20.1 Complex Functions as Mappings

INTRODUCTION

In Chapter 17 we emphasized the algebraic definitions and properties of complex functions. In order to give a geometric interpretation of a complex function w = f(z), we place a z-plane and a w-plane side by side and imagine that a point z = x + iy in the domain of the definition of f has mapped (or transformed) to the point w = f(z) in the second plane. Thus the complex function w = f(z) = u(x, y) + iv(x, y) may be considered as the planar transformation

and w = f(z) is called the image of z under f.

FIGURE 20.1.1 indicates the images of a finite number of complex numbers in the region R. More useful information is obtained by finding the image of the region R together with the images of a family of curves lying inside R. Common choices for the curves are families of lines, families of circles, and the system of level curves for the real and imaginary parts of f.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a dot labeled z subscript 1 in the first quadrant, a dot labeled z subscript 2 in the second quadrant, and a dot labeled z subscript 3 in the fourth quadrant. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows two dots labeled w subscript 1 and w subscript 2 in its the second quadrant, and a dot labeled w subscript 3 in its the fourth quadrant. Arrows labeled f from z subscript 1, z subscript 2, and z subscript 3 are point at w subscript 1, w subscript 2, and w subscript 3, respectively.

FIGURE 20.1.1 w1, w2, w3 are images of z1, z2, z3

Images of Curves

Note that if z(t) = x(t) + iy(t), a ≤ t ≤ b, describes a curve C in the region, then w = f(z(t)), a ≤ t ≤ b, is a parametric representation of the corresponding curve C′ in the w-plane. In addition, a point z on the level curve u(x, y) = a will be mapped to a point w that lies on the vertical line u = a, and a point z on the level curve v(x, y) = b will be mapped to a point w that lies on the horizontal line v = b.

EXAMPLE 1 The Mapping f(z) = ez

The horizontal strip 0 ≤ y ≤ π lies in the fundamental region of the exponential function f(z) = ez. A vertical line segment x = a in this region can be described by z(t) = a + it, 0 ≤ t ≤ π, and so w = f(z(t)) = eaeit. Thus the image is a semicircle with center at w = 0 and with radius r = ea. Similarly, a horizontal line y = b can be parametrized by z(t) = t + ib, −∞ < t < ∞, and so w = f(z(t)) = eteib. Since Arg w = b and w = et, the image is a ray emanating from the origin, and since 0 ≤ Arg w ≤ π, the image of the entire horizontal strip is the upper half-plane v ≥ 0. Note that the horizontal lines y = 0 and y = π are mapped onto the positive and negative u-axis, respectively. See FIGURE 20.1.2 for the mapping by f(z) = ez.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a rectangular mesh spanning the first and second quadrant and intersecting the positive vertical axis at pi i, and the region bound by the mesh is shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The positive horizontal axis is labeled Arg w = 0 and the negative horizontal axis is labeled Arg w = pi. The graph shows five concentric semicircles centered at the origin. Two lines rising lines each in the first and second intersect the semicircles. The area bound by the first and second quadrants is shaded.

FIGURE 20.1.2 Images of vertical and horizontal lines in Example 1

From w = exeiy, we can conclude that w = ex and y = Arg w. Hence, z = x + iy = logew + i Arg w = Ln w. The inverse function f−1(w) = Ln w therefore maps the upper half-plane v ≥ 0 back to the horizontal strip 0 ≤ y ≤ π. ≡

EXAMPLE 2 The Mapping f(z) = 1/z

The complex function f(z) = 1/z has domain z ≠ 0 and real and imaginary parts u(x, y) = x/(x2 + y2) and v(x, y) = −y/(x2 + y2), respectively. When a ≠ 0, a level curve u(x, y) = a can be written as

The level curve is therefore a circle with its center on the x-axis and passing through the origin. A point z on this circle other than zero is mapped to a point w on the line u = a. Likewise, the level curve v(x, y) = b, b ≠ 0, can be written as

and a point z on this circle is mapped to a point w on the line v = b. FIGURE 20.1.3 shows the mapping by f(z) = 1/z. Figure 20.1.3(a) shows the two collections of circular level curves, and Figure 20.1.3(b) shows their corresponding images in the w-plane.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows 4 pairs of circles, each comprising two differently sized circles, tangential to the origin and centered on each of the four axes. The two pairs of circles on the vertical axes are of one color and two pairs of circles on the horizontal axes are of another color. The pair of circles at the right is labeled a = 1 over 2, at the left is labeled is labeled a = minus 1 over 2, at the top is labeled b = minus 1 over 2, and at the bottom is labeled b = 1 over 2. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a point labeled 2 at the extreme right of the positive horizontal axis, a point labeled minus 2 at the extreme left of the horizontal axis, a point labeled 2 at the top of the positive vertical axis, and a point labeled minus 2 at the down of the negative vertical axis. The graph shows two horizontal lines intersecting the positive vertical axis and two horizontal lines intersecting the negative vertical axis. The graph also shows two vertical lines intersecting the positive horizontal axis and two vertical lines intersecting the negative horizontal axis intersecting the horizontal lines at all the quadrants.

FIGURE 20.1.3 Images of circles in Example 2

Since w = 1/z, we have z = 1/w. Thus f−1(w) = 1/w, and so f = f−1. We can therefore conclude that f maps the horizontal line y = b to the circle u2 + (v + b)2 = ( b)2, and f maps the vertical line x = a to the circle . ≡

Translation and Rotation

The elementary linear function f(z) = z + z0 may be interpreted as a translation in the z-plane. To see this, we let z = x + iy and z0 = h + ik. Since w = f(z) = (x + h) + i(y + k), the point (x, y) has been translated h units in the horizontal direction and k units in the vertical direction to the new position at (x + h, y + k). In particular, the origin O has been mapped to z0 = h + ik.

The elementary function g(z) = z may be interpreted as a rotation through θ0 degrees, for if z = reiθ, then w = g(z) = . Note that if the complex mapping h(z) = z + z0 is applied to a region R that is centered at the origin, the image region R′ can be obtained by first rotating R through θ0 degrees and then translating the center to the new position z0. See FIGURE 20.1.4 for the mapping by h(z) = z + z0.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a closed curve covering all quadrants. R is labeled inside the curve and the region bound by the curve is shaded. In the second graph, the horizontal axis is rotated vertically with origin labeled z subscript 0. The graph shows a closed curve covering all quadrants. R prime is labeled inside the curve and the region bound by the curve is shaded. In the third graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows two dashed lines from the first quadrant to the third quadrant and from the second quadrant to fourth quadrant intersecting at the origin. The graph also shows a closed curve covering all quadrants. The angle formed in the positive horizontal axis is labeled theta subscript 0 and the region bound by the curve is shaded.

FIGURE 20.1.4 Translation and rotation

EXAMPLE 3 Rotation and Translation

Find a complex function that maps the horizontal strip −1 ≤ y ≤ 1 onto the vertical strip 2 ≤ x ≤ 4.

SOLUTION

If the horizontal strip −1 ≤ y ≤ 1 is rotated through 90°, the vertical strip −1 ≤ x ≤ 1 results, and the vertical strip 2 ≤ x ≤ 4 can be obtained by shifting this vertical strip 3 units to the right. See FIGURE 20.1.5. Since eiπ/2 = i, we obtain h(z) = iz + 3 as the desired complex mapping.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The horizontal axis ranges from minus 4 to 4 and the vertical axis ranges from minus 4 to 4. The graph shows a horizontal line from minus 4 to 4 intersecting the positive vertical axis and a horizontal line from minus 4 to 4 intersecting the negative vertical axis. The region bound by the two horizontal lines is shaded. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The horizontal axis ranges from minus 4 to 4. The vertical axis ranges from minus 4 to 4. The graph shows two vertical lines intersecting the positive horizontal axis at (2, 0) and (4, 0), respectively, from the first quadrant to the fourth quadrant. The region bound by the two horizontal lines is shaded.

FIGURE 20.1.5 Image of horizontal strip in Example 3 ≡

Magnification

A magnification is a complex function of the form f(z) = αz, where α is a fixed positive real number. Note that w = αz = αz, and so f changes the length (but not the direction) of the complex number z by a fixed factor α. If g(z) = az + b and a = , then the vector z is rotated through θ0 degrees, magnified by a factor of r0, and then translated using b.

EXAMPLE 4 Contraction and Translation

Find a complex function that maps the disk z ≤ 1 onto the disk w − (1 + i) ≤ .

SOLUTION

We must first contract the radius of the disk by a factor of and then translate its center to the point 1 + i. Therefore, maps z ≤ 1 to the disk w − (1 + i) ≤ . ≡

Power Functions

A complex function of the form f(z) = zα, where α is a fixed positive real number, is called a real power function. FIGURE 20.1.6 shows the effect of the complex function f(z) = zα on the angular wedge 0 ≤ Arg z ≤ θ0. If z = reiθ, then w = f(z) = rαeiαθ. Hence, 0 ≤ Arg w ≤ αθ0 and the opening of the wedge is changed by a factor of α. It is not difficult to show that a circular arc with center at the origin is mapped to a similar circular arc, and rays emanating from the origin are mapped to similar rays.

Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows an increasing line from the origin to right and a horizontal line passes though the horizontal axis forming an acute angle labeled theta subscript 0 in the first quadrant. The region bound by the lines is labeled R. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled y. The graph also shows a line with negative slope in the second quadrant and a horizontal line passes through the horizontal axis in the first quadrant forming an obtuse angle labeled alpha theta subscript 0. The region bound by the lines is labeled R prime.

FIGURE 20.1.6 R′ is the image of the angular wedge R

EXAMPLE 5 The Power Function f(z) = z1/4

Find a complex function that maps the upper half-plane y ≥ 0 onto the wedge 0 ≤ Arg w ≤ π/4.

SOLUTION

The upper half-plane y ≥ 0 can also be described by the inequality 0 ≤ Arg z ≤ π. We must therefore find a complex mapping that reduces the angle θ0 = π by a factor of α = . Hence, f(z) = z1/4. ≡

Successive Mappings

To find a complex mapping between two regions R and R′, it is often convenient to first map R onto a third region R″ and then find a complex mapping from R″ onto R′. More precisely, if ζ = f(z) maps R onto R″, and w = g(ζ) maps R″ onto R′, then the composite function w = g(f(z)) maps R onto R′. See FIGURE 20.1.7 for a diagram of successive mappings.

Three graphs. The first graph labeled z plane shows a shaded oval labeled R partially intersecting the horizontal axis. The second graph labeled w plane shows a shaded triangle labeled R prime partially intersecting the vertical axis. The third graph labeled zeta plane shows a shaded rectangle labeled R double prime. The z plane and w plane are placed in parallel and zeta plane is placed below them. A downward arrow, labeled f, flows from R to R double prime and another downward arrow, labeled g, flows from R double prime to R prime.

FIGURE 20.1.7 R′ is image of R under successive mappings

EXAMPLE 6 Successive Mappings

Find a complex function that maps the horizontal strip 0 ≤ y ≤ π onto the wedge 0 ≤ Arg w ≤ π/4.

SOLUTION

We saw in Example 1 that the complex function f(z) = ez mapped the horizontal strip 0 ≤ y ≤ π onto the upper half-plane 0 ≤ Arg ζ ≤ π. From Example 5, the upper half-plane 0 ≤ Arg ζ ≤ π is mapped onto the wedge 0 ≤ Arg w ≤ π/4 by g(ζ) = ζ1/4. It therefore follows that the composite function w = g(f(z)) = g(ez) = ez/4 maps the horizontal strip 0 ≤ y ≤ π onto the wedge 0 ≤ Arg w ≤ π/4. ≡

EXAMPLE 7 Successive Mappings

Find a complex function that maps the wedge π/4 ≤ Arg z ≤ 3π/4 onto the upper half-plane v ≥ 0.

SOLUTION

We first rotate the wedge π/4 ≤ Arg z ≤ 3π/4 so that it is in the standard position shown in Figure 20.1.6. If ζ = f(z) = e−iπ/4z, then the image of this wedge is the wedge R″ defined by 0 ≤ Arg ζ ≤ π/2. The real power function w = g(ζ) = ζ2 expands the opening of R″ by a factor of two to give the upper half-plane 0 ≤ Arg w ≤ π as its image. Therefore, w = g(f(z)) = (e−π/4z)2 = −iz2 is the desired mapping. ≡

In Sections 20.2–20.4, we will expand our knowledge of complex mappings and show how they can be used to solve Laplace’s equation in the plane.

20.1 Exercises Answers to selected odd-numbered problems begin on page ANS-48.

In Problems 1–10, a curve in the z-plane and a complex mapping w = f(z) are given. In each case, find the image curve in the w-plane.

  1. y = x under w = 1/z
  2. y = 1 under w = 1/z
  3. Hyperbola xy = 1 under w = z2
  4. Hyperbola x2 − y2 = 4 under w = z2
  5. Semicircle z = 1, y > 0, under w = Ln z
  6. Ray θ = π/4 under w = Ln z
  7. Ray θ = θ0 under w = z1/2
  8. Circular arc r = 2, 0 ≤ θ ≤ π/2, under w = z1/2
  9. Curve ex cos y = 1 under w = ez
  10. Circle z = 1 under w = z + 1/z

In Problems 11–20, a region R in the z-plane and a complex mapping w = f(z) are given. In each case, find the image region R′ in the w-plane.

  1. First quadrant under w = 1/z
  2. Strip 0 ≤ y ≤ 1 under w = 1/z
  3. Strip π/4 ≤ y ≤ π/2 under w = ez
  4. Rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ π, under w = ez
  5. Circle z = 1 under w = z + 4i
  6. Circle z = 1 under w = 2z − 1
  7. Strip 0 ≤ y ≤ 1 under w = iz
  8. First quadrant under w = (1 + i)z
  9. Wedge 0 ≤ Arg z ≤ π/4 under w = z3
  10. Wedge 0 ≤ Arg z ≤ π/4 under w = z1/2

In Problems 21–30, find a complex mapping from the given region R in the z-plane to the image region R′ in the w-plane.

  1. Strip 1 ≤ y ≤ 4 to the strip 0 ≤ u ≤ 3
  2. Strip 1 ≤ y ≤ 4 to the strip 0 ≤ v ≤ 3
  3. Disk z − 1 ≤ 1 to the disk w ≤ 2
  4. Strip −1 ≤ x ≤ 1 to the strip −1 ≤ v ≤ 1
  5. Wedge π/4 ≤ Arg z ≤ π/2 to the upper half-plane v ≥ 0
  6. Strip 0 ≤ y ≤ 4 to the upper half-plane v ≥ 0
  7. Strip 0 ≤ y ≤ π to the wedge 0 ≤ Arg w ≤ 3π/2
  8. Wedge 0 ≤ Arg z ≤ 3π/2 to the half-plane u ≥ 2
  9. Two graphs. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a horizontal line labeled y = pi from the second quadrant to the first quadrant intersecting the positive vertical axis. The region from the horizontal line covering all quadrants is shaded and labeled R. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line labeled v = pi from the second quadrant to the first quadrant intersecting the positive vertical axis and another horizontal line passing through the horizontal axis parallel to v = pi. The region bound by the horizontal lines covering the first and the second quadrants is shaded and labeled R prime.

    FIGURE 20.1.8 Regions R and R′ for Problem 29

  10. In the first graph, the horizontal axis is labeled x and the vertical axis is labeled y. The graph shows a vertical line labeled i passing through the positive vertical axis from its top and turning left at point (0, i) to become a horizontal line, and a horizontal line passing through the positive and negative horizontal axes. The lines together bind a shaded area labeled R, which looks like a mirrored L covering the first quadrant and the lower half of the second quadrant. In the second graph, the horizontal axis is labeled u and the vertical axis is labeled v. The graph shows a horizontal line labeled v = 1 from the second quadrant to the first quadrant with the positive vertical axis at its center and another horizontal line passing through the positive horizontal axis from the origin to extreme right. The graph also shows a vertical line passing through the negative vertical axis and ending at origin. The region bound by the horizontal and vertical lines covering the lower half of the first and the second and the complete third quadrant is shaded and labeled R prime.

    FIGURE 20.1.9 Regions R and R′ for Problem 30

  11. Project The mapping in Problem 10 is a special case of the mapping w = z + k2/z, where k is a positive constant, called the Joukowski transformation after the Russian mathematician and engineer Nikolai Yegorovich Zhukovsky (1847–1921) who published it in 1910.
    1. Show that the Joukowski transformation maps any circle x2 + y2 = R2 into the ellipse
    2. What is the image of the circle when R = k?
    3. The importance of the transformation w = z + k2/z does not lie in its effect on circles z = R centered at the origin but on off-centered circles with center on the real axis. Show that the Joukowski transformation can be written

      With k = 1, this particular transformation maps a circle passing through z = −1 and containing the point z = 1 into a closed curve with a sharply pointed trailing edge. This kind of curve, which resembles a cross section of an airplane wing, is known as a Joukowski airfoil.

      Write a report on the use of the Joukowski transformation in the study of the flow of air around an airfoil. There is a lot of information on this topic on the Internet.