Module

for

Bilinear Transformations

Mobius Transformations

10.2  Bilinear Transformations - Mobius Transformations

Another important class of elementary mappings was studied by August Ferdinand Möbius (1790-1868).  These mappings are conveniently expressed as the quotient of two linear expressions and are commonly known as linear fractional or bilinear transformations.  They arise naturally in mapping problems involving the function  arctan(z).  In this section, we show how they are used to map a disk one-to-one and onto a half-plane.  An important property is that these transformations are conformal in the entire complex plane except at one point. (see Section 10.1)

Let    denote four complex constants with the restriction that .  Then the function

(10-13)

is called  a bilinear transformation, a Möbius transformation, or a linear fractional transformation.
If the expression for S(z) in Equation (10-13) is multiplied through by the quantity  ,  then the resulting expression has the bilinear form  .
We collect terms involving z and write  .  Then, for values of    the inverse transformation is given by

(10-14)            .

We can extend    to mappings in the extended complex plane.  The value    should be chosen to equal the limit of   as  .  Therefore we define

,

and the inverse is  .  Similarly, the value    is obtained by

,

and the inverse is  .  With these extensions we conclude that the transformation    is a one-to-one mapping of the extended complex z-plane onto the extended complex w-plane.

We now show that a bilinear transformation carries the class of circles and lines onto itself.  If S(z) is an arbitrary bilinear transformation given by Equation (10-13) and  ,  then S(z)  reduces to a linear transformation, which carries lines onto lines and circles onto circles.  If  ,  then we can write S(z) in the form

(10-15)

The condition    precludes the possibility that S(z) reduces to a constant.  Equation (10-15) indicates that S(z) can be considered as a composition of functions.
It is a linear mapping  ,  followed by the reciprocal transformation  ,  followed by  .  In Section 2.1 we showed that each function in this composition maps the class of circles and lines onto itself; it follows that the bilinear transformation S(z) has this property.  A half-plane can be considered to be a family of parallel lines and a disk as a family of circles.  Therefore we conclude that a bilinear transformation maps the class of half-planes and disks onto itself.  Example 10.3 illustrates this idea.

Example 10.3.  Show that    maps the unit disk    one-to-one and onto the upper half-plane  .

Solution.  We first consider the unit circle  ,  which forms the boundary of the disk and find its image in the w plane.
If we write  ,  then we see that  ,  ,  , and .
Using Equation (10-14), we find that the inverse is given by

(10-16)        .

If  ,  then Equation (10-16) implies that the images of points on the unit circle satisfy which yields the equation

(10-17)        .

Squaring both sides of Equation (10-17), we obtain

which is the equation of the u axis in the w plane.

The circle C divides the z plane into two portions, and its image is the u axis, which divides the w plane into two portions.  The image of the point    is  ,  so we expect that the interior of the circle C is mapped onto the portion of the w plane that lies above the u axis.  To show that this outcome is true, we let  .  Then Equation (10-16) implies that the image values must satisfy the inequality  ,  which we write as

.

If we interpret as the distance from    and as the distance from  ,  then a geometric argument shows that the image point w must lie in the upper half-plane  ,  as shown in Figure 10.5.  As S(z) is one-to-one and onto in the extended complex plane, it follows that S(z) maps the disk onto the half-plane.

Figure 10.5  The image    under  ,  the points
are mapped onto the points  ,  respectively.

Explore Solution 10.3.

The general formula for a bilinear transformation (Equation (10-13)) appears to involve four independent coefficients:  .  But as S(z) is not identically constant, either    or  ,  we can express the transformation with three unknown coefficients and write either

or     ,

respectively.  Doing so permits us to determine a unique a bilinear transformation if three distinct image values  ,   , and    are specified.  To determine such a mapping, we can conveniently use an implicit formula involving z and w.

Theorem 10.3 (The Implicit Formula). There exists a unique bilinear transformation that maps three distinct points   onto three distinct points , respectively.  An implicit formula for the mapping is given by the equation

(10-18)            .

Proof.

Example 10.4.  Construct the bilinear transformation  w = S(z)  that maps the points    onto the points  ,  respectively.

Solution.  We use the implicit formula, Equation (10-18), and write

.

