INTRODUCTION

We saw in the last section that if the complex function f has an isolated singularity at the point z₀, then f has a Laurent series representation

which converges for all z near z₀. More precisely, the representation is valid in some deleted neighborhood of z₀, or punctured open disk, 0 < |z − z₀| < R. In this section our entire focus will be on the coefficient a₋₁ and its importance in the evaluation of contour integrals.

Residue

The coefficient a₋₁ of 1/(z − z₀) in the Laurent series given above is called the residue of the function f at the isolated singularity z₀. We shall use the notation

to denote the residue of f at z₀. Recall, if the principal part of the Laurent series valid for 0 < |z − z₀| < R contains a finite number of terms with a_−n the last nonzero coefficient, then z₀ is a pole of order n; if the principal part of the series contains an infinite number of terms with nonzero coefficients, then z₀ is an essential singularity.

EXAMPLE 1 Residues

Later on in this section we will see why the coefficient a₋₁ is so important. In the meantime we are going to examine ways of obtaining this complex number when z₀ is a pole of a function f without the necessity of expanding f in a Laurent series at z₀. We begin with the residue at a simple pole.

PROOF:

Since z = z₀ is a simple pole, the Laurent expansion of f about that point has the form

By multiplying both sides by z − z₀ and then taking the limit as z → z₀, we obtain (z − z₀) f(z) = [a₋₁ + a₀(z − z₀) + a₁(z − z₀)² + …] = a₋₁ = Res (f(z), z₀). ≡

PROOF:

Since f is assumed to have a pole of order n, its Laurent expansion for 0 < |z − z₀| < R must have the form

We multiply the last expression by (z − z₀)ⁿ:

and then differentiate n − 1 times:

(3)

Since all the terms on the right side after the first involve positive integer powers of z − z₀, the limit of (3) as z → z₀ is

Solving the last equation for a₋₁ gives (2). ≡

Note that (2) reduces to (1) when n = 1.

EXAMPLE 2 Residue at a Pole

The function f(z) = has a simple pole at z = 3 and a pole of order 2 at z = 1. Use Theorems 19.5.1 and 19.5.2 to find the residue at each pole.

SOLUTION

Since z = 3 is a simple pole, we use (1):

Now at the pole of order 2 it follows from (2) that

When f is not a rational function, calculating residues by means of (1) can sometimes be tedious. It is possible to devise alternative residue formulas. In particular, suppose a function f can be written as a quotient f(z) = g(z)/h(z), where g and h are analytic at z = z₀. If g(z₀) ≠ 0 and if the function h has a zero of order 1 at z₀, then f has a simple pole at z = z₀ and

(4)

To see this last result, we use (1) and the facts that h(z₀) = 0 and that lim_z→z0(h(z) − h(z₀))/(z − z₀) is a definition of the derivative h′(z₀):

Analogous formulas for residues at poles of order greater than 1 are complicated and will not be given.

EXAMPLE 3 Using (4) to Compute a Residue

The polynomial z⁴ + 1 can be factored as (z − z₁)(z − z₂)(z − z₃)(z − z₄), where z₁, z₂, z₃, and z₄ are the four distinct roots of the equation z⁴ + 1 = 0. It follows from Theorem 19.4.1 that the function

has four simple poles. Now from (10) of Section 17.2 we have z₁ = e^πi/4, z₂ = e^3πi/4, z₃ = e^5πi/4, z₄ = e^7πi/4. To compute the residues, we use (4) and Euler’s formula:

Residue Theorem

We come now to the reason for the importance of the residue concept. The next theorem states that under some circumstances, we can evaluate complex integrals dz by summing the residues at the isolated singularities of f within the closed contour C.

