Example 7.35.
The goal of this example is to construct an conformal map that
takes a half disk
D
+
=
b

z

<
1
,
Im
z >
0
B
(7
.
73)
to the full unit disk
D
=
{
ζ

<
1
}
. The answer is
not
ζ
=
z
2
because the image
of
D
+
omits the positive real axis, resulting in a disk that has a slit cut out of it:
{
ζ

<
1
,
0
<
ph
ζ <
2
π
}
. To obtain the entire disk as the image of the conformal map,
we must think a little harder. The Frst observation is that the map
z
= (
w
−
1)
/
(
w
+ 1)
that we analyzed in Example 7.24 takes the right half plane
R
=
{
Re
w >
0
}
to the unit
disk. Moreover, it maps the upper right quadrant
Q
=
{
0
<
ph
w <
1
2
π
}
to the half disk
(7.73). Its inverse,
w
=
z
+ 1
z
−
1
(7
.
74)
will therefore map the half disk,
z
∈
D
+
, to the upper right quadrant
w
∈
Q
.
On the other hand, we just constructed a conformal map (7.72) that takes the upper
right quadrant
Q
to the unit disk
D
. Therefore, if compose the two maps — replacing
z
by
w
in (7.72) and then using (7.74) — we obtain the desired conformal map:
ζ
=
i
w
2
+ 1
i
w
2
−
1
=
i
p
z
+ 1
z
−
1
P
2
+ 1
i
p
z
+ 1
z
−
1
P
2
−
1
=
( i + 1)(
z
2
+ 1) + 2( i
−
1)
z
( i
−
1)(
z
2
+ 1) + 2( i + 1)
z
.
The formula can be further simpliFed by multiplying numerator and denominator by i +1,
and so
ζ
=
−
i
z
2
+ 2 i
z
+ 1
z
2
−
2 i
z
+ 1
.
The leading factor
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 Fall '10
 Olver
 Differential Equations, Equations, Partial Differential Equations, Conformal map, Riemann mapping theorem, Peter J. Olver

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