Expanding this equation, collecting terms involving w and zw on the left and then simplify.

Therefore the desired bilinear transformation is

.

Explore Solution 10.4.

Example 10.5.  Find the bilinear transformation  w = S(z)  that maps the points    onto the points  ,  respectively.

Solution.  Again, we use the implicit formula, Equation (10-18), and write

Using the fact that  ,  we rewrite this equation as

.

We now expand the equation and obtain

which can be solved for w in terms of z, giving the desired solution

.

Explore Solution 10.5.

We let D be a region in the z plane that is bounded by either a circle or a straight line C.  We further let be three distinct points that lie on C and have the property that an observer moving along C from through finds the region D to be on the left.  If C is a circle and D is the interior of C, then we say that C is positively oriented.  Conversely, the ordered triple uniquely determines a region that lies to the left of C.

We let G be a region in the w plane that is bounded by either a circle of a straight line K.  We further let be three distinct points that lie on K such that an observer moving along K from through finds the region G to be on the left.  Because a bilinear transformation is a conformal mapping that maps the class of circles and straight lines onto itself, we can use the implicit formula to construct a bilinear transformation that is a one-to-one mapping of D onto G.

Example 10.6.  Show that  the mapping    maps the disk    one-to-one and onto the upper half plane  .

Solution.  For convenience, we choose the ordered triple  , which gives the circle    a positive orientation and the disk D a left orientation.  From Example 10.5, the corresponding image points are

Because the ordered triple of points  ,  lie on the u axis, it follows that the image of circle C is the u axis.  The points   give the upper half-plane    a left orientation.  Therefore maps the disk D onto the upper half-plane G.  To check our work, we choose a point that lies in D and find the half-plane in which its image, lies.  The choice    yields  .  Hence the upper half-plane is the correct image.  This situation is illustrated in Figure 10.6.

Figure 10.6  The bilinear mapping  .

Explore Solution 10.6.

Corollary 10.1 (The Implicit Formula with a point at Infinity).  In equation (10-18) the point at infinity can be introduced as one of the prescribed points in either the z plane or the  w plane.

Proof.

Case 1.  If  ,  then we can write    and substitute this expression into Equation (10-18) to obtain    which can be rewritten as   and simplifies to obtain

.

Case 2.  If  ,  then we can write   and substitute this expression into Equation (10-18) to obtain   which can be rewritten as    and simplifies to obtain

(10-21)        .

Equation (10-21) is sometimes used to map the crescent-shaped region that lies between the tangent circles onto an infinite strip.

Example 10.7.  Find the bilinear transformation that maps the crescent-shaped region that lies inside the disk   and outside the circle   onto a horizontal strip.

Solution.  For convenience we choose    and the image values  ,  respectively.  The ordered triple    gives the circle    a positive orientation and the disk    has a left orientation.  The image points    all lie on the extended u axis, and they determine a left orientation for the upper half-plane  .  Therefore we can use the second implicit formula (Equation (10-21)) to write

,

which determines a mapping of the disk    onto the upper half-plane  .  Use the fact that   to simplify the preceding equation and get

which can be written in the form

.

A straightforward calculation shows that the points    are mapped onto the points

respectively.  The points    lie on the horizontal line    in the upper half-plane.  Therefore the crescent-shaped region is mapped onto the horizontal strip  ,  as shown in Figure 10.7.

Figure 10.7  The mapping  .

Explore Solution 10.7.

Lines of Flux

In electronics, images of certain lines represent lines of electric flux, which comprise the trajectory of an electron placed in an electrical field. Consider the bilinear transformation

and    .

The half rays  ,  where c is a constant, that meet at the origin represent the lines of electric flux produced by a source located at   (and a sink at ).  The preimage of this family of lines is a family of circles that pass through the points .  We visualize these circles as the lines of electric flux from one point charge to another.  The limiting case as a is called a dipole and is discussed in Exercise 6, Section 11.11.  The graphs for , , and are shown in Figure 10.8.

Figure 10.8  Images of under the mapping .

Conformal Mapping

Mobius - Bilinear Transformation

Smith Chart

The Next Module is

Mapping Involving Elementary Functions

(c) 2012 John H. Mathews, Russell W. Howell