INTRODUCTION

If a complex function f fails to be analytic at a point z = z₀, then this point is said to be a singularity or a singular point of the function. For example, the complex numbers z = 2i and z = −2i are singularities of the function f(z) = z/(z² + 4) because f is discontinuous at each of these points. Recall from Section 17.6 that the principal value of the logarithm, Ln z, is analytic at all points except those points on the branch cut consisting of the nonpositive x-axis; that is, the branch point z = 0 as well as all negative real numbers are singular points of Ln z. In this section we will be concerned with a new kind of “power series” expansion of f about an isolated singularity z₀. This new series will involve negative as well as nonnegative integer powers of z − z₀.

Isolated Singularities

Suppose that z = z₀ is a singularity of a complex function f. The point z = z₀ is said to be an isolated singularity of the function f if there exists some deleted neighborhood, or punctured open disk, 0 < | z − z₀ | < R of z₀ throughout which f is analytic. For example, we have just seen that z = 2i and z = −2i are singularities of f(z) = z/(z² + 4). Both 2i and −2i are isolated singularities since f is analytic at every point in the neighborhood defined by |z − 2i| < 1 except at z = 2i and at every point in the neighborhood defined by |z − (−2i)| < 1 except at z = −2i. In other words, if f is analytic in the deleted neighborhood, 0 < |z − 2i| < 1 and 0 < | z + 2i | < 1. On the other hand, the branch point z = 0 is not an isolated singularity of Ln z since every neighborhood of z = 0 must contain points on the negative x-axis. We say that a singular point z = z₀ of a function f is nonisolated if every neighborhood of z₀ contains at least one singularity of f other than z₀. For example, the branch point z = 0 is a nonisolated singularity of Ln z since every neighborhood of z = 0 contains points on the negative real axis.

A New Kind of Series

If z = z₀ is a singularity of a function f, then certainly f cannot be expanded in a power series with z₀ as its center. However, about an isolated singularity z = z₀ it is possible to represent f by a new kind of series involving both negative and nonnegative integer powers of z − z₀; that is,

Using summation notation, the last expression can be written as the sum of two series

(1)

The two series on the right-hand side in (1) are given special names. The part with negative powers of z − z₀, that is,

,

is called the principal part of the series (1) and will converge for |1/(z − z₀)| < r* or equivalently for |z − z₀| > 1/r* = r. The part consisting of the nonnegative powers of z − z₀,

,

is called the analytic part of the series (1) and will converge for |z − z₀| < R. Hence, the sum of these parts converges when z both |z − z₀| > r and |z − z₀| < R; that is, when z is a point in an annular domain defined by r < |z − z₀| < R.

By summing over negative and nonnegative integers, (1) can be written compactly as

The next example illustrates a series of the form (1) in which the principal part of the series consists of a finite number of nonzero terms, but the analytic part consists of an infinite number of nonzero terms.

EXAMPLE 1 A New Kind of Series

The function f(z) = (sin z)/z³ is not analytic at z = 0 and hence cannot be expanded in a Maclaurin series. However, sin z is an entire function, and from (13) of Section 19.2 we know that its Maclaurin series,

converges for all z. Dividing this power series by z³ gives the following series with negative and nonnegative integer powers of z:

(2)

This series converges for all z except z = 0; that is, for 0 < |z|. ≡

A series representation of a function f that has the form given in (1)—and (2) is such an example—is called a Laurent series or a Laurent expansion of f after the French mathematician and military engineer Pierre Alphonse Laurent (1813–1854). Basically, (1) is a generalization of a Taylor series expansion, but Laurent’s 1843 paper in which this series was discussed was not published in his lifetime.

Let f be analytic within the annular domain D defined by r < |z − z₀| < R. Then f has the series representation

(3)

valid for r < |z − z₀| < R. The coefficients a_k are given by

(4)

where C is a simple closed curve that lies entirely within D and has z₀ in its interior (see FIGURE 19.3.1).

PROOF:

Let C₁ and C₂ be concentric circles with center z₀ and radii r₁ and R₂, where r < r₁ < R₂ < R. Let z be a fixed point in D that also satisfies r₁ < |z − z₀| < R₂. See FIGURE 19.3.2. By introducing a cross cut between C₂ and C₁, we find from Cauchy’s integral formula that

(5)

Proceeding as in the proof of Theorem 19.2.4, we can write

(6)

where (7)

Now using (5) and (8) of Section 19.1, we have

(8)

where (9)

and

Now let d denote the distance from z to z₀; that is, |z − z₀| = d, and let M denote the maximum value of |f(z)| on the contour C₁. Since |s − z₀| = r₁,

|z − s| = |z − z₀ − (s − z₀)| ≥ |z − z₀| − |s − z₀| = d − r₁.

The ML-inequality then gives

Because r₁ < d, (r₁/d)ⁿ → 0 as n → ∞ and so |R_n(z)| → 0 as n → ∞. Thus we have shown that

(10)

where the coefficients a_−k are given in (9). Combining (6) and (10), we see that (5) yields

(11)

Finally, by summing over nonnegative and negative integers, we can write (11) as f(z) = (z − z₀)^k. However, (7) and (9) can be written as a single integral:

where, in view of (3) of Section 18.2, we have replaced the contours C₁ and C₂ by any simple closed contour C in D with z₀ in its interior. ≡

In the case when a_−k = 0 for k = 1, 2, 3, . . ., the Laurent series (3) is a Taylor series. Because of this, a Laurent expansion is a generalization of a Taylor series.

The annular domain in Theorem 19.3.1 defined by r < |z − z₀| < R need not have the “ring” shape illustrated in Figure 19.3.2. Some other possible annular domains are (i) r = 0, R finite; (ii) r ≠ 0, R → ∞; and (iii) r = 0, R → ∞. In the first case, the series converges in the annular domain defined by 0 < |z − z₀| < R. This is the interior of the circle |z − z₀| = R except the point z₀. In the second case, the annular domain is defined by r < |z − z₀|; in other words, the domain consists of all points exterior to the circle |z − z₀| = r. In the third case, the domain is defined by 0 < |z − z₀|. This represents the entire complex plane except the point z₀. The series we obtained in (2) is valid on this last type of domain.

In actual practice, the formula in (4) for the coefficients of a Laurent series is seldom used. As a consequence, finding the Laurent series of a function in a specified annular domain is generally not an easy task. We often use the geometric series (5) and (6) of Section 19.1 or, as we did in Example 1, other known series. But regardless of how a Laurent expansion of a function f is obtained in a specified annular domain, it is the Laurent series; that is, the series we obtain is unique.

EXAMPLE 2 Laurent Expansions

Expand f(z) = in a Laurent series valid for the following annular domains.

(a) 0 < |z| < 1

(b) 1 < |z|

(c) 0 < |z − 1| < 1

(d) 1 < |z − 1|

SOLUTION

The four specified annular domains are shown in FIGURE 19.3.3. In parts (a) and (b), we want to represent f in a series involving only negative and nonnegative integer powers of z, whereas in parts (c) and (d) we want to represent f in a series involving negative and nonnegative integer powers of z − 1.

(a) By writing

we can use (5) of Section 19.1:

The series in the brackets converges for |z| < 1, but after this expression is multiplied by 1/z, the resulting series

converges for 0 < |z| < 1.

(b) To obtain a series that converges for 1 < |z| we start by constructing a series that converges for |1/z| < 1. To this end we write the given function f as

and again use (5) of Section 19.1 with z replaced by 1/z:

The series in the brackets converges for |1/z| < 1 or equivalently for 1 < |z|. Thus the required Laurent series is

(c) This is basically the same problem as part (a) except that we want all powers of z − 1. To that end we add and subtract 1 in the denominator and use (6) of Section 19.1 with z replaced by z − 1:

The series in brackets converges for |z − 1| < 1, and so the last series converges for 0 < |z − 1| < 1.

(d) Proceeding as in part (b), we write

Because the series within the brackets converges for |1/(z − 1)| < 1, the final series converges for 1 < |z − 1|. ≡

EXAMPLE 3 Laurent Expansions

Expand f(z) = in a Laurent series valid for (a) 0 < |z − 1| < 2 and (b) 0 < |z − 3| < 2.

SOLUTION

(a) As in parts (c) and (d) of Example 2, we want only powers of z − 1, and so we need to express z − 3 in terms of z − 1. This can be done by writing

and then using (5) of Section 19.1 with z replaced by (z − 1)/2:

(12)

(b) To obtain powers of z − 3 we write z − 1 = 2 + (z − 3) and

At this point we can expand in a power series using the general binomial theorem:

The binomial series is valid for |(z − 3)/2| < 1 or |z − 3| < 2. Multiplying this series by 1/4(z − 3) gives a Laurent series that is valid for 0 < |z − 3| < 2:

≡

EXAMPLE 4 A Laurent Expansion

Expand f(z) = in a Laurent series valid for 0 < |z| < 1.

SOLUTION

By (5) of Section 19.1 we can write

We then multiply the series by 8 + 1/z and collect like terms:

The geometric series converges for |z| < 1. After multiplying by 8 + 1/z, the resulting Laurent series is valid for 0 < |z| < 1. ≡

In the preceding examples, the point at the center of the annular domain of validity for each Laurent series was an isolated singularity of the function f. A reexamination of Theorem 19.3.1 shows that this need not be the case.

EXAMPLE 5 A Laurent Expansion

Expand f(z) = in a Laurent series valid for 1 < |z − 2| < 2.

SOLUTION

The specified annular domain is shown in FIGURE 19.3.4. The center of this domain, z = 2, is a point of analyticity of the function f. Our goal now is to find two series involving integer powers of z − 2: one converging for 1 < |z − 2| and the other converging for |z − 2| < 2. To accomplish this, we start with the decomposition of f into partial fractions:

(13)

Now,

This series converges for |(z − 2)/2| < 1 or |z − 2| < 2. Furthermore,

converges for |1/(z − 2)| < 1 or 1 < |z − 2|. Substituting these two results in (13) then gives

This representation is valid for 1 < |z − 2| < 2. ≡

EXAMPLE 6 A Laurent Expansion

Expand f(z) = e^3/z in a Laurent series valid for 0 < |z|.

SOLUTION

From (12) of Section 19.2 we know that for all finite z,

(14)

By replacing z in (14) by 3/z, z ≠ 0, we obtain the Laurent series

This series is valid for 0 < |z|. ≡

REMARKS

In conclusion, we point out a result that will be of particular importance to us in Sections 19.5 and 19.6. Replacing the complex variable s with the usual symbol z, we see that when k = −1, (4) for the Laurent series coefficients yields

or, more importantly, the integral can be written as

(15)

19.3 Exercises Answers to selected odd-numbered problems begin on page ANS-47.

In Problems 1–6, expand the given function in a Laurent series valid for the indicated annular domain.

1. f(z) = , 0 < |z|
2. f(z) = , 0 < |z|
3. f(z) = , 0 < |z|
4. f(z) = , 0 < |z|
5. f(z) = , 0 < |z − 1|
6. f(z) = z cos , 0 < |z|

In Problems 7–12, expand f(z) = in a Laurent series valid for the indicated annular domain.

7. 0 < |z| < 3
8. |z| > 3
9. 0 < |z − 3| < 3
10. |z − 3| < 3
11. 1 < |z − 4| < 4
12. 1 < |z + 1| < 4

In Problems 13–16, expand f(z) = in a Laurent series valid for the indicated annular domain.

13. 1 < |z| < 2
14. |z| < 2
15. 0 < |z − 1| < 1
16. 0 < |z − 2| < 1

In Problems 17–20, expand f(z) = in a Laurent series valid for the indicated annular domain.

17. 0 < |z + 1| < 3
18. |z + 1| < 3
19. 1 < |z| < 2
20. 0 < |z − 2| < 3

In Problems 21 and 22, expand f(z) = in a Laurent series valid for the indicated annular domain.

21. 0 < |z| < 1
22. |z| > 1

In Problems 23 and 24, expand f(z) = in a Laurent series valid for the indicated annular domain.

23. 0 < |z − 2| < 1
24. 0 < |z − 1| < 1

In Problems 25 and 26, expand f(z) = in a Laurent series valid for the indicated annular domain.

25. 0 < |z| < 1
26. 0 < |z − 1| < 1

In Problems 27 and 28, expand f(z) = in a Laurent series valid for the indicated annular domain.

27. 1 < |z − 1|
28. 0 < |z − 2|