INTRODUCTION

In this section we shall concentrate on contour integrals where the contour C is a simple closed curve with a positive (counterclockwise) orientation. Specifically, we shall see that when f is analytic in a special kind of domain D, the value of the contour integral f(z) dz is the same for any simple closed curve C that lies entirely within D. This theorem, called the Cauchy–Goursat theorem, is one of the fundamental results in complex analysis. Preliminary to discussing the Cauchy–Goursat theorem and some of its ramifications, we first need to distinguish two kinds of domains in the complex plane: simply connected and multiply connected.

Simply and Multiply Connected Domains

In the discussion that follows, we shall concentrate on contour integrals where the contour C is a simple closed curve with a positive (counterclockwise) orientation. Before doing this, we need to distinguish two kinds of domains. A domain D is said to be simply connected if every simple closed contour C lying entirely in D can be shrunk to a point without leaving D. In other words, in a simply connected domain, every simple closed contour C lying entirely within it encloses only points of the domain D. Expressed yet another way, a simply connected domain has no “holes” in it. The entire complex plane is an example of a simply connected domain. A domain that is not simply connected is called a multiply connected domain; that is, a multiply connected domain has “holes” in it. See FIGURE 18.2.1. As in Section 9.9, we call a domain with one “hole” doubly connected, a domain with two “holes” triply connected, and so on.

Cauchy’s Theorem

In 1825, the French mathematician Louis-Augustin Cauchy proved one of the most important theorems in complex analysis. Cauchy’s theorem says:

The proof of this theorem is an immediate consequence of Green’s theorem and the Cauchy–Riemann equations. Since f′ is continuous throughout D, the real and imaginary parts of f(z) = u + iv and their first partial derivatives are continuous throughout D. By (2) of Section 18.1 we write f(z) dz in terms of real-line integrals and use Green’s theorem on each line integral:

(1)

Now since f is analytic, the Cauchy–Riemann equations, ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x, imply that the integrands in (1) are identically zero. Hence, we have f(z) dz = 0.

In 1883, the French mathematician Edouard Goursat (1858–1936) proved Cauchy’s theorem without the assumption of continuity of f′ . The resulting modified version of Cauchy’s theorem is known as the Cauchy–Goursat theorem.

Since the interior of a simple closed contour is a simply connected domain, the Cauchy–Goursat theorem can be stated in the slightly more practical manner:

EXAMPLE 1 Applying the Cauchy–Goursat Theorem

Evaluate e^z dz, where C is the curve shown in FIGURE 18.2.2.

SOLUTION

The function f(z) = e^z is entire and C is a simple closed contour. It follows from the form of the Cauchy–Goursat theorem given in (2) that . ≡

EXAMPLE 2 Applying the Cauchy–Goursat Theorem

EXAMPLE 3 Applying the Cauchy–Goursat Theorem

Cauchy–Goursat Theorem for Multiply Connected Domains

If f is analytic in a multiply connected domain D, then we cannot conclude that f(z) dz = 0 for every simple closed contour C in D. To begin, suppose D is a doubly connected domain and C and C₁ are simple closed contours such that C₁ surrounds the “hole” in the domain and is interior to C. See FIGURE 18.2.4(a). Suppose, also, that f is analytic on each contour and at each point interior to C but exterior to C₁. When we introduce the cut AB shown in Figure 18.2.4(b), the region bounded by the curves is simply connected. Now the integral from A to B has the opposite value of the integral from B to A, and so from (2) we have f(z) dz + f(z) dz = 0 or

(3)

The last result is sometimes called the principle of deformation of contours, since we can think of the contour C₁ as a continuous deformation of the contour C. Under this deformation of contours, the value of the integral does not change. Thus, on a practical level, (3) allows us to evaluate an integral over a complicated simple closed contour by replacing that contour with one that is more convenient.

EXAMPLE 4 Applying Deformation of Contours

Evaluate , where C is the outer contour shown in FIGURE 18.2.5.

SOLUTION

In view of (3), we choose the more convenient circular contour C₁ in the figure. By taking the radius of the circle to be r = 1, we are guaranteed that C₁ lies within C. In other words, C₁ is the circle |z − i| = 1, which can be parameterized by x = cos t, y = 1 + sin t, 0 ≤ t ≤ 2π, or equivalently by z = i + e^it, 0 ≤ t ≤ 2π. From z − i = e^it and dz = ie^it dt we obtain

≡

The result in Example 4 can be generalized. Using the principle of deformation of contours (3) and proceeding as in the example, we can show that if z₀ is any constant complex number interior to any simple closed contour C, then

(4)

The fact that the integral in (4) is zero when n is an integer ≠ 1 follows only partially from the Cauchy–Goursat theorem. When n is zero or a negative integer, 1/(z − z₀)ⁿ is a polynomial (for example, n = −3, 1/(z − z₀)⁻³ = (z − z₀)³) and therefore entire. Theorem 18.2.1 then implies dz/(z − z₀)ⁿ = 0. It is left as an exercise to show that the integral is still zero when n is a positive integer different from one. See Problem 22 in Exercises 18.2.

EXAMPLE 5 Applying Formula (4)

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

SOLUTION

Since the denominator factors as z² + 2z − 3 = (z − 1)(z + 3), the integrand fails to be analytic at z = 1 and z = −3. Of these two points, only z = 1 lies within the contour C, which is a circle centered at z = 2 of radius r = 2. Now by partial fractions,

and so (5)

In view of the result given in (4), the first integral in (5) has the value 2πi. By the Cauchy–Goursat theorem, the value of the second integral is zero. Hence, (5) becomes

≡

If C, C₁, and C₂ are the simple closed contours shown in FIGURE 18.2.6 and if f is analytic on each of the three contours as well as at each point interior to C but exterior to both C₁ and C₂, then by introducing cuts, we get from Theorem 18.2.1 that f(z) dz + f(z) dz + f(z) dz = 0. Hence,

The next theorem summarizes the general result for a multiply connected domain with n “holes”:

EXAMPLE 6 Applying Theorem 18.2.2

Evaluate where C is the circle |z| = 3.

SOLUTION

In this case the denominator of the integrand factors as z² + 1 = (z − i)(z + i). Consequently, the integrand 1/(z² + 1) is not analytic at z = i and z = −i. Both of these points lie within the contour C. Using partial fraction decomposition once more, we have

and

We now surround the points z = i and z = −i by circular contours C₁ and C₂, respectively, that lie entirely within C. Specifically, the choice |z − i| = for C₁ and |z + i| = for C₂ will suffice. See FIGURE 18.2.7. From Theorem 18.2.2 we can then write

(7)

Because 1/(z + i) is analytic on C₁ and at each point in its interior and because 1/(z − i) is analytic on C₂ and at each point in its interior, it follows from (4) that the second and third integrals in (7) are zero. Moreover, it follows from (4), with n = 1, that

Thus (7) becomes ≡

REMARKS

Throughout the foregoing discussion we assumed that C was a simple closed contour; in other words, C did not intersect itself. Although we shall not give the proof, it can be shown that the Cauchy–Goursat theorem is valid for any closed contour C in a simply connected domain D. As shown in FIGURE 18.2.8, the contour C is closed but not simple. Nevertheless, if f is analytic in D, then f(z) dz = 0.

18.2 Exercises Answers to selected odd-numbered problems begin on page ANS-46.

In Problems 1–8, prove that f(z) dz = 0, where f is the given function and C is the unit circle |z| = 1.

1. f(z) = z³ − 1 + 3i
2. f(z) = z² +
3. f(z) =
4. f(z) =
5. f(z) =
6. f(z) =
7. f(z) = tan z
8. f(z) =
9. Evaluate where C is the contour shown in FIGURE 18.2.9.

   

10. Evaluate where C is the contour shown in FIGURE 18.2.10.

    

In Problems 11–20, use any of the results in this section to evaluate the given integral along the indicated closed contour(s).

11. 
12. 
13. 
14. 
15. (a) |z| = , (b) |z| = 2, (c) |z − 3i| = 1
16. (a) |z| = 1, (b) |z − 2i| = 1, (c) |z| = 4
17. (a) |z − 5| = 2, (b) |z| = 9
18. (a) |z| = 5, (b) |z − 2i| =
19. 
20. 
21. Evaluate where C is the closed contour shown in FIGURE 18.2.11. [Hint: Express C as the union of two closed curves C₁ and C₂.]

    

22. Suppose z₀ is any constant complex number interior to any simple closed contour C. Show that

    

In Problems 23 and 24, evaluate the given integral by any means.

23. C is the unit circle |z| = 1
24. (z³ + z² + Re(z)) dz, C is the triangle with vertices z = 0, z = 1 + 2i, z = 1