Exercise 5. Find
the image of the line
under
the reciprocal transformation
.
Make sketches and indicate the points
in
the z-plane and their images
in the w-plane.
Solution 5.
See text and/or instructor's solution manual.
Answer. The circle ![]()
.
The points
are
mapped onto
.
![[Graphics:../Images/ComplexFunReciprocalModHome_gr_188.gif]](../Images/ComplexFunReciprocalModHome_gr_188.gif)
Graph of the line
, using
the parametric equations
, and
the image circle
, using
the parametric equations
, and
.
Solution using algebra.
The inverse mapping
is ![]()
.
Now use the substitution
and
get:
Solution. The inverse mapping is ![]()
.
Now use the substitution
and
get:
This is an equation of the circle ![]()
.
We are done.
![[Graphics:../Images/ComplexFunReciprocalModHome_gr_204.gif]](../Images/ComplexFunReciprocalModHome_gr_204.gif)
The
line
and
the image circle
, divide
the z-plane and w-plane
into two pieces.
Solution using point images.
The points
lie
on the line
in
the z-plane.
Their image points are
, and
these three points
determine
the circle
![]()
in
the w-plane.
If we observe that the center of the circle
is
, then
an easy calculation shows that
,
, and
,
and we have verified this observation.
Aside. We can let Mathematica double check our work.
![[Graphics:../Images/ComplexFunReciprocalModHome_gr_218.gif]](../Images/ComplexFunReciprocalModHome_gr_218.gif)
![[Graphics:../Images/ComplexFunReciprocalModHome_gr_220.gif]](../Images/ComplexFunReciprocalModHome_gr_220.gif)
The
line
and
the image circle
.
The
points
and
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell