Solution 4.
Solution. Method
I. In Example
10.3 we showed that
maps
the disk
onto
the upper half plane
.
So all that is needed is to to determine the image of the upper half
plane
under
the mapping
.
The boundary of upper half plane
is
the real axis
,
and we can give the upper half plane
a
left orientation by using the points
.
The image points
,
lie on the imaginary axis
, and
give the right half plane
a
left orientation.
Therefore, the image of the upper half plane
under
is
.
Furthermore, as a double-check we can choose the
point
in
the upper half plane
,
then
lies
in the right half plane
,
which leads us to conclude that the image of the upper half
plane
is
the right half plane
.
Now intersect the
images of the disk
and
upper half plane
.
The image of the portion of the disk
that
lies in the upper half plane
under
is the intersection of the upper half-plane
and
the right half plane
, which
is the first quadrant
.
We are done.
Solution. Method
II. In Example
10.3 we showed that
maps
the disk
onto
the upper half plane
.
So all that is needed is to to determine the image of the upper half
plane
under
the mapping
.
The inverse transformation is
and
we get
Then
implies
that
which
in turn implies that
.
Hence the image of the upper half plane
under
the mapping
is
the right half plane
.
Now intersect the
images of the disk
and
upper half plane
. Therefore,
the image of the portion of the disk
that
lies in the upper half plane
under
is the intersection of the upper half-plane
and
the right half plane
, which
is the first quadrant
.
We are really done.
Aside. We can let Mathematica double check our work.
We are really really done.
Aside. We can look
at some graphs of the mapping
.
![[Graphics:../Images/MobiusTranformationModHome_gr_187.gif]](../Images/MobiusTranformationModHome_gr_187.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_189.gif]](../Images/MobiusTranformationModHome_gr_189.gif)
The
image of
under
is
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell