THEOREM 1 Laurent's Theorem

Let f(z) be analytic in a domain containing two concentric circles C₁ and C₂ with center z₀ and the annulus between them (blue in Fig. 370). Then f(z) can be represented by the Laurent series

consisting of nonnegative and negative powers. The coefficients of this Laurent series are given by the integrals

taken counterclockwise around any simple closed path C that lies in the annulus and encircles the inner circle, as in Fig. 370. [The variable of integration is denoted by z* since z is used in (1).]

This series converges and represents f(z) in the enlarged open annulus obtained from the given annulus by continuously increasing the outer circle C₁ and decreasing C₂ until each of the two circles reaches a point where f(z) is singular.

In the important special case that z₀ is the only singular point of f(z) inside C₂, this circle can be shrunk to the point z₀, giving convergence in a disk except at the center. In this case the series (or finite sum) of the negative powers of (1) is called the principal part of f(z) at z₀ [or of that Laurent series (1)].

PROOF

(a) The nonnegative powers are those of a Taylor series.

To see this, we use Cauchy's integral formula (3) in Sec. 14.3 with z* (instead of z) as the variable of integration and z instead of z₀. Let g(z) and h(z) denote the functions represented by the two terms in (3), Sec. 14.3. Then

Here z is any point in the given annulus and we integrate counterclockwise over both C₁ and C₂, so that the minus sign appears since in (3) of Sec. 14.3 the integration over C₂ is taken clockwise. We transform each of these two integrals as in Sec. 15.4. The first integral is precisely as in Sec. 15.4. Hence we get exactly the same result, namely, the Taylor series of g(z),

with coefficients [see (2), Sec. 15.4, counterclockwise integration]

Here we can replace C₁ by C (see Fig. 370), by the principle of deformation of path, since z₀, the point where the integrand in (5) is not analytic, is not a point of the annulus. This proves the formula for the a_n in (2).

(b) The negative powers in (1) and the formula for b_n in (2) are obtained if we consider h(z). It consists of the second integral times −1/(2πi) in (3). Since z lies in the annulus, it lies in the exterior of the path C₂. Hence the situation differs from that for the first integral. The essential point is that instead of [see in Sec. 15.4]

Consequently, we must develop the expression 1/(z* − z) in the integrand of the second integral in (3) in powers of (z* − z₀)/(z − z₀) (instead of the reciprocal of this) to get a convergent series. We find

Compare this for a moment with (7) in Sec. 15.4, to really understand the difference. Then go on and apply formula (8), Sec. 15.4, for a finite geometric sum, obtaining

Multiplication by −f(z*)/2πi and integration over C₂ on both sides now yield

with the last term on the right given by

As before, we can integrate over C instead of C₂ in the integrals on the right. We see that on the right, the power 1/(z − z₀)ⁿ is multiplied by b_n as given in (2). This establishes Laurent's theorem, provided

(c) Convergence proof of (8). Very often (1) will have only finitely many negative powers. Then there is nothing to be proved. Otherwise, we begin by noting that f(z*)/(z − z*) in (7) is bounded in absolute value, say,

because f(z*) is analytic in the annulus and on C₂, and z* lies on C₂ and z outside, so that z − z* ≠ 0. From this and the ML-inequality (Sec. 14.1) applied to (7) we get the inequality (L = 2πr₂ = length of C₂, r₂ = |z* −z₀| = radius of C₂ = const)

From (6b) we see that the expression on the right approaches zero as n approaches infinity. This proves (8). The representation (1) with coefficients (2) is now established in the given annulus.

(d) Convergence of (1) in the enlarged annulus. The first series in (1) is a Taylor series [representing g(z)]; hence it converges in the disk D with center z₀ whose radius equals the distance of the singularity (or singularities) closest to z₀. Also, g(z) must be singular at all points outside C₁ where f(z) is singular.

