Solution 7.
Answer.
.
Solution. Method
I. Use the implicit
formula
.
Substitute the values given above and get
,
,
then simplify and get
.
Solving for w we obtain
Therefore,
.
We are done.
Aside. We can let Mathematica double check our work.
We are really done.
Solution. Method
II. The general form of a bilinear
transformation is
, and
it is not the case that both
.
So the desired formula must have one of the following two forms:
either
or
.
Let us assume that the first form
is
the one that works out.
Then we can set up three equations to solve
for
:
,
then simplify these equations get the system of
equations
Add row 1 to row 3 and get
![[Graphics:../Images/MobiusTranformationModHome_gr_325.gif]](../Images/MobiusTranformationModHome_gr_325.gif)
Divide row 2 by 1 and subtract it from row 1 to get
![[Graphics:../Images/MobiusTranformationModHome_gr_326.gif]](../Images/MobiusTranformationModHome_gr_326.gif)
Use
to
rewrite the second equation as
and
then obtain
.
Use
to
rewrite the third equation as
and
then obtain
.
Substituting these into
produces
the desired result:
.
We are really really done.
Aside. We can let Mathematica double check our work.
We are really really really done.
Aside. We can look
at some graphs of the mapping
.
![[Graphics:../Images/MobiusTranformationModHome_gr_345.gif]](../Images/MobiusTranformationModHome_gr_345.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_348.gif]](../Images/MobiusTranformationModHome_gr_348.gif)
The
image of the left half-plane
under
is
the disk
.
![[Graphics:../Images/MobiusTranformationModHome_gr_353.gif]](../Images/MobiusTranformationModHome_gr_353.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_355.gif]](../Images/MobiusTranformationModHome_gr_355.gif)
The
image of the disk
under
is
the upper half-plane
.
![[Graphics:../Images/MobiusTranformationModHome_gr_360.gif]](../Images/MobiusTranformationModHome_gr_360.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_362.gif]](../Images/MobiusTranformationModHome_gr_362.gif)
The
image of the right half-plane
under
is
the disk
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell