INTRODUCTION

In many applications that involve boundary-value problems associated with Laplace’s equation, it is necessary to find a conformal mapping that maps a disk onto the half-plane v ≥ 0. Such a mapping would have to map the circular boundary of the disk to the boundary line of the half-plane. An important class of elementary conformal mappings that map circles to lines (and vice versa) are the fractional transformations. In this section we will define and study this special class of mappings.

Linear Fractional Transformation

If a, b, c, and d are complex constants with ad − bc ≠ 0, then the complex function defined by

is called a linear fractional transformation. Since

T is conformal at z provided Δ = ad − bc ≠ 0 and z ≠ −d/c. (If Δ = 0, then T′(z) = 0 and T(z) would be a constant function.) Linear fractional transformations are circle preserving in a sense that we will make precise in this section, and, as we saw in Example 7 of Section 20.2, they can be useful in solving Dirichlet problems in regions bounded by circles.

Note that when c ≠ 0, T(z) has a simple pole at z₀ = −d/c and so

We will write T(z₀) = ∞ as shorthand for this limit. In addition, if c ≠ 0, then

and we write T(∞) = a/c.

EXAMPLE 1 A Linear Fractional Transformation

Circle-Preserving Property

If c = 0, the linear fractional transformation reduces to a linear function T(z) = Az + B. We saw in Section 20.1 that such a complex mapping can be considered as the composite of a rotation, magnification, and translation. As such, a linear function will map a circle in the z-plane to a circle in the w-plane. When c ≠ 0, we can divide cz + d into az + b to obtain

(1)

If we let A = (bc − ad)/c and B = a/c, T(z) can be written as the composite of transformations:

(2)

A general linear fractional transformation can therefore be written as the composite of two linear functions and the inversion w = 1/z. Note that if z − z₁ = r and w = 1/z, then

(3)

It is not hard to show that the set of all points w that satisfy

(4)

is a line when λ = 1 and is a circle when λ > 0 and λ ≠ 1. It follows from (3) that the image of the circle z − z₁ = r under the inversion w = 1/z is a circle except when r = 1/w₁ = z₁. In the latter case, the original circle passes through the origin and the image is a line. See Figure 20.1.3. From (2), we can deduce the following theorem.

PROOF:

We have shown that a linear function maps a circle to a circle, whereas an inversion maps a circle to a circle or a line. It follows from (2) that a circle in the z-plane will be mapped to either a circle or a line in the w-plane. If the original circle passes through a pole z₀, then T(z₀) = ∞, and so the image is unbounded. Therefore, the image of such a circle must be a line. If the original circle does not pass through z₀, then the image is bounded and must be a circle. ≡

EXAMPLE 2 Images of Circles

Find the images of the circles z = 1 and z = 2 under T(z) = (z + 2)/(z − 1). What are the images of the interiors of these circles?

SOLUTION

The circle z = 1 passes through the pole z₀ = 1 of the linear fractional transformation and so the image is a line. Since T(−1) = − and T(i) = − − i, we can conclude that the image is the line u = −. The image of the interior z < 1 is either the half-plane u < − or the half-plane u > − . Using z = 0 as a test point, T(0) = −2, and so the image is the .

The circle z = 2 does not pass through the pole and so the image is a circle. For z = 2,

Therefore, is a point on the image circle and so the image circle is symmetric with respect to the u-axis. Since T(−2) = 0 and T(2) = 4, the center of the circle is w = 2 and the image is the circle w − 2 = 2. See FIGURE 20.3.1. The image of the interior z < 2 is either the interior or the exterior of the image circle w − 2 = 2. Since T(0) = −2, we can conclude that the image is |w − 2| > 2.

Constructing Special Mappings

In order to use linear fractional transformations to solve Dirichlet problems, we must construct special functions that map a given circular region R to a target region R′ in which the corresponding Dirichlet problem is solvable. Since a circular boundary is determined by three of its points, we must find a linear fractional transformation w = T(z) that maps three given points z₁, z₂, and z₃ on the boundary of R to three points w₁, w₂, and w₃ on the boundary of R′. In addition, the interior of R′ must be the image of the interior of R. See FIGURE 20.3.2.

Matrix Methods

Matrix methods can be used to simplify many of the computations. We can associate the matrix

with T(z) = (az + b)/(cz + d).^* If T₁(z) = (a₁z + b₁)/(c₁z + d₁) and T₂(z) = (a₂z + b₂)/(c₂z + d₂), then the composite function T(z) = T₂(T₁(z)) is given by T(z) = (az + b)/(cz + d), where

(5)

If w = T(z) = (az + b)/(cz + d), we can solve for z to obtain z = (dw − b)/(−cw + a). Therefore the inverse of the linear fractional transformation T is T⁻¹(w) = (dw − b)/(−cw + a) and we associate the adjoint of the matrix A,

, (6)

with T⁻¹. The matrix adj A is the adjoint matrix of A (see Section 8.6), the matrix for T.

EXAMPLE 3 Using Matrices to Find an Inverse Transform

Triples to Triples

The linear fractional transformation

has a zero at z = z₁, a pole at z = z₃, and T(z₂) = 1. Therefore, T(z) maps three distinct complex numbers z₁, z₂, and z₃ to 0, 1, and ∞, respectively. The term is called the cross-ratio of the complex numbers z, z₁, z₂, and z₃.