The second series in (1), representing h(z), is a power series in Z = 1/(z − z₀). Let the given annulus be r₂ < |z − z₀| < r₁, where r₁ and r₂ are the radii of C₁ and C₂, respectively (Fig. 370). This corresponds to 1/r₂ > |Z| > 1/r₁. Hence this power series in Z must converge at least in the disk |Z| < 1/r₂. This corresponds to the exterior |z − z₀| > r₂ of C₂, so that h(z) is analytic for all z outside C₂. Also, h(z) must be singular inside C₂ where f(z) is singular, and the series of the negative powers of (1) converges for all z in the exterior E of the circle with center z₀ and radius equal to the maximum distance from z₀ to the singularities of f(z) inside C₂. The domain common to D and E is the enlarged open annulus characterized near the end of Laurent's theorem, whose proof is now complete.

EXAMPLE 1 Use of Maclaurin Series

Find the Laurent series of z⁻⁵ sin z with center 0.

Solution. By (14), Sec. 15.4, we obtain

Here the “annulus” of convergence is the whole complex plane without the origin and the principal part of the series at 0 is

EXAMPLE 2 Substitution

Find the Laurent series of z²e^1/z with center 0.

Solution. From (12) in Sec. 15.4 with z replaced by 1/z we obtain a Laurent series whose principal part is an infinite series,

EXAMPLE 3 Development of 1/(1 − z)

Develop 1/(1 − z) (a) in nonnegative powers of z, (b) in negative powers of z.

Solution.

EXAMPLE 4 Laurent Expansions in Different Concentric Annuli

Find all Laurent series of 1/(z³ − z⁴) with center 0.

Solution. Multiplying by 1/z³, we get from Example 3

EXAMPLE 5 Use of Partial Fractions

Find all Taylor and Laurent series of with center 0.

Solution. In terms of partial fractions,

(a) and (b) in Example 3 take care of the first fraction. For the second fraction,

(I) From (a) and (c), valid for |z| < 1 (see Fig. 371),

Fig. 371. Regions of convergence in Example 5

(II) From (c) and (b), valid for 1 < |z| < 2

(III) From (d) and (b), valid for |z| > 2,

PROBLEM SET 16.1

1–8 LAURENT SERIES NEAR A SINGULARITY AT 0

Expand the function in a Laurent series that converges for 0 < |z| < R and determine the precise region of convergence. Show the details of your work.

9–16 LAURENT SERIES NEAR A SINGULARITY AT z₀

Find the Laurent series that converges for 0 < |z − z₀| < R and determine the precise region of convergence. Show details.

• 9.
• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16.
• 17. CAS PROJECT. Partial Fractions. Write a program for obtaining Laurent series by the use of partial fractions. Using the program, verify the calculations in Example 5 of the text. Apply the program to two other functions of your choice.
• 18. TEAM PROJECT. Laurent Series. (a) Uniqueness. Prove that the Laurent expansion of a given analytic function in a given annulus is unique.

  (b) Accumulation of singularities. Does tan (1/z) have a Laurent series that converges in a region 0 < |z| < R? (Give a reason.)

  (c) Integrals. Expand the following functions in a Laurent series that converges for |z| > 0:

19–25 TAYLOR AND LAURENT SERIES

Find all Taylor and Laurent series with center z₀. Determine the precise regions of convergence. Show details.

• 19.
• 20.
• 21.
• 22.
• 23.
• 24.
• 25.

EXAMPLE 1 Poles. Essential Singularities

The function

has a simple pole at z = 0 and a pole of fifth order at z = 2. Examples of functions having an isolated essential singularity at z = 0 are

and

Section 16.1 provides further examples. In that section, Example 1 shows that z⁻⁵ sin z has a fourth-order pole at 0. Furthermore, Example 4 shows that 1/(z³ − z⁴) has a third-order pole at 0 and a Laurent series with infinitely many negative powers. This is no contradiction, since this series is valid for |z| > 1; it merely tells us that in classifying singularities it is quite important to consider the Laurent series valid in the immediate neighborhood of a singular point. In Example 4 this is the series (I), which has three negative powers.

EXAMPLE 2 Behavior Near a Pole

f(z) = 1/z² has a pole at z = 0, and |f(z)| → ∞ as z → 0 in any manner. This illustrates the following theorem.

THEOREM 1 Poles

If f(z) is analytic and has a pole at z = z₀, then |f(z)| → ∞ as z → z₀ in any manner.

EXAMPLE 3 Behavior Near an Essential Singularity

The function f(z) = e^1/z has an essential singularity at z = 0. It has no limit for approach along the imaginary axis; it becomes infinite if z → 0 through positive real values, but it approaches zero if z → 0 through negative real values. It takes on any given value c = c₀e^i∞ ≠ 0 in an arbitrarily small -neighborhood of z = 0. To see the latter, we set z = re^iθ, and then obtain the following complex equation for r and θ, which we must solve:

Equating the absolute values and the arguments, we have e^{(cos θ)/r} = c₀, that is

respectively. From these two equations and cos² θ + sin²θ = r²(ln c₀)² + α²r² = 1 we obtain the formulas

Hence r can be made arbitrarily small by adding multiples of 2π to α, leaving c unaltered. This illustrates the very famous Picard's theorem (with z = 0 as the exceptional value).

THEOREM 2 Picard's Theorem

If f(z) is analytic and has an isolated essential singularity at a point z₀, it takes on every value, with at most one exceptional value, in an arbitrarily small -neighborhood of z₀.

EXAMPLE 4 Zeros

The function 1 + z² has simple zeros at ±i. The function (1 − z⁴)² has second-order zeros at ±1 and ±i. The function (z − a)³ has a third-order zero at z = a. The function e^z has no zeros (see Sec. 13.5). The function sin z has simple zeros at 0, ±π, ±2π, …, and sin²z has second-order zeros at these points. The function 1 − cos z has second-order zeros at 0, ±2π, ±4π, …, and the function (1 − cos z)² has fourth-order zeros at these points.

THEOREM 3 Zeros

The zeros of an analytic function f(z)( 0) are isolated; that is, each of them has a neighborhood that contains no further zeros of f(z)

PROOF

The factor (z − z₀)ⁿ in (3) is zero only at z = z₀. The power series in the brackets […] represents an analytic function (by Theorem 5 in Sec. 15.3), call it g(z). Now g(z₀) = a_n ≠ 0, since an analytic function is continuous, and because of this continuity, also g(z) ≠ 0 in some neighborhood of z = z₀. Hence the same holds of f(z).

THEOREM 4 Poles and Zeros

Let f(z) be analytic at z = z₀ and have a zero of nth order at z = z₀. Then 1/f(z) has a pole of nth order at z = z₀; and so does h(z)/f(z), provided h(z) is analytic at z = z₀ and h(z₀) ≠ 0.

EXAMPLE 5 Functions Analytic or Singular at Infinity. Entire and Meromorphic Functions

The function f(z) = 1/z² is analytic at ∞ since g(w) = f(w) = w² is analytic at w = 0, and f(z) has a second-order zero at ∞. The function f(z) = z³ is singular at ∞ and has a third-order pole there since the function g(w) = f(1/w) = 1/w³ has such a pole at w = 0. The function e^z has an essential singularity at ∞ since e^1/w has such a singularity at w = 0. Similarly, cos z and sin z have an essential singularity at ∞.

Recall that an entire function is one that is analytic everywhere in the (finite) complex plane. Liouville's theorem (Sec. 14.4) tells us that the only bounded entire functions are the constants, hence any nonconstant entire function must be unbounded. Hence it has a singularity at ∞, a pole if it is a polynomial or an essential singularity if it is not. The functions just considered are typical in this respect.

An analytic function whose only singularities in the finite plane are poles is called a meromorphic function. Examples are rational functions with nonconstant denominator, tan z, cot z, sec z, and csc z.

PROBLEM SET 16.2

1–10 ZEROS

Determine the location and order of the zeros.

1. 
2. (z⁴ − 81)³
3. (z + 81i)⁴
4. tan²2z
5. z⁻²sin²πz
6. cosh⁴z
7. z⁴ + (1 − 8i)z² − 8i
8. (sin z − 1)³
9. sin 2z cos 2z
10. (z² − 8)³(exp(z²) − 1)
11. Zeros. If f(z) is analytic and has a zero of order n at z = z₀, show that f²(z) has a zero of order 2n at z₀.
12. TEAM PROJECT. Zeros. (a) Derivative. Show that if f(z) has a zero of order n > 1 at z = z₀, then f′(z) has a zero of order n − 1 at z₀.
    • (b) Poles and zeros. Prove Theorem 4.
    • (c) Isolated k-points. Show that the points at which a nonconstant analytic function f(z) has a given value k are isolated.
    • (d) Identical functions. If f₁(z) and f₂(z) are analytic in a domain D and equal at a sequence of points z_n in D that converges in D, show that f₁(z) ≡ f₂(z) in D.

13–22 SINGULARITIES

Determine the location of the singularities, including those at infinity. For poles also state the order. Give reasons.

• 13.
• 14.
• 15. z exp (1/z − 1 − i)²)
• 16. tan πz
• 17. cot²z
• 18. z³ exp
• 19. 1/(e^z − e^2z)
• 20. 1/(cos z − sin z)
• 21. e^1/(z−1)/(e^z − 1)
• 22. (z − π)⁻¹ sin z
• 23. Essential singularity. Discuss e^1/z2 in a similar way as e^1/z is discussed in Example 3 of the text.
• 24. Poles. Verify Theorem 1 for f(z) = z⁻³ − z⁻¹. Prove Theorem 1.
• 25. Riemann sphere. Assuming that we let the image of the x-axis be the meridians 0° and 180°, describe and sketch (or graph) the images of the following regions on the Riemann sphere: (a) |z| > 100, (b) the lower half-plane, (c)

EXAMPLE 1 Evaluation of an Integral by Means of a Residue

Integrate the function f(z) = z⁻⁴ sin z counterclockwise around the unit circle C.

Solution. From (14) in Sec. 15.4 we obtain the Laurent series

which converges for |z| > 0 (that is, for all z ≠ 0). This series shows that f(z) has a pole of third order at z = 0 and the residue From (1) we thus obtain the answer

EXAMPLE 2 CAUTION! Use the Right Laurent Series!

Integrate f(z) = 1/(z³ − z⁴) clockwise around the circle C:

Solution. z³ − z⁴ = z³(1 − z) shows that f(z) is singular at z = 0 and z = 1. Now z = 1 lies outside C. Hence it is of no interest here. So we need the residue of f(z) at 0. We find it from the Laurent series that converges for 0 < |z| < 1. This is series (I) in Example 4, Sec. 16.1,

We see from it that this residue is 1. Clockwise integration thus yields

we would have obtained the wrong answer, 0, because this series has no power 1/z.

PROOF

We prove (3). For a simple pole at z = z₀ the Laurent series (1), Sec. 16.1,

Here b₁ ≠ 0. (Why?) Multiplying both sides by z − z₀ and then letting z → z₀, we obtain the formula (3):

where the last equality follows from continuity (Theorem 1, Sec. 15.3).

We prove (4). The Taylor series of q(z) at a simple zero z₀ is

Substituting this into f = p/q and then f into (3) gives

z − z₀ cancels. By continuity, the limit of the denominator is q′(z₀) and (4) follows.

EXAMPLE 3 Residue at a Simple Pole

f(z) = (9z + i)/(z³ + z) has a simple pole at i because z² + 1 = (z + i)(z − i), and (3) gives the residue

By (4) with p(i) = 9i + i and q′(z) = 3z² + 1 we confirm the result,

PROOF

We prove (5). The Laurent series of f(z) converging near z₀ (except at z₀ itself) is (Sec. 16.2)

where b_m ≠ 0. The residue wanted is b₁. Multiplying both sides by (z − z₀)^m gives

We see that b₁ is now the coefficient of the power (z − z₀)^m−1 of the power series of g(z) = (z − z₀)^mf(z). Hence Taylor's theorem (Sec. 15.4) gives (5):

EXAMPLE 4 Residue at a Pole of Higher Order

f(z) = 50z/(z³ + 2z² − 7z + 4) has a pole of second order at z = 1 because the denominator equals (z + 4)(z − 1)² (verify!). From (5*) we obtain the residue

THEOREM 1 Residue Theorem

Let f(z) be analytic inside a simple closed path C and on C, except for finitely many singular points z₁, z₂, …, z_k inside C. Then the integral of f(z) taken counterclockwise around C equals 2πi times the sum of the residues of f(z) at z₁, …, z_k:

PROOF

We enclose each of the singular points z_j in a circle C_j with radius small enough that those k circles and C are all separated (Fig. 373 where k = 3). Then f(z) is analytic in the multiply connected domain D bounded by C and C₁, …, C_k and on the entire boundary of D. From Cauchy's integral theorem we thus have

the integral along C being taken counterclockwise and the other integrals clockwise (as in Figs. 354 and 355, Sec. 14.2). We take the integrals over C₁, …, C_k to the right and compensate the resulting minus sign by reversing the sense of integration. Thus,

where all the integrals are now taken counterclockwise. By (1) and (2),

so that (8) gives (6) and the residue theorem is proved.

EXAMPLE 5 Integration by the Residue Theorem. Several Contours

Evaluate the following integral counterclockwise around any simple closed path such that (a) 0 and 1 are inside C, (b) 0 is inside, 1 outside, (c) 1 is inside, 0 outside, (d) 0 and 1 are outside.

Solution. The integrand has simple poles at 0 and 1, with residues [by (3)]

[Confirm this by (4).] Answer: (a) 2πi(−4 + 1) = −6πi, (b) −8πi, (c) 2πi, (d) 0.

EXAMPLE 6 Another Application of the Residue Theorem

Integrate (tan z)/(z² − 1) counterclockwise around the circle C: .

Solution. tan z is not analytic at ±π/2, ±3π/2, …, but all these points lie outside the contour C. Because of the denominator z² − 1 = (z − 1)(z + 1) the given function has simple poles at ±1. We thus obtain from (4) and the residue theorem

EXAMPLE 7 Poles and Essential Singularities

Evaluate the following integral, where C is the ellipse 9x² + y² = 9 (counterclockwise, sketch it).

Solution. Since z⁴ − 16 = 0 at ±2i and ±2, the first term of the integrand has simple poles at ±2i inside C, with residues [by (4); note that e^2πi = 1]

and simple poles at ±2, which lie outside C, so that they are of no interest here. The second term of the integrand has an essential singularity at 0, with residue π²/2 as obtained from

Answer: by the residue theorem.

PROBLEM SET 16.3

1. Verify the calculations in Example 3 and find the other residues.
2. Verify the calculations in Example 4 and find the other residue.

3–12 RESIDUES

Find all the singularities in the finite plane and the corresponding residues. Show the details.

• 3.
• 4.
• 5.
• 6. tan z
• 7. cot πz
• 8.
• 9.
• 10.
• 11.
• 12. e^1/(1−z)
• 13. CAS PROJECT. Residue at a Pole. Write a program for calculating the residue at a pole of any order in the finite plane. Use it for solving Probs. 5–10.

14–25 RESIDUE INTEGRATION

Evaluate (counterclockwise). Show the details.

