Exercise 13.  Find the image of the half-plane  [Graphics:Images/ComplexFunReciprocalModHome_gr_543.gif]  under the mapping  [Graphics:Images/ComplexFunReciprocalModHome_gr_544.gif].  

Make sketches of the domain set and range set.

Hint.  Consider the points  [Graphics:Images/ComplexFunReciprocalModHome_gr_545.gif]  in the z-plane and their images  [Graphics:Images/ComplexFunReciprocalModHome_gr_546.gif]  in the w-plane.

Solution 13.

See text and/or instructor's solution manual.

Answer.  The disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_547.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_548.gif].

Solution using algebra.

     The inverse mapping is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_549.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_550.gif].   

Now use the substitution  [Graphics:../Images/ComplexFunReciprocalModHome_gr_551.gif]  and get:

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_552.gif]  

This is an equation of the disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_553.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_554.gif].

We are done.   

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_555.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_556.gif]

                    The half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_557.gif]  and the image disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_558.gif].

                    The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_559.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_560.gif].

Solution using point images.  

     The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_561.gif]   lie on the line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_562.gif]  in the z-plane.

The image points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_563.gif]  lie on the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_564.gif]  in the w-plane,

and this is shown by observing that the center of the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_565.gif]  is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_566.gif],  and then making the computations:  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_567.gif],  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_568.gif],  and  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_569.gif].

This verifies our claim that the image of the line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_570.gif]  is the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_571.gif].  

Furthermore, we observe that the point  [Graphics:../Images/ComplexFunReciprocalModHome_gr_572.gif]  is mapped onto the point  [Graphics:../Images/ComplexFunReciprocalModHome_gr_573.gif]   so the image region lies inside the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_574.gif].

Therefore, the image of the half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_575.gif]  is the disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_576.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_577.gif].

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexFunReciprocalModHome_gr_578.gif]

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_579.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_580.gif]

  

                    The half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_581.gif]  and the image disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_582.gif].

                    The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_583.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_584.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2008 John H. Mathews, Russell W. Howell