# Laurent Series.

The series is convergent and as as its sum if both and are convergent and if its sum is equal to the sum of the sums of these series.

Theorem 8.2.1   Take .
1. The series is convergent in the domain , where

2. If , then the function is analytic in the annulus .

Theorem 8.2.2 (Laurent expansion of a function)   Suppose that the function is analytic on the annulus . Then has an expansion as a Laurent series in .

This series expansion of is unique and its coefficients are given by:

where is any simple smooth loop included in and enclosing the inner boundary circle of .

Corollary 8.2.3 (Generalization)   Let be analytic in the annulus . Then has a unique Laurent expansion . The coefficients are given by the formula

where is a any simple smooth loop included in and enclosing the inner boundary circle of .

Example 8.2.4

Yes, this can happen: sometimes the Laurent expansion contains a finite numbers of terms (when?).

Example 8.2.5

This series converges for . The convergence domain in the plane is the open unit disk centered at the origin; cf Fig 4(a).

Example 8.2.6

This series is convergent when , i.e. when (cf Fig 4(a)).

Example 8.2.7   Let . We wish to expand as a Laurent series convergent on an annulus.

We decompose as a sum of partial fractions:

Now, we expand both partial fractions as Laurent series about 0:

1. This series is convergent for .

2. This series is convergent for ,i.e. .
The intersection of the two domains of convergence is empty, so we went in a wrong direction. Let us try in another way:

1. This series is convergent for , i.e. .

2. This series is convergent for , i.e. .
The intersection of these convergence domains is the annulus displayed on Fig 4(b). On this annulus, a Laurent series expansion of is:

Noah Dana-Picard 2007-12-24