Likewise, the complex mapping S(w) = sends w₁, w₂, and w₃ to 0, 1, and ∞, and so S⁻¹ maps 0, 1, and ∞ to w₁, w₂, and w₃. It follows that the linear fractional transformation w = S⁻¹(T(z)) maps the triple z₁, z₂, and z₃ to w₁, w₂, and w₃. From w = S⁻¹(T(z)), we have S(w) = T(z) and we can conclude that

(7)

In constructing a linear fractional transformation that maps the triple z₁, z₂, and z₃ to w₁, w₂, and w₃, we can use matrix methods to compute w = S⁻¹(T(z)). Alternatively, we can substitute into (7) and solve the resulting equation for w.

EXAMPLE 4 Constructing a Linear Fractional Transformation

Construct a linear fractional transformation that maps the points 1, i, and −1 on the circle z = 1 to the points −1, 0, and 1 on the real axis.

SOLUTION

Substituting into (7), we have

or

Solving for w, we obtain w = −i(z − i)/(z + i). Alternatively, we could use the matrix method to compute w = S⁻¹(T(z)). ≡

When z_k = ∞ plays the role of one of the points in a triple, the definition of the cross-ratio is changed by replacing each factor that contains z_k by 1. For example, if z₂ = ∞, both z₂ − z₃ and z₂ − z₁ are replaced by 1, giving (z − z₁)/(z − z₃) as the cross-ratio.

EXAMPLE 5 Constructing a Linear Fractional Transformation

Construct a linear fractional transformation that maps the points ∞, 0, and 1 on the real axis to the points 1, i, and −1 on the circle w = 1.

SOLUTION

Since z₁ = ∞, the terms z − z₁ and z₂ − z₁ in the cross product are replaced by 1. It follows that

or

If we use the matrix method to find w = S⁻¹(T(z)), then

and so ≡

EXAMPLE 6 Solving a Dirichlet Problem

Solve the Dirichlet problem in FIGURE 20.3.3(a) using conformal mapping by constructing a linear fractional transformation that maps the given region into the upper half-plane.

SOLUTION

The boundary circles z = 1 and z − = each pass through z = 1. We can therefore map each boundary circle to a line by selecting a linear fractional transformation that has z = 1 as a pole. If we further require that T(i) = 0 and T(−1) = 1, then

Since T(0) = 1 + i and T ( + i) = −1 + i, T maps the interior of the circle z = 1 onto the upper half-plane and maps the circle z − = onto the line v = 1. Figure 20.3.3(b) shows the transferred boundary conditions.

The harmonic function U(u, v) = v is the solution to the simplified Dirichlet problem in the w-plane, and so, by Theorem 20.2.2, u(x, y) = U(T(z)) is the solution to the original Dirichlet problem in the z-plane.

Since the imaginary part of T(z) = (1 − i) is , the solution is given by

The level curves u(x, y) = c can be written as

and are therefore circles that pass through z = 1. See FIGURE 20.3.4. These level curves may be interpreted as the isotherms of the steady-state temperature distribution induced by the boundary temperatures. ≡

20.3 Exercises Answers to selected odd-numbered problems begin on page ANS-49.

In Problems 1–4, a linear fractional transformation is given.

1. T(z) =
2. T(z) =
3. T(z) =
4. T(z) =

In Problems 5–8, use the matrix method to compute S⁻¹(w) and S⁻¹(T(z)) for each pair of linear fractional transformations.

5. T(z) = and S(z) =
   
   
6. T(z) = and S(z) =
   
   
7. T(z) = and S(z) =
   
   
8. T(z) = and S(z) =

In Problems 9–16, construct a linear fractional transformation that maps the given triple z₁, z₂, and z₃ to the triple w₁, w₂, and w₃.

9. −1, 0, 2 to 0, 1, ∞
10. i, 0, −i to 0, 1, ∞
11. 0, 1, ∞ to 0, i, 2
12. 0, 1, ∞ to 1 + i, 0, 1 − i
13. −1, 0, 1 to i, ∞, 0
14. −1, 0, 1 to ∞, −i, 1
15. 1, i, −i to −1, 0, 3
16. 1, i, −i to i, −i, 1
17. Use the results in Example 2 and the harmonic function U = (log_e r)/(log_e r₀) to solve the Dirichlet problem in FIGURE 20.3.5. Explain why the level curves must be circles.

    

18. Use the linear fractional transformation that maps −1, 1, 0 to 0, 1, ∞ to solve the Dirichlet problem in FIGURE 20.3.6. Explain why, with one exception, all level curves must be circles. Which level curve is a line?

    

19. Derive the conformal mapping H-1 in the conformal mappings in Appendix D.
20. Derive the conformal mapping H-5 in the conformal mappings in Appendix D by first mapping 1, i, −1 to ∞, i, 0.
21. Show that the composite of two linear fractional transformations is a linear fractional transformation and verify (5).
22. If w₁ ≠ w₂ and λ > 0, show that the set of all points w that satisfy w − w₁ = λw − w₂ is a line when λ = 1 and is a circle when λ ≠ 1. [Hint: Write as w − w₁² = λ²w − w₂² and expand.]

*The matrix A is not unique since the numerator and denominator in T(z) can be multiplied by a nonzero constant.