
ad
)
W
, which is a linear map
Hence, linear fractional transformations are compositions of up to 3 known transforma
tions which map lines and circles to circles and lines. Hence, they do the same.
Definition 39
(fixed point)
If
f
is a map and
f
(
z
)
=
z
, then
z
is called a fixed point of
f
.
Example 34
Find all the fixed points of the linear fractional transformation
w
=
az
+
v
cz
+
d
, where
ad

bc
=
0
.
We are looking for
z
s.t.
az
+
b
cz
+
d
=
z
Case 1:
c
=
0
a
d
z
+
b
d
=
z
⇒
a
d

1
z
+
b
d
=
0
•
If
a
d
=
1
, then we have a unique solution
z
=

b d
a d

1
•
If
a
d
=
1
and
b
=
0
, then
0
+
b
z
=
z
. This is always true, i.e.
∀
z
∈
C
,
z
is a fixed point
and we have
w
=
az
+
0
0
z
+
d
=
a
d
z
=
z
⇒
this is the identity transformation.
•
If
a
d
=
1
and
b
=
0
, then there is no solution, i.e. no fixed points.
Case 2:
c
=
0
.
az
+
b
cz
+
d
=
z
⇒
az
+
b
=
z
(
cz
+
d
)
=
cz
2
+
dz
⇒
0
=
cz
2
+
(
d

a
)
z

b
⇒
z
=

(
d

a
)
±
(
d

a
)
2
+
4
bc
2
c
So
w
has up to 2 fixed points.
The conclusion is that if
f
is a linear fractional transformation and
f
=
Id
, it has at most
2 fixed points.
Corollary 6
If
f
is a linear fractional transformation and
f
(
z
1
)
=
z
1
,
f
(
z
2
)
=
z
2
,
f
(
z
3
)
=
z
3
, with
z
1
,z
2
,z
3
distinct, then
f
(
z
)
=
z
,
∀
z
∈
C
.
Example 35
Given
z
1
,z
2
,z
3
∈
C
distinct and
w
1
,w
2
,w
3
∈
C
distinct, find all linear fractional transfor
mations
T
with
T
(
z
i
)
=
w
i
57
Consider the following transformation
T
. We claim that it is the one we are looking for.
w

w
1
w

w
2
w
1

w
2
w
1

w
3
=
z

z
2
z

z
3
z
1

z
2
z
1

z
3
⇒
w

w
1
w

w
2
.
z
1

z
2
z
1

z
3
=
z

z
2
z

z
3
.
w
1

w
2
w
1

w
3
⇒
(
w

w
2
)(
z
1

z
2
)(
z

z
3
)(
w
1

w
3
)
=
(
z

z
2
)(
w
1

w
2
)(
w

w
3
)(
z
1

z
3
)
•
Plugging in
z
=
z
3
into the LHS, we get 0, so the RHS must also be 0. But since
z
i
are distinct, it must be true that if
z
3
=
0
, then
z
2
=
0
and
z
1
=
0
and since
w
1
=
w
2
,
the only term in the RHS that can be 0 is
(
w

w
3
)
. So we conclude that
w
=
w
3
,
i.e.
z
=
z
3
⇒
w
=
w
3
.
•
Plugging in
z
=
z
2
into the LHS, we get 0, so the RHS must also be 0. Since
w
1
=
w
3
and the
z
i
are also distinct, the only term that we can get to be 0 from the RHS is
(
w

w
2
)
. So we conclude that
z
=
z
2
⇒
w
=
w
2
•
Plugging in
z
=
z
1
in the LHS, we get 0. We get
(
w

w
2
)(
z
1

z
2
)(
z
1

z
3
)(
w
1

w
3
)
=
(
z
1

z
2
)(
w
1

w
2
)(
w

w
3
)(
z
1

z
3
)
=
0
⇒
(
w

w
2
)(
w
1

w
3
)
=
(
w
1

w
2
)(
w

w
3
)
⇒
w

w
2
=
w
1

w
2
w
1

w
3
(
w

w
3
)
This is linear, so it has only one solution.
This is true for
w
=
w
1
, which is the
unique solution. So once again we get that
z
=
z
1
⇒
w
=
w
1
Note:
T
is unique. Suppose that it wasn’t and that we could find another transformation
S
(
z
i
)
=
w
i
.
Consider
S
○
T

1
(
w
i
)
=
S
(
T

1
(
w
i
))
=
S
(
w
i
)
=
w
i
. So
S
○
T

1
has three fixed points and it
is a linear fractional transformation
⇒
S
○
T

1
=
Id
⇒
S
=
T
.
Example 36
Determine all linear fractional transformations
f
∶
H
→
D
(
0
,
1
)
, where
H
=
{
z
∈
C
∶
I
(
z
)
>
0
}
and
D
(
0
,
1
)
=
{
w
∈
C
∶
w
<
1
}
.
Solution:
We are looking for a transformation of the form
f
(
z
)
=
az
+
b
cz
+
d
with
ad

bc
=
0
.
l
H
D
(0,1)
f(l)
We are looking for linear fractional transformations of the form
f
(
z
)
=
az
+
b
cz
+
d
with
ad

bc
=
0
.
First of all
H
is bounded by the real axis, call it
l
. Thus by the Inertia principle and since
58
we are looking for a bijective mapping, as l.f.t. are continuous, we deduce that
f
(
l
)
is a
circle, which is the boundary of the disk
D
(
0
,
1
)
, the unit circle.
You've reached the end of your free preview.
Want to read all 167 pages?
 Winter '18
 Mr Ali
 lim, Complex number, Limit of a sequence, = Z