Math 814 HW 3
October 16, 2007
p. 54: 9, 14, 18, 24, 25, 26
p.54, Exercise 9.
If
Tz
=
az
+
b
cz
+
d
, find necessary and sufficient conditions for
T
to
preserve the unit circle.
T
preserves the unit circle iff

ae
iθ
+
b

=

ce
iθ
+
d

,
for all
θ
∈
[0
,
2
π
)
. Squaring both sides and multiplying out, we get

a

2
+

b

2
+
a
¯
be
iθ
+ ¯
abe

iθ
=

c

2
+

d

2
+
c
¯
de
iθ
+ ¯
cde

iθ
.
Comparing coefficients, we get two equations:

a

2
+

b

2
=

c

2
+

d

2
,
a
¯
b
=
c
¯
d.
Dividing the first by

d

2
and using the second, we get

c

2

b

2
+

b

2

d

2
=

c

2

d

2
+ 1
,
which can be written

c

2
 
b

2

b

2
=

c

2
 
b

2

d

2
.
If the numerators are zero, this means

c

=

b

,
hence

a

=

d

.
If the numerators are nonzero, we have

b

=

d

,
hence

a

=

c

.
1
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In the first case, there are
u, v
with

u

=

v

= 1
such that
c
=
ub,
d
=
va.
Then we have
a
¯
b
=
c
¯
d
=
ub
¯
v
¯
a,
so
ub
¯
b
=
va
¯
a
.
This last number, call it
λ
, also has

λ

= 1
, and we have
a
b
c
d
=
1
0
0
λ
a
b
¯
b
¯
a
.
(1)
In the second case, there are
u, v
with

u

=

v

= 1
such that
d
=
ub,
c
=
va.
This time, we have
a
¯
b
=
c
¯
d
=
va
¯
u
¯
b,
so
v
¯
u
= 1
. But
¯
u
=
u

1
, so
v
=
u
, and we have
ad

bc
=
aub

bua
= 0
, which
is illegal for a M¨obius transformation. So the second case does not occur, and all
such transformations
T
are of the form (1).
p.54, Exercise 14.
Let
G
be the region between two circles inside one another,
tangent at the point
a
. Map
G
conformally to the open unit disk.
The map
Tz
= (
z

a
)

1
sends
a
to
∞
, hence the circles go to lines
,
. If
and
are not parallel, they must meet at some point
z
∈
C
. Then
T

1
z
would be
a point on both circles other than
a
. Since there is no such point, the lines
,
are
parallel. The original region
G
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 Three '09
 Cong
 Math, Unit Circle, AZ, M¨ bius, M¨ bius transformation

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