ACM95a/100a
Last updated December 2, 2009
Problem Set VIII Solutions
Prepared by Faisal Amlani
1. (20 pts.) Find a M¨obius transformation mapping the lower halfplane to the disk

w
+ 1

<
1.
[Hint: Do it in steps.]
SOLUTION
:
The conformal transformation mapping one region to another is never unique. One possible
mapping begins with
w
1
=
f
(
z
) =
z
+
i
z

i
.
Checking the image of the boundary of the lower half plane, we see that
w
1
(

1) =

i, w
1
(0) =

1
, w
1
(1) =
i.
w
1
maps the line through

1
,
0
,
1 through a line or circle connecting

i,

1
, i.
Since a line
cannot connect these three points, the map can only be a circle
=
⇒
w
1
maps the real axis
boundary to the unit circle centered at the origin. Since the function must map the lower half
plane either to the interior or exterior of this circle, and since
w
1
(

i
) = 0
,
we must have that
w
1
maps the lower half plane to the interior. Now to recenter at
z
=

1
,
we merely translate
with
w
=
g
(
w
1
) =
w
1

1
.
Putting it together gives the map
w
=
g
(
w
1
) =
w
1

1 =
f
(
z
)

1 =
z
+
i
z

i

1 =
2
i
z

i
.
2. (20 pts.) Map the region between the two cirles

z

= 2 and

z

1

= 1 conformally onto the
upper half plane. [Hint: Use a M¨obius transformation to map the point 2 to
∞
. Show that the
image region is a
strip
. Then use the exponential map, after modifying (translating, rotating) the
strip, if necessary.]
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SOLUTION
:
Consider a map
w
1
=
f
(
z
) =
1
2

z
.
This clearly maps the point
z
= 2 to
∞
.
To show that the image is a strip, we can check
where key points on each circle are mapped:
2
→
∞
0
→
1
/
2
1 +
i
→
1
2
(1 +
i
)
1

i
→
1
2
(1

i
)

2
→
1
/
4
2
i
→
1
4
(1 +
i
)

2
i
→
1
4
(1

i
)
.
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 Fall '09
 PROF
 Upper halfplane, half plane, upper half plane, Modular form

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