• 14. C: |z − 2 − i| = 3.2
• 15. C: |z − 0.2| = 0.2
• 16. the unit circle
• 17. C: |z − πi/2| = 4.5
• 18. C: |z − 1| = 2
• 19. C: |z − 2i| = 2
• 20. C: |z − i| = 3
• 21.
• 22. C the unit circle
• 23. C the unit circle
• 24. C: |z| = 1.5
• 25. |z| = π

EXAMPLE 1 An Integral of the Type (1)

Show by the present method that

Solution. We use cos θ = (z + 1/z) and dθ = dz/iz. Then the integral becomes

We see that the integrand has a simple pole at outside the unit circle C, so that it is of no interest here, and another simple pole at (where ) inside C with residue [by (3), Sec. 16.3]

Answer: (Here −2/i is the factor in front of the last integral.)

EXAMPLE 2 An Improper Integral from 0 to ∞

Using (7), show that

Fig. 375. Example 2

Solution. Indeed, f(z) = 1/(1 + z⁴) has four simple poles at the points (make a sketch)

The first two of these poles lie in the upper half-plane (Fig. 375). From (4) in the last section we find the residues

(Here we used e^πi = −1 and e^−2πi = 1.) By (1) in Sec. 13.6 and (7) in this section,

Since 1/(1 + x⁴) is an even function, we thus obtain, as asserted,

EXAMPLE 3 An Application of (10)

Show that

Solution. In fact, e^isz/(k² + z²) has only one pole in the upper half-plane, namely, a simple pole at z = ik, and from (4) in Sec. 16.3 we obtain

Thus

Since e^isz = cos sx + i sin sx, this yields the above results [see also (15) in Sec. 11.7.]

THEOREM 1 Simple Poles on the Real Axis

If f(z) has a simple pole at z = a on the real axis, then (Fig. 376)

PROOF

By the definition of a simple pole (Sec. 16.2) the integrand f(z) has for 0 < |z − a| < R the Laurent series

Here g(z) is analytic on the semicircle of integration (Fig. 376)

and for all z between C₂ and the x-axis, and thus bounded on C₂, say, |g(z)| M. By integration,

The second integral on the right cannot exceed Mπr in absolute value, by the ML-inequality (Sec. 14.1), and ML = Mπr → 0 as r → 0.

EXAMPLE 4 Poles on the Real Axis

Find the principal value

Solution. Since

the integrand f(x), considered for complex z, has simple poles at

and at z = −i in the lower half-plane, which is of no interest here. From (14) we get the answer

PROBLEM SET 16.4

1–9 INTEGRALS INVOLVING COSINE AND SINE

Evaluate the following integrals and show the details of your work.

10–22 IMPROPER INTEGRALS: INFINITE INTERVAL OF INTEGRATION

Evaluate the following integrals and show details of your work.

• 10.
• 11.
• 12.
• 13.
• 14.
• 15.
• 16.
• 17.
• 18.
• 19.
• 20.
• 21.
• 22.

23–28 IMPROPER INTEGRALS: POLES ON THE REAL AXIS

Find the Cauchy principal value (showing details):

• 23.
• 24.
• 25.
• 26.
• 27. CAS EXPERIMENT. Simple Poles on the Real Axis. Experiment with integrals f(x) dx, f(x) = [(x − a₁)(x − a₂) … (x − a_k)]⁻¹, a_j real and all different, k > 1. Conjecture that the principal value of these integrals is 0. Try to prove this for a special k, say, k = 3. For general k.
• 28. TEAM PROJECT. Comments on Real Integrals.

  (a) Formula (10) follows from (9). Give the details.

  (b) Use of auxiliary results. Integrating e^−z2 around the boundary C of the rectangle with vertices −a, a, a + ib, −a + ib, letting a − ∞, and using

  

  show that

  

  (This integral is needed in heat conduction in Sec. 12.7.)

  (c) Inspection. Solve Probs. 13 and 17 without calculation.