Extra Problems Sheet 6
Stephen Taylor
May 10, 2005
1. Show that a M¨
obius transformation
T
(
z
) can have at most two fixed points
in the complex plane unless
T
(
z
)
≡
z
.
For
a
0
, b
0
, c
0
, d
0
∈
C
, let
T
(
z
)
≡
a
0
z
+
b
0
c
0
z
+
d
0
be a M¨
obius transformation. Since we may assume
a
0
= 0 we multiply
T
(
z
) by
a

1
0
/a

1
0
to obtain
T
(
z
) =
z
+
b
0
/a
0
c
0
/a
0
z
+
d
0
/a
0
≡
z
+
a
bz
+
c
from which we observe that the transformation has three degrees of freedom.
This follows analytically from the fact that
z
=
z
+
b
cz
+
d
→
z
=
(1

d
)
±
(
d

1)
2

4
cb
2
c
are the fixed points of the transformation.
2. Discuss the image of the circle

z

2

= 1 and its interior under the map
T
(
z
) =
z

1
.
Since this map is a M¨
obius transformation if maps circles to circles. So it
suffices to check the image of three points on the given domain to obtain the
image of the entire map. So we note
3
→
1
/
3
2 +
i
→
1
/
5(2

i
)
1
→
1
. So the image of the circle is the unit circle determined by the points
1
/
3
,
1
/
5(2

i
), and 1. Moreover the point 2
→
1
/
2 which is in the interior of
the circle in the
w
plane. So we conclude the mapping takes the interior of the
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 Three '09
 Cong
 Math, three degrees, Unit disk, Stephen Taylor, M¨bius transformation

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