The two functional equations, 57.1.24 and 57.1.17 along with some other properties
presented above are of fundamental importance. For convenience, they are summarized
here in the following lemma.
Lemma 57.1.18The following functional equations hold for λ.
It follows λ is real on the boundary of Ω in the above picture. This proves the
corollary.
Now, following Alfors [?], cut off Ω by considering the horizontal line segment,
z = a + ib_{0} where b_{0} is very large and positive and a ∈
[0,1]
. Also cut Ω off by the
images of this horizontal line, under the transformations z =
1
τ
and z = 1 −
1
τ
.
These are arcs of circles because the two transformations are fractional linear
transformations. It is left as an exercise for you to verify these arcs are situated as
shown in the following picture. The important thing to notice is that for b_{0} large
the points of these circles are close to the origin and
(1,0)
respectively. The
following picture is a summary of what has been obtained so far on the mapping by
λ.
PICT
In the picture, the descriptions are of λ acting on points of the indicated
boundary of Ω. Consider the oriented contour which results from λ
(z)
as z
moves first up l_{2} as indicated, then along the line z = a + ib and then down
l_{1} and then along C_{1} to C and along C till C_{2} and then along C_{2} to l_{2}. As
indicated in the picture, this involves going from a large negative real number to a
small negative real number and then over a smooth curve which stays small
to a real positive number and from there to a real number near 1. λ
(z)
stays
fairly near 1 on C_{1} provided b_{0} is large so that the circle, C_{1} has very small
radius. Then along C, λ
(z)
is real until it hits C_{2}. What about the behavior of λ
on C_{2}? For z ∈ C_{2}, it follows from the definition of C_{2} that z = 1 −
-1
τ
where
τ is on the line, a + ib_{0}. Therefore, by Lemma 57.1.20, 57.1.17, and 57.1.24
These points are essentially on a large half circle in the upper half plane which has radius
approximately
eπb0
16
.
Now let w ∈ ℂ with Im
(w)
≠0. Then for b_{0} large enough, the motion over
the boundary of the truncated region indicated in the above picture results
in λ tracing out a large simple closed curve oriented in the counter clockwise
direction which includes w on its interior if Im
(w)
> 0 but which excludes w if
Im
(w )
< 0.
Theorem 57.1.22Let Ω be the domain described above. Then λ maps Ω one toone and onto the upper half plane of ℂ,
{z ∈ ℂ such that Im (z) > 0}
. Also, theline λ
(l1)
=
(0,1)
,λ
(l2)
=
(− ∞, 0)
, and λ
(C)
=
(1,∞ )
.
Proof: Let Im
(w )
> 0 and denote by γ the oriented contour described above and
illustrated in the above picture. Then the winding number of λ ∘ γ about w equals 1.
Thus
1 ∫ 1
--- -----dz = 1.
2πi λ∘γ z − w
But, splitting the contour integrals into l_{2},the top line, l_{1},C_{1},C, and C_{2} and changing
variables on each of these, yields
1 ∫ λ′(z)
1 = 2πi λ-(z)−-w-dz
γ
and by the theorem on counting zeros, Theorem 51.6.1 on Page 5656, the function,
z → λ
(z)
− w has exactly one zero inside the truncated Ω. However, this shows this
function has exactly one zero inside Ω because b_{0} was arbitrary as long as it is
sufficiently large. Since w was an arbitrary element of the upper half plane, this
verifies the first assertion of the theorem. The remaining claims follow from the
above description of λ, in particular the estimate for λ on C_{2}. This proves the
theorem.
Note also that the argument in the above proof shows that if Im
(w )
< 0, then w is
not in λ
(Ω)
. However, if you consider the reflection of Ω about the y axis, then it will
follow that λ maps this set one to one onto the lower half plane. The argument will
make significant use of Theorem 51.6.3 on Page 5663 which is stated here for
convenience.
Theorem 57.1.23Let f : B
(a,R )
→ ℂ be analytic and let
f (z) − α = (z − a)m g(z),∞ > m ≥ 1
where g
(z)
≠0 in B
(a,R )
. (f
(z)
− α has a zero of order m at z = a.) Then there existε,δ > 0 with the property that for each z satisfying 0 <
|z − α|
< δ, there existpoints,
{a ,⋅⋅⋅,a } ⊆ B (a,ε) ,
1 m
such that
−1
f (z)∩ B(a,ε) = {a1,⋅⋅⋅,am }
and each a_{k}is a zero of order 1 for the function f
(⋅)
− z.
Corollary 57.1.24Let Ω be the region above. Consider the set of points, Q =
Ω∪ Ω^{′}∖
{0,1}
described by the following picture.
PICT
Then λ
(Q )
= ℂ∖
{0,1}
. Also λ^{′}
(z)
≠0 for every z in ∪_{k=−∞}^{∞}
(Q + 2k)
≡ H.
Proof: By Theorem 57.1.22, this will be proved if it can be shown that
λ
′
(Ω )
=
{z ∈ ℂ : Im (z) < 0}
. Consider λ_{1} defined on Ω^{′} by
----------
λ1(x+ iy) ≡ λ(− x+ iy).
Claim:λ_{1} is analytic.
Proof of the claim: You just verify the Cauchy Riemann equations. Letting
λ