PROOF:

Suppose C₁, C₂, . . ., C_n are circles centered at z₁, z₂, . . ., z_n, respectively. Suppose further that each circle C_k has a radius r_k small enough so that C₁, C₂, . . ., C_n are mutually disjoint and are interior to the simple closed curve C. See FIGURE 19.5.1. Recall that (15) of Section 19.3 implies dz = 2πi Res(f(z), z_k), and so Theorem 18.2.2 gives

≡

EXAMPLE 4 Evaluation by the Residue Theorem

Evaluate where

(a) the contour C is the rectangle defined by x = 0, x = 4, y = −1, y = 1,

(b) the contour C is the circle |z| = 2.

SOLUTION

(a) Since both poles z = 1 and z = 3 lie within the square, we have from (5) that

We found these residues in Examples 2 and 3, and so

(b) Since only the pole z = 1 lies within the circle |z| = 2, we have from (5) that

≡

EXAMPLE 5 Evaluation by the Residue Theorem

Evaluate where the contour C is the circle |z − i| = 2.

SOLUTION

By writing z² + 4 = (z − 2i)(z + 2i), we see that the integrand has simple poles at −2i and 2i. Now since only 2i lies within the contour C, it follows from (5) that

But

Hence, ≡

EXAMPLE 6 Evaluation by the Residue Theorem

EXAMPLE 7 Evaluation by the Residue Theorem

Evaluate dz, where the contour C is the circle |z| = 2.

SOLUTION

The integrand tan z = sin z/cos z has simple poles at the points where cos z = 0. We saw in Section 17.7 that the only zeros for cos z are the real numbers z = (2n + 1)π/2, n = 0, ±1, ±2, . . . . Since only −π/2 and π/2 are within the circle |z| = 2, we have

Now from (4) with g(z) = sin z, h(z) = cos z, and h′(z) = −sin z, we see that

Therefore, ≡

EXAMPLE 8 Evaluation by the Residue Theorem

REMARKS

In the application of the limit formulas (1) and (2) for computing residues, the indeterminate form 0/0 may result. Although we are not going to prove it, it should be pointed out that L’Hôpital’s rule is valid in complex analysis. If f(z) = g(z)/h(z), where g and h are analytic at z = z₀, g(z₀) = 0, h(z₀) = 0, and h′(z₀) ≠ 0, then

19.5 Exercises Answers to selected odd-numbered problems begin on page ANS-48.

In Problems 1–6, use a Laurent series to find the indicated residue.

1. f(z) = ; Res (f(z), 1)
2. f(z) = ; Res (f(z), 0)
3. f(z) = ; Res (f(z), 0)
4. f(z) = (z + 3)² sin ; Res (f(z), −3)
5. f(z) = ; Res (f(z), 0)
6. f(z) = ; Res (f(z), 2)

In Problems 7–16, use (1), (2), or (4) to find the residue at each pole of the given function.

7. f(z) =
8. f(z) =
9. f(z) =
10. f(z) =
11. f(z) =
12. f(z) =
13. f(z) =
14. f(z) =
15. f(z) = sec z
16. f(z) =

In Problems 17–20, use Cauchy’s residue theorem, where appropriate, to evaluate the given integral along the indicated contours.

17. 

    (a) |z| =

    (b) |z| =

    (c) |z| = 3

18. 

    (a) |z| = 1

    (b) |z − 2i| = 1

    (c) |z − 2i| = 4

19. 

    (a) |z| = 5

    (b) |z + i| = 2

    (c) |z − 3| = 1

20. 

    (a) |z − 2i| = 1

    (b) |z − 2i| = 3

    (c) |z| = 5

In Problems 21–32, use Cauchy’s residue theorem to evaluate the given integral along the indicated contour.

21. C: |z − 3i| = 3
22. C: |z − 2| =
23. C: |z| = 2
24. C is the ellipse 16x² + y² = 4
25. C: |z| = 2
26. C: |z| = 3
27. C: |z − 1| = 2
28. C: |z| =
29. C is the rectangle defined by x = , x = π , y = −1, y = 1
30. C is the rectangle defined by x = −2, x = 1, y = −, y = 1
31. C: |z − 3| = 1
32. C: |z − 1| = 1