Solution 6.
Answer.
.
Solution. Method
I. Use the implicit
formula
.
Substitute the values given above and get
,
then simplify and get
.
Solving for w we obtain
![[Graphics:../Images/MobiusTranformationModHome_gr_249.gif]](../Images/MobiusTranformationModHome_gr_249.gif)
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 last equation is easy to solve and we
get
and
then the first equation yields
.
Use these values to rewrite the second 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_281.gif]](../Images/MobiusTranformationModHome_gr_281.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_283.gif]](../Images/MobiusTranformationModHome_gr_283.gif)
The
image of the right half-plane
under
is
the upper half-plane
.
![[Graphics:../Images/MobiusTranformationModHome_gr_288.gif]](../Images/MobiusTranformationModHome_gr_288.gif)
![[Graphics:../Images/MobiusTranformationModHome_gr_290.gif]](../Images/MobiusTranformationModHome_gr_290.gif)
The
image of the disk
under
is
the region
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell