Exercise 9. Show
that the function
maps
the rectangle
one-to-one
and onto
the portion of the upper half plane
that
lies inside the ellipse
.
Solution 9.
Hint. Extend the results in Example 10.13. You can use the information in Figure 10.17.
Solution. Using
Equation (5-33), we
write
.
If
, then
the image of the vertical line
is
the curve in the w plane given by the
parametric equations
and
, for
.
The inequalities
imply
that the image points lie in the upper
half-plane
.
Eliminate y from the parametric
equations and the result is
.
When
the
image curve approaches the ray
.
When
the
image curve approaches the ray
.
Hence,
the image of
under
the
upper half-plane
.
Next, using
Equation (5-33), we
write
.
If
, then
the image of the horizontal segment
is
the curve in the w-plane given by the
parametric equations
,
We rewrite them as
.
We now eliminate x from the equations
by squaring and using the trigonometric
identity
. The
result is the single equation
.
This curve is an ellipse in the w-plane
that passes through the points
and
and
has foci at the points
.
The inequality
implies
that
and
, and
we can conclude
that the ellipse
lies
inside the bounding ellipse
.
Hence,
the image of
, lies
inside the bounding ellipse
.
Therefore,
the image of the rectangle
under
the transformation
is
,
which is the portion of the upper half plane
that
lies inside the ellipse
.
We are done.
Aside. We can look
at some graphs of the mapping
.
![[Graphics:../Images/MapTrigonometricFunModHome_gr_698.gif]](../Images/MapTrigonometricFunModHome_gr_698.gif)
![[Graphics:../Images/MapTrigonometricFunModHome_gr_700.gif]](../Images/MapTrigonometricFunModHome_gr_700.gif)
The
mapping
.
Observe
the points
and their images
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell