INTRODUCTION

In the last two sections we saw the importance of the Cauchy–Goursat theorem in the evaluation of contour integrals. In this section we are going to examine several more consequences of the Cauchy–Goursat theorem. Unquestionably, the most significant of these is the following result:

After establishing this proposition we shall use it to further show that

The ramifications of these two results alone will keep us busy not only for the remainder of this section but in the next chapter as well.

First Formula

We begin with the Cauchy integral formula. The idea in the next theorem is this: If f is analytic in a simply connected domain and z₀ is any point D, then the quotient f(z)/(z − z₀) is not analytic in D. As a consequence, the integral of f(z)/(z − z₀) around a simple closed contour C that contains z₀ is not necessarily zero but has, as we shall now see, the value 2πi f(z₀). This remarkable result indicates that the values of an analytic function f at points inside a simple closed contour C are determined by the values of f on the contour C.

PROOF:

Let D be a simply connected domain, C a simple closed contour in D, and z₀ an interior point of C. In addition, let C₁ be a circle centered at z₀ with radius small enough that it is interior to C. By the principle of deformation of contours, we can write

(2)

We wish to show that the value of the integral on the right is 2πi f(z₀). To this end we add and subtract the constant f(z₀) in the numerator:

(3)

Now from (4) of Section 18.2 we know that

Thus, (3) becomes

(4)

Since f is continuous at z₀ for any arbitrarily small ϵ > 0, there exists a δ > 0 such that |f(z) − f(z₀)| < ϵ whenever |z − z₀| < δ. In particular, if we choose the circle C₁ to be |z − z₀| = δ/2 < δ, then by the ML-inequality (Theorem 18.1.3) the absolute value of the integral on the right side of (4) satisfies

In other words, the absolute value of the integral can be made arbitrarily small by taking the radius of the circle C₁ to be sufficiently small. This can happen only if the integral is zero. The Cauchy integral formula (1) follows from (4) by dividing both sides by 2πi. ≡

The Cauchy integral formula (1) can be used to evaluate contour integrals. Since we often work problems without a simply connected domain explicitly defined, a more practical restatement of Theorem 18.4.1 is

EXAMPLE 1 Using Cauchy’s Integral Formula

EXAMPLE 2 Using Cauchy’s Integral Formula

Evaluate dz, where C is the circle |z − 2i| = 4.

SOLUTION

By factoring the denominator as z² + 9 = (z − 3i)(z + 3i), we see that 3i is the only point within the closed contour at which the integrand fails to be analytic. See FIGURE 18.4.1.

Now by writing

,

we can identify f(z) = z/(z + 3i). This function is analytic at all points within and on the contour C. From the Cauchy integral formula we then have

≡

EXAMPLE 3 Flux and Cauchy’s Integral Formula

The complex function f(z) = k/(), where k = a + ib and z₁ are complex numbers, gives rise to a flow in the domain z ≠ z₁. If C is a simple closed contour containing z = z₁ in its interior, then from the Cauchy integral formula we have

Thus the circulation around C is 2πb and the net flux across C is 2πa. If z₁ were in the exterior of C, both the circulation and net flux would be zero by Cauchy’s theorem.

Note that when k is real, the circulation around C is zero but the net flux across C is 2πk. The complex number z₁ is called a source for the flow when k > 0 and a sink when k < 0. Vector fields corresponding to these two cases are shown in FIGURE 18.4.2(a) and 18.4.2(b). ≡

Second Formula

We can now use Theorem 18.4.1 to prove that an analytic function possesses derivatives of all orders; that is, if f is analytic at a point z₀, then f′, f″, f‴, and so on, are also analytic at z₀. Moreover, the values of the derivatives f⁽ⁿ⁾(z₀), n = 1, 2, 3, … , are given by a formula similar to (1).

PARTIAL PROOF:

We will prove (6) only for the case n = 1. The remainder of the proof can be completed using the principle of mathematical induction.

We begin with the definition of the derivative and (1):

Before proceeding, let us set up some preliminaries. Since f is continuous on C, it is bounded; that is, there exists a real number M such that |f(z)| ≤ M for all points z on C. In addition, let L be the length of C and let δ denote the shortest distance between points on C and the point z₀. Thus for all points z on C, we have

Furthermore, if we choose |Δz| ≤ δ/2, then

Now,

Because the last expression approaches zero as Δz → 0, we have shown that

≡

If f(z) = u(x, y) + iv(x, y) is analytic at a point, then its derivatives of all orders exist at that point and are continuous. Consequently, from

we can conclude that the real functions u and v have continuous partial derivatives of all orders at a point of analyticity.

Like (1), (6) can sometimes be used to evaluate integrals.

EXAMPLE 4 Using Cauchy’s Integral Formula for Derivatives

Evaluate dz, where C is the circle |z| = 1.

SOLUTION

Inspection of the integrand shows that it is not analytic at z = 0 and z = −4, but only z = 0 lies within the closed contour. By writing the integrand as

we can identify z₀ = 0, n = 2, and f(z) = (z + 1)/(z + 4). By the Quotient Rule, f″(z) = −6/(z + 4)³ and so by (6) we have

≡

EXAMPLE 5 Using Cauchy’s Integral Formula for Derivatives

Evaluate dz, where C is the contour shown in FIGURE 18.4.3.

SOLUTION

Although C is not a simple closed contour, we can think of it as the union of two simple closed contours C₁ and C₂ as indicated in Figure 18.4.3. By writing

we are in a position to use both (1) and (6).

To evaluate I₁, we identify z₀ = 0 and f(z) = (z³ + 3)/(z − i)². By (1) it follows that

To evaluate I₂ we identify z₀ = i, n = 1, f(z) = (z³ + 3)/z, and f′(z) = (2z³ − 3)/z². From (6) we obtain

Finally we get

≡

Liouville’s Theorem

If we take the contour C to be the circle |z − z₀| = r, it follows from (6) and the ML-inequality that

(7)

where M is a real number such that |f(z)| ≤ M for all points z on C. The result in (7), called Cauchy’s inequality, is used to prove the next result.

PROOF:

Suppose f is an entire function and is bounded; that is, |f(z)| ≤ M for all z. Then for any point z₀, (7) gives |f′(z₀)| ≤ M/r. By making r arbitrarily large, we can make |f′(z₀)| as small as we wish. This means f′(z₀) = 0 for all points z₀ in the complex plane. Hence f must be a constant. ≡

Fundamental Theorem of Algebra

Liouville’s theorem enables us to prove, in turn, a result that is learned in elementary algebra:

This result is known as the Fundamental Theorem of Algebra. To prove it, let us suppose that P(z) ≠ 0 for all z. This implies that the reciprocal of P, f(z) = 1/P(z), is an entire function. Now since |f(z)| → 0 as |z| → ∞, the function f must be bounded for all finite z. It follows from Liouville’s theorem that f is a constant and therefore P is a constant. But this is a contradiction to our underlying assumption that P was not a constant polynomial. We conclude that there must exist at least one number z for which P(z) = 0.

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

In Problems 1–24, use Theorems 18.4.1 and 18.4.2, when appropriate, to evaluate the given integral along the indicated closed contour(s).

1. 
2. 
3. 
4. 
5. 
6. 
7. (a) (b)
8. (a) (b)
9. 
10. 
11. 
12. 
13. 
14. 
15. (a) (b) (c) (d)
16. (a) (b) (c) (d)
17. (a) (b)
18. (a) (b)
19. 
20. 
21. 
22. 
23. is given in FIGURE 18.4.4

    

24. C is given in FIGURE 18.4.5