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In this
session we consider the transformation
where a,b,c,d are complex constants satisfying
ad-bc≠0.
This
transformation is clearly a combination of translation,
rotation, stretching or
contraction, and inversion.
This
transformation is called bilinear because it is
equivalent to
 ,
where the
left-hand side is seen to be linear in z and w.
If we
expect circles to map into circles; that is,
we notice
the “problem” in the variable term | and
we now set our objective to show that this special
conformal transformation maps “lines and circles” into
“lines and circles” in the extended complex plane
 
as we shall see shortly. If c=0, then
implies
 .
If
then
implies that
and
implies that
.
Notice that this correspondence is continuous in the sense
that as
then
and as
 ,
then
The inverse transformation
shows a one-to-one correspondence from the extended
complex plane onto the extended complex plane.
Theorem. The
equation of any line or circle in the complex plane can be
written in the form
 +C=0,
where A
and C are real numbers with AC<|B|
Proof.
For A=0
the equation reduces to
 +C=0.
Letting z=x+iy and B=
we get ( Then
 which
represents a line.
Now the
equation of a circle with center
 and
radius
is
or
 .
Thus
where
For
the equation above clearly represents a circle.
In the
extended complex plane, lines can regarded as circles
through
As a
matter of fact, in the case
replacing z by
 transforms
the resulting equation into
which for
is a circle. Hence we group circles and lines in one
category.
Since
z=w-b
it follows that under these three transformations “lines
and circles” are mapped into “lines and circles”. Now we
show that this is also true for any bilinear
transformation:
If
 then
implies that the result is true in this case.
If
then by dividing the numerator and denominator in the
general bilinear transformation by
 ,
we may write the transformation as
Making the substitution
 =z+d
yields
which can be regarded as an inversion
followed by a magnification and rotation
( ),
followed by a translation
 .
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