CHAPTER 5

Complex Integration

BY

Dr. Pulak Sahoo

Assistant Professor Department of University Of Kalyani West Bengal, India E-mail : [email protected]

1 Module-4: Winding Number

1 Introduction

We now prove an important result which is quite different from the theory of functions of a real variable.

Theorem 1. If f(z) is analytic in a domain D then all the derivatives of f(z) exists and are analytic in D.

Proof. Let z0 be an arbitrary in D and we consider the C having center at z0 and radius r, so small that C lies entirely within D (see Fig. 1). Then for n = 0, 1, 2,...,

Fig. 1: and for any point α inside C, we have Z (n) n! f(z) f (α) = n+1 dz. 2πi C (z − α)

Thus f(z) has a derivative of all order in a neighbourhood of z0. Since z0 is arbitrary, we can conclude that all the derivatives of f(z) exists and are analytic functions in D. This proves the theorem.

2 Note 1. In basic calculus we have learnt that the existence of the derivative of f(x) does not guarantee the continuity of the derivative f 0(x). We consider the

f(x) = x5/3

which has a first derivative for all x ∈ R and 5 f 0(x) = x2/3. 3

See that f 0(x) does not have a first derivative at x = 0 and therefore f(x) does not have a second derivative at the origin. Similarly, f(x) = x7/3 has a first and second derivative in R but has no third derivative at x = 0. Thus Theorem 1 does not hold in the case of functions of a real variable. This result gives an essential difference between functions of a real variable and func- tions of a complex variable.

Corollary 1. If f = u + iv is analytic in a domain D, then all partial derivatives of u and v exist and are continuous in D.

Proof. Let f = u + iv be analytic in D. Then

0 f (z) = ux + ivx = vy − iuy. f 0(z) being analytic in D, it is continuous in D and hence each of the first order partial derivatives of u and v exist and are continuous in D. Also, because of analyticity of f 0(z) in D, we obtain from above equation

00 f (z) = uxx + ivxx

= (vx)y − i(ux)y

= (vy)x − i(uy)x.

This process may be continued to conclude that u and v have continuous partial deriva- tives of all orders at each point where the function f = u + iv is analytic.

R log z Example 1. Evaluate C (z−1)3 dz, where C is the circle | z − 2 |= 3/2.

Solution. The center of the circle C : | z − 2 |= 3/2 is z = 2 and its radius is 3/2. The log z function f(z) = (z−1)3 is not analytic at the point z = 1 which lies inside C. Therefore,

3 from Cauchy’s formula for nth derivative (here n = 2) we obtain

Z 00 2! log z f (1) = 3 dz, where f(z) = log z 2πi C (z − 1) Z log z 00 i.e. 3 dz = πif (1) = −πi. C (z − 1)

We might expect a function f(z) to be analytic at a point α if

Z f(z)dz = 0 C along every simple closed rectifiable in which α is an interior point. Unfortunately, this converse of Cauchy’s fundamental theorem is not true. As for example,

Z 1 2 dz = 0 C z

along every simple closed rectifiable curve C having the origin as an interior point. This is

1 because f(z) = z2 is analytic in the region between C and some circle | z |= ε contained in C. Therefore by Cauchy’s theorem for multiply connected regions, we have

Z 1 Z 1 Z 2π iεeiθ 2 dz = 2 dz = 2 2iθ dθ C z |z|=ε z 0 ε e i Z 2π = e−iθdθ = 0. ε 0

But we do have a partial converse to Cauchy’s fundamental theorem even when the domain D is not simply connected.

Theorem 2. (Morera’s Theorem) R Let f(z) be continuous in a simply connected domain D. If Γ f(z)dz = 0 along every simple closed rectifiable curve Γ contained in D, then f(z) is analytic in D.

Proof. Let α be a fixed point and z be a variable point in D. Then the value of the R z integral α f(w)dw is independent of the path so long the path lies in D (see Fig. 2). Let Z z g(z) = f(w)dw. α

Let h be a such that z + h lies inside D. Then

Z z+h Z z Z z+h g(z + h) − g(z) = f(w)dw − f(w)dw = f(w)dw. α α z

4 Fig. 2:

Hence

g(z + h) − g(z) 1 Z z+h | − f(z) | = | f(w)dw − f(z) | h h z 1 Z z+h ≤ | f(w) − f(z) || dw | . (1) | h | z Since f(z) is continuous at z, given ε > 0 we can find a δ > 0 such that | f(w)−f(z) |< ε whenever | w − z |< δ. Choosing | h |< δ we see that | f(w) − f(z) |< ε, for all point w lying on the straight line joining z to z + h. Therefore using ML-formula we obtain from (1)

g(z + h) − g(z) 1 | − f(z) | ≤ · ε· | h | = ε. h | h |

Proceeding to the limit as h → 0 we get

g(z + h) − g(z) lim = f(z) h→0 h i.e. g0(z) = f(z).

This shows that g(z) is analytic in D. Thus f(z), being the derivative of an , is itself analytic in D. This proves the theorem.

Note 2. We note that even if f(z) is analytic in a domain D, we are not guaranteed that R C f(z)dz = 0 along every simple closed rectifiable curve C contained in D. The function

5 1 f(z) = z is analytic in the annulus bounded by the | z |= 1/3 and | z |= 3/2. But R |z|=1(1/z)dz = 2πi even though the circle is contained in the annulus. Note, however, that the interior of | z |= 1 is not contained in the annulus, so that Cauchy’s theorem is not applicable. For a simply connected domain, it is true that the integral around every simple closed contour in the domain is zero. In view of Morera’s theorem, we can say that a necessary and sufficient condition for a to be analytic in a simply connected domain is that the integral be independent of the path of integration.

2 Winding Number or Index of a Curve

Now we introduce the concept of winding number (or index) of a closed curve with

respect to a point. Suppose that γ is a simple closed curve in the complex C and a is a given point in C \ γ. Then there is a useful formula that measures how often γ winds around a. For example, if γ : = {z : z − a = reiθ, 0 ≤ θ ≤ 2kπ}, then γ encircles the point a k times in positive sense. Also, Z dz Z 2kπ ireiθ = iθ dθ = 2kπi γ z − a 0 re 1 Z dz i.e. = k. 2πi γ z − a From this we also see that if γ encircles the point a k times in negative sense, then 1 Z dz = −k. 2πi γ z − a

1 R dz In both the cases, either in positive sense or in negative sense, 2πi γ z−a is an . Now we give the analytic definition of winding number.

Definition 1. Let γ be a closed rectifiable curve and a be a point not on the curve γ. Then, the index or winding number n(γ, a) of γ with respect to the point a is given by the integral 1 Z 1 n(γ, a) = dz. 2πi γ z − a Geometrically speaking, the winding number counts the number of rounds of a curve around a point. It is a positive integer for positive oriented and a negative integer for curves with negative orientation. If the point a is enclosed by a simple closed rectifiable

6 curve, then n(γ, a) = 1 and, if a is outside the curve, then n(γ, a) = 0 (see Fig. 3). Figure 3 shows how the index number changes with respect to the orientation and nature

Fig. 3: of the curves. Here we see that, for first one n(γ, a) = 1, n(γ, b) = 0; for second one n(γ, a) = −1, n(γ, b) = 0 and for last n(γ, a) = 1, n(γ, b) = 0, n(γ, c) = 2. Properties of the Winding Number

(i) For every closed rectifiable curve γ in C and every a ∈ C \ γ, n(γ, a) is an integer. (ii) If γ is a closed rectifiable curve in C, then the mapping a → n(γ, a) is a continuous function of a at any a 6∈ γ.

(iii) If γ is the sum of two closed rectifiable curves γ1 and γ2 in a , then for every a 6∈ γ, we have

n(γ, a) = n(γ1, a) + n(γ2, a).

Example 2. Let C be a closed rectifiable curve which contains the origin inside it. Show that an 1 Z eaz = n+1 dz. n! 2πi C z Solution. Let α be an arbitrary point lying inside C. Then by Cauchy’s integral formula for n-th derivative we obtain Z (n) n! f(z) f (α) = n+1 dz. 2πi C (z − α) Since C contains the origin within it, we can take α = 0. Therefore we have Z (n) n! f(z) f (0) = n+1 dz. 2πi C z Letting f(z) = eaz we obtain from above Z az n n! e a = n+1 dz 2πi C z an 1 Z eaz i.e. = n+1 dz. n! 2πi C z

7 Example 3. If f is analytic within and on a simple closed rectifiable curve γ and α is a point interior to γ, prove that

1 Z f (n)(z) Z f(z) dz = n+1 dz. n! γ z − α γ (z − α)

Solution. Since f(z) is analytic within and on γ, f (n)(z) is also analytic within and on γ. Also α is an interior point of γ. Therefore using Cauchy’s integral formula for the function f (n)(z) we obtain

1 Z f (n)(z) f (n)(α) = dz. 2πi γ z − α Again Z (n) n! f(z) f (α) = n+1 dz. 2πi C (z − α) Comparing this two we obtain the required result.

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