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
Let denote four complex constants with the restriction that . Then the function
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
We can extend to
mappings in the extended complex plane. The
be chosen to equal the limit of as . Therefore
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
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
If we write , then we see that , , , and .
Using Equation (10-14), we find that the inverse is given by
If , then
Equation (10-16) implies that the images
of points on the unit circle satisfy
which yields the equation
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.
10.5 The image under , the
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
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
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
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.
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
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
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
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
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 .
Exercises for Section 10.2. Bilinear Transformations
Mobius - Bilinear Transformation
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Mapping Involving Elementary Functions
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This material is coordinated with our book Complex Analysis for Mathematics and Engineering.
(c) 2012 John H. Mathews, Russell W. Howell