Exercise 12. Find
the image of the upper half-plane
under
.
Solution 12.
Answer. The
image of the upper half-plane
under ![]()
is the horizontal strip
slit
along the ray
.
Short
Solution. The image
of
under
is
, then
the image of
under
is
the horizontal strip
slit
along the ray
.
Solution. The
mapping
can
be written as a composition
,
where
, and
.
The upper half-plane
can
be expressed in the form
.
The
mapping
can be expressed as
,
or
and
.
Then
implies
implies
.
We can solve
and
get
.
Then
implies
implies
which
in turn implies
.
Thus, the image of
under
is
,
which is the Z-plane slit along the
ray
.
However, the
mapping at hand is
and
it shifts the above region S to
the right by the amount
.
Thus, the image of
under
is
,
which is the Z-plane slit along the
ray
.
It helps if we
determine the image of the set
.
Use
and
the formulas
and
.
Then
implies
that
which
in turn implies that
.
Also,
implies
that
.
Hence,
the image
under
the mapping ![]()
is the horizontal strip
. However,
the required image will be missing a certain ray.
Recall that the
domain of
is
the Z-plane slit along the
ray
.
But our set T is the Z-plane
slit along the ray
.
So the image of T will be the horizontal
strip ![]()
minus the image of the interval
.
Here we have
implies
that
which
in turn implies that
.
Also, we have
so
that
.
Thus, the image of the segment
under
is the ray
.
Hence,
the image of the set
under
the mapping
is the horizontal strip
slit
along the ray
.
Therefore,
the image of the upper half-plane
under ![]()
is the horizontal strip
slit
along the ray
.
We are done.
![[Graphics:../Images/MapElementaryFunModHome_gr_656.gif]](../Images/MapElementaryFunModHome_gr_656.gif)
![[Graphics:../Images/MapElementaryFunModHome_gr_657.gif]](../Images/MapElementaryFunModHome_gr_657.gif)
The
mapping
, followed
by the mapping
.
Observe
the points
and their images
and
The image of the branch cut for
has
four pieces which are shown with dashed lines and rays in the
w-plane.
The point
can
be explained in more detail, and depends on which quadrant you
approach
,
this requires visualizing the four values of zero:
.
We are really done.
Aside. If it is important then we can include the following details.
Here
the values
,
,
,
,
,
are
calculated as limits.
,
,
,
,
,
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell