**for**

**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.

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) .

**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

.

**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

.

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 .

**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.

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 .

**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 .

**Exercises
for Section 10.2. Bilinear
Transformations****
**

__Mobius
- Bilinear Transformation__

**The Next Module
is**

**Mapping
Involving Elementary Functions**

**Return to the Complex
Analysis Modules **

__Return
to the Complex Analysis Project__

This material is coordinated with our book Complex Analysis for Mathematics and Engineering.

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