Laurent Series Representations

Suppose f(z) is not analytic in , but is analytic in . For

example, the function is not analytic when but is analytic for . This function does not have a Maclaurin series representation. If we use the Maclaurin series for

, however, and formally divide each term in that series by , we obtain the representation

that is valid for all z such that .

The question arises as to whether it might be possible to generalize the Taylor series method to functions analytic in an ring

.

Perhaps we can represent these functions with a series that employs negative powers of z in

some way as we did with . As you will see shortly, we can indeed. We begin by defining a series that allows for negative powers of z.

Definition Laurent Series: Let be a complex number for . The doubly

infinite series , called a Laurent series, is defined by

,

provided the series on the right-hand side of this equation converge.

Remark Recall that is a simplified expression for the sum . At

times it will be convenient to write as

,

rather than using the expression above

Definition: Given , we define the ring centered at with radii r and R by

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.

The closed ring centered at with radii r and R is denoted by

.

The figure below illustrates these terms.

The closed ring . The shaded portion is the open ring .

Theorem Suppose that the Laurent series converges on a ring . Then the series converges uniformly on any closed sub-ring where .

The main result of this section specifies how functions analytic in a ring can be expanded in a Laurent series. In it, we will use symbols of the form , which - we remind you - designate the positively oriented circle with radius and center . That is, , oriented counterclockwise.

Theorem (Laurent's Theorem). Suppose , and that f(z) is analytic in the ring . If is any number such that , then for all

the function value has the Laurent series representation

,

where for , the coefficients are given by

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and .

Remark. What happens to the Laurent series if f(z) is analytic in the disk ? If we look at

the equations we see that the coefficient for the positive power equals by using Cauchy's integral formula for derivatives. Hence, the series iinvolving the positive powers of is actually the Taylor series for f(z). The Cauchy - Goursat theorem shows us that the coefficients for the negative powers of equal zero. In this case, therefore, there are no negative powers involved, and the Laurent series reduces to the Taylor series.

The theorem beow delineates two important aspects of the Laurent series.

Theorem (Uniqueness and differentiation of Laurent Series). Suppose that is analytic in the ring , and has the Laurent series representation

for all .

(i) If for all , then for all n. (In other words, the Laurent series for f(z) in a given ring is unique.)

(ii) For all , the derivatives for may be obtained by termwise differentiation of its Laurent series.

The uniqueness of the Laurent series is an important property because the coefficients in the Laurent expansion of a function are seldom found by using Equation (7-23). The following examples illustrate some methods for finding Laurent series coefficients.

Example. Find three different Laurent series representations for the function involving powers of z.

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Solution. The function f(z) has singularities at and is analytic in the disk , in the ring , and in the region . We want to find a different Laurent series for f(z) in each of the three domains D, A, and R. We start by writing f(z) in its partial fraction form:

.We use the theorems and corollaries to obtain the following representations for the terms on the right side of the equation above:

1

2

3

4

As we saw in class, representations 1 and 3 are both valid in the disk , and thus we have

valid for ,

which is a Laurent series that reduces to a Maclaurin series.

In the ring , representations 2 and 3 are valid; hence we get

valid for .

Finally, in the region we use Representations 2 and 4 to obtain

valid for .

Example Find the Laurent series representation for that involves powers of z.

Solution. We know that , and hence the Maclaurin series for is

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, then we can write

or in another way we can write

We formally divide each term by to obtain the Laurent series

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