Exercise 14.  Show that the half-plane  [Graphics:Images/ComplexFunReciprocalModHome_gr_585.gif]  is mapped onto the disk  [Graphics:Images/ComplexFunReciprocalModHome_gr_586.gif]  by the reciprocal transformation.  

Make sketches of the domain set and range set.

Hint.  Consider the points  [Graphics:Images/ComplexFunReciprocalModHome_gr_587.gif]  in the z-plane and their images  [Graphics:Images/ComplexFunReciprocalModHome_gr_588.gif]  in the w-plane.

Solution 14.

See text and/or instructor's solution manual.

Solution using algebra.

      The inverse mapping is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_589.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_590.gif].   

Now use the substitution  [Graphics:../Images/ComplexFunReciprocalModHome_gr_591.gif]  and get:

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_592.gif]  

This is an equation of the disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_593.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_594.gif].

We are done.   

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_595.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_596.gif]

  

                    The half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_597.gif]  and the image disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_598.gif].

                    The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_599.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_600.gif].

Solution using point images.  

     The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_601.gif]   lie on the line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_602.gif]  in the z-plane.

The image points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_603.gif]  lie on the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_604.gif]  in the w-plane,

and this is shown by observing that the center of the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_605.gif]  is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_606.gif],  and then making the computations:  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_607.gif],  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_608.gif],  and  

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_609.gif].

This verifies our claim that the image of the line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_610.gif]  is the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_611.gif].  

Furthermore, we observe that the point  [Graphics:../Images/ComplexFunReciprocalModHome_gr_612.gif]  is mapped onto the point  [Graphics:../Images/ComplexFunReciprocalModHome_gr_613.gif]   so the image region lies inside the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_614.gif].

Therefore, the image of the half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_615.gif]  is the  disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_616.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_617.gif].

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexFunReciprocalModHome_gr_618.gif]

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_619.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_620.gif]

  

                    The half-plane  [Graphics:../Images/ComplexFunReciprocalModHome_gr_621.gif]  and the image disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_622.gif].

                    The points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_623.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_624.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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