write . Because is analytic on this is a power series with no principal part.

2) about One way to do it is to revert to partial fractions:

So . Note the second sum is analytic near so we can write a power series expansion:

which is the regular part of the Laurent series.

The total series is

and the principal part has only one non-zero term, as expected.

The punctured disk of convergence has radius 2, as expected. I’ll leave it as an exercise to find the Laurent series about but the result will look *very* similar.

Now for the Laurent series centered at but valid for . This will have an infinite number of terms with negative exponent but…this series is NOT centered at either singularity.

(the even powers cancel each other)

Which equals

Try graphing versus up to, say the 14’th power, 16’th power…22’nd power, etc. on the range, say, . That gives you an idea of the convergence.