Solution 5.
See text and/or instructor's solution manual.
Answer.
Solution. The
transformation
maps
the point
onto
the point
.
For the function
we
have
. Since
,
by Theorem 10.1 in Section
10.1 , the mapping is conformal at
and
"angles are preserved."
The
ray
at
make
an angle
. In
Section
10.2, we proved that a bilinear transformation maps the
class of half-planes and disks onto itself.
Hence, the image of a line through the origin under the Möbius
transformation
is
a "circle." Since
, and
the
image of the ray
will be an arc
of
a circle that passes through the points
.
Therefore, the
arc
is
inclined at the angle
at
the point
.
We are done.
Aside. The image of
the point
will be the point
on
the arc
as
shown by the calculation
We are really done.
Aside. We can let Mathematica double check our work.
We are really really done.
Aside. We can explore some graphs.
![[Graphics:../Images/JoukowskiTransModHome_gr_280.gif]](../Images/JoukowskiTransModHome_gr_280.gif)
The
image of the ray
under
is
an arc
of
a circle that passes through
and
.
If
the ray
make an angle
at
then
the arc
is
inclined at the angle
at
the point
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell