18.1 Contour Integrals

INTRODUCTION

In Section 9.8 we saw that the definition of the definite integral starts with a real function y = f(x) that is defined on an interval [a, b] on the x-axis. Because a planar curve is the two-dimensional analogue of an interval, we then generalized the definition of the definite integral to integrals of real functions of two variables defined on a curve C in the Cartesian plane. We shall see in this section that a complex integral is defined in a manner that is quite similar to that of a line integral in the Cartesian plane. In case you have not studied Sections 9.8 and 9.9, a review of those sections is recommended.

A Definition

Integration in the complex plane is defined in a manner similar to that of a line integral in the plane. In other words, we shall be dealing with an integral of a complex function f(z) that is defined along a curve C in the complex plane. These curves are defined in terms of parametric equations x = x(t), y = y(t), a ≤ t ≤ b, where t is a real parameter. By using x(t) and y(t) as real and imaginary parts, we can also describe a curve C in the complex plane by means of a complex-valued function of a real variable t: z(t) = x(t) + iy(t), a ≤ t ≤ b. For example, x = cos t, y = sin t, 0 ≤ t ≤ 2π, describes a unit circle centered at the origin. This circle can also be described by z(t) = cos t + i sin t, or even more compactly by z(t) = eit, 0 ≤ t ≤ 2π. The same definitions of smooth curve, piecewise-smooth curve, closed curve, and simple closed curve given in Section 9.8 carry over to this discussion. As before, we shall assume that the positive direction on C corresponds to increasing values of t. In complex variables, a piecewise-smooth curve C is also called a contour or path. An integral of f(z) on C is denoted by ∫C f(z) dz or f(z) dz if the contour C is closed; it is referred to as a contour integral or simply as a complex integral.

  1. Let f(z) = u(x, y) + iv(x, y) be defined at all points on a smooth curve C defined by x = x(t), y = y(t), a ≤ t ≤ b.
  2. Divide C into n subarcs according to the partition a = t0 < t1 < . . . < tn = b of [a, b]. The corresponding points on the curve C are z0 = x0 + iy0 = x(t0) + iy(t0), z1 = x1 + iy1 = x(t1) + iy(t1), . . ., zn = xn + iyn = x(tn) + iy(tn). Let Δzk = zk − zk−1, k = 1, 2, . . ., n.
  3. Let ∥P∥ be the norm of the partition, that is, the maximum value of |Δzk|.
  4. Choose a sample point on each subarc. See FIGURE 18.1.1.
  5. Form the sum Δzk.
An image of an S-shaped curve labeled C with 9 dots and 3 differently colored dots in it. The first dot is labeled z subscript n, the third dot is labeled z subscript n minus 1, the eighth dot is labeled z subscript 2, the tenth dot is labeled z subscript 1, and the dot at the end is labeled z subscript 0. The differently colored dots are labeled z asterisk subscript n at the top and z asterisk subscript 2 and z asterisk subscript 1 at the bottom.

FIGURE 18.1.1 Sample points are the red dots

DEFINITION 18.1.1 Contour Integral

Let f be defined at points of a smooth curve C defined by x = x(t), y = y(t), a ≤ t ≤ b. The contour integral of f along C is

(1)

The limit in (1) exists if f is continuous at all points on C and C is either smooth or piecewise smooth. Consequently we shall, hereafter, assume these conditions as a matter of course.

Method of Evaluation

We shall turn now to the question of evaluating a contour integral. To facilitate the discussion, let us suppress the subscripts and write (1) in the abbreviated form

This means (2)

In other words, a contour integral ∫Cf(z) dz is a combination of two real-line integrals ∫C u dx − v dy and ∫C v dx + u dy. Now, since x = x(t) and y = y(t), a ≤ t ≤ b, the right side of (2) is the same as

But if we use z(t) = x(t) + iy(t) to describe C, the last result is the same as when separated into two integrals. Thus we arrive at a practical means of evaluating a contour integral:

THEOREM 18.1.1 Evaluation of a Contour Integral

If f is continuous on a smooth curve C given by z(t) = x(t) + iy(t), a ≤ t ≤ b, then

(3)

If f is expressed in terms of the symbol z, then to evaluate f(z(t)) we simply replace the symbol z by z(t). If f is not expressed in terms of z, then to evaluate f(z(t)), we replace x and y wherever they appear by x(t) and y(t), respectively.

EXAMPLE 1 Evaluating a Contour Integral

Evaluate dz, where C is given by x = 3t, y = t2, −1 ≤ t ≤ 4.

SOLUTION

We write z(t) = 3t + it2 so that z′(t) = 3 + 2it and f(z(t)) = = 3t − it2. Thus,

≡

EXAMPLE 2 Evaluating a Contour Integral

Evaluate dz, where C is the circle x = cos t, y = sin t, 0 ≤ t ≤ 2π.

SOLUTION

In this case z(t) = cos t + i sin t = eit, z′(t) = ieit, and f(z) = 1/z = e−it. Hence,

≡

For some curves, the real variable x itself can be used as the parameter. For example, to evaluate on the line segment we write

and integrate in the usual manner

Properties

The following properties of contour integrals are analogous to the properties of line integrals.

THEOREM 18.1.2 Properties of Contour Integrals

Suppose f and g are continuous in a domain D and C is a smooth curve lying entirely in D. Then

(i) kf(z) dz = k f(z) dz, k a constant

(ii) [f(z) + g(z)] dz = f(z) dz + g(z) dz

(iii) f(z) dz = f(z) dz + f(z) dz, where C is the union of the smooth curves C1 and C2

(iv) f(z) dz = − f(z) dz, where −C denotes the curve having the opposite orientation of C.

The four parts of Theorem 18.1.2 also hold when C is a piecewise-smooth curve in D.

EXAMPLE 3 Evaluating a Contour Integral

Evaluate ∫C (x2 + iy2) dz, where C is the contour shown in FIGURE 18.1.2.

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point at the center labeled 1 plus i and another point vertically above that point labeled 1 plus 2 i. A line, c subscript 2, connects the two points while another line, C subscript 1, connects 1 + i with the origin.

FIGURE 18.1.2 Piecewise-smooth contour in Example 3

SOLUTION

In view of Theorem 18.1.2(iii) we write

Since the curve C1 is defined by y = x, it makes sense to use x as a parameter. Therefore, z(x) = x + ix, z′(x) = 1 + i, f(z(x)) = x2 + ix2, and

The curve C2 is defined by x = 1, 1 ≤ y ≤ 2. Using y as a parameter, we have z(y) = 1 + iy, z′(y) = i, and f(z(y)) = 1 + iy2. Thus,

Finally, we have ≡

There are times in the application of complex integration that it is useful to find an upper bound for the absolute value of a contour integral. In the next theorem we shall use the fact that the length of a plane curve is s = dt. But if z′(t) = x′(t) + iy′(t), then |z′(t)| = and consequently .

THEOREM 18.1.3 A Bounding Theorem

If f is continuous on a smooth curve C and if |f(z)| ≤ M for all z on C, then where L is the length of C.

PROOF:

From the triangle inequality (6) of Section 17.1 we can write

(4)

Now, |Δzk| can be interpreted as the length of the chord joining the points zk and zk−1. Since the sum of the lengths of the chords cannot be greater than the length of C, (4) becomes . Hence, as ∥P∥ → 0, the last inequality yields |∫C f(z) dz| ≤ ML. ≡

Theorem 18.1.3 is used often in the theory of complex integration and is sometimes referred to as the ML-inequality.

EXAMPLE 4 A Bound for a Contour Integral

Find an upper bound for the absolute value of dz, where C is the circle |z| = 4.

SOLUTION

First, the length s of the circle of radius 4 is 8π. Next, from the inequality (7) of Section 17.1, it follows that |z + 1| ≥ |z| − 1 = 4 − 1 = 3, and so

(5)

In addition, |ez| = |ex(cos y + i sin y)| = ex. For points on the circle |z| = 4, the maximum that x can be is 4, and so (5) becomes

Hence from Theorem 18.1.3 we have

≡

Circulation and Net Flux

Let T and N denote the unit tangent vector and the unit normal vector to a positively oriented simple closed contour C. When we interpret the complex function f(z) = u(x, y) + iv(x, y) as a vector, the line integrals

(6)

and (7)

have special interpretations. The line integral in (6) is called the circulation around C and measures the tendency of the flow to rotate the curve C. See Section 9.8 for the derivation. The net flux across C is the difference between the rate at which fluid enters and the rate at which fluid leaves the region bounded by C. The net flux across C is given by the line integral in (7), and a nonzero value for · N ds indicates the presence of sources or sinks for the fluid inside the curve C. Note that

and so

(8)

(9)

Thus, both of these key quantities may be found by computing a single complex integral.

EXAMPLE 5 Net Flux

Given the flow f(z) = (1 + i)z, compute the circulation around, and the net flux across, the circle C: |z| = 1.

SOLUTION

Since = (1 − i) and z(t) = eit, 0 ≤ t ≤ 2π, we have

Using (8) and (9), the circulation around C is 2π and the net flux across C is 2π. See FIGURE 18.1.3. ≡

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a circle centered at the origin labeled C. There are numerous arrows moving anticlockwise inside and outside the circle.

FIGURE 18.1.3 Flow f(z) + (1 + i)z

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

In Problems 1–16, evaluate the given integral along the indicated contour.

  1. ∫C(z + 3) dz, where C is x = 2t, y = 4t − 1, 1 ≤ t ≤ 3
  2. ∫C(2 − z) dz, where C is x = −t, y = t2 + 2, 0 ≤ t ≤ 2
  3. ∫C z2 dz, where C is z(t) = 3t + 2it, −2 ≤ t ≤ 2
  4. ∫C(3z2 − 2z) dz, where C is z(t) = t + it2, 0 ≤ t ≤ 1
  5. ∫C dz, where C is the right half of the circle |z| = 1 from z = −i to z = i
  6. ∫C|z|2 dz, where C is x = t2, y = 1/t, 1 ≤ t ≤ 2
  7. (z) dz, where C is the circle |z| = 1
  8. dz, where C is the circle | z + i | = 1, 0 ≤ t ≤ 2π
  9. ∫C (x2 + iy3) dz, where C is the straight line from z = 1 to z = i
  10. ∫C(x3 − iy3) dz, where C is the lower half of the circle |z| = 1 from z = −1 to z = 1
  11. ∫C ez dz, where C is the polygonal path consisting of the line segments from z = 0 to z = 2 and from z = 2 to z = 1 + πi
  12. ∫C sin z dz, where C is the polygonal path consisting of the line segments from z = 0 to z = 1 and from z = 1 to z = 1 + i
  13. ∫C Im(z − i) dz, where C is the polygonal path consisting of the circular arc along |z| = 1 from z = 1 to z = i and the line segment from z = i to z = −1
  14. ∫C dz, where C is the left half of the ellipse x2/36 + y2/4 = 1 from z = 2i to z = −2i
  15. ez dz, where C is the square with vertices z = 0, z = 1, z = 1 + i, and z = i
  16. ∫C f(z) dz, where f(z) = and C is the parabola y = x2 from z = −1 + i to z = 1 + i

In Problems 17–20, evaluate the given integral along the contour C given in FIGURE 18.1.4.

A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point at the center labeled 1 plus i. A vertical line with an up arrow connects 1 + i to the horizontal axis at (1, 0), and a slanting line with a down arrow connects it to the origin. A horizontal line with a right arrow starts from the origin and ends at (1, 0). The three lines form a right-angled triangle.

FIGURE 18.1.4 Contour in Problems 17–20

In Problems 21–24, evaluate ∫C(z2 − z + 2) dz from i to 1 along the indicated contours.

  1. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point on the horizontal axis labeled 1 and a point on the vertical axis labeled i. A decreasing line with a down arrow starts from (0, i) and at (1, 0).

    FIGURE 18.1.5 Contour in Problem 21

  2. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point labeled 1 plus i at the center and two points i and 1 marked on the vertical and horizontal axis, respectively. A line flows from i to i + 1 and continues from i + 1 to 1. The two lines form a square with the vertical and horizontal axis.

    FIGURE 18.1.6 Contour in Problem 22

  3. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point on the horizontal axis labeled 1 and a point on the vertical axis labeled i. A decreasing curve with a down arrow from i to 1 is labeled y = 1 minus x^2.

    FIGURE 18.1.7 Contour in Problem 23

  4. A graph. The horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a point on the horizontal axis labeled 1 and a point on the vertical axis labeled i. A decreasing curve with a down arrow from i to 1 is labeled x^2 plus y^2 = 1.

    FIGURE 18.1.8 Contour in Problem 24

In Problems 25–28, find an upper bound for the absolute value of the given integral along the indicated contour.

  1. where C is the circle |z| = 5
  2. , where C is the right half of the circle |z| = 6 from z = −6i to z = 6i
  3. ∫C(z2 + 4) dz, where C is the line segment from z = 0 to z = 1 + i
  4. dz, where C is one quarter of the circle |z| = 4 from z = 4i to z = 4
    1. Use Definition 18.1.1 to show for any smooth curve C between z0 and zn that ∫C dz = zn − z0.
    2. Use the result in part (a) to verify the answer to Problem 14.
  6. Use Definition 18.1.1 to show for any smooth curve C between z0 and zn that . [Hint: The integral exists, so choose and .]
  7. Use the results of Problems 29 and 30 to evaluate where C is

    (a) the straight line from 1 + i to 2 + 3i, and

    (b) the closed contour x4 + y4 = 4.

In Problems 32–35, compute the circulation and net flux for the given flow and the indicated closed contour.

  1. f(z) = 1/z, where C is the circle |z| = 2
  2. f(z) = 2z, where C is the circle |z| = 1
  3. f(z) = 1/, where C is the circle |z − 1| = 2
  4. f(z) = , where C is the square with vertices z = 0, z = 1, z = 1 + i, z = i