Exercise 1.  Find the image of the horizontal line  [Graphics:Images/ComplexFunReciprocalModHome_gr_6.gif] under the reciprocal transformation  [Graphics:Images/ComplexFunReciprocalModHome_gr_7.gif].  

Make sketches and indicate the points  [Graphics:Images/ComplexFunReciprocalModHome_gr_8.gif]  in the z-plane and their images  [Graphics:Images/ComplexFunReciprocalModHome_gr_9.gif]  in the w-plane.

Solution 1.

See text and/or instructor's solution manual.

Answer.  The circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_10.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_11.gif].  

The points [Graphics:../Images/ComplexFunReciprocalModHome_gr_12.gif]  are mapped onto  [Graphics:../Images/ComplexFunReciprocalModHome_gr_13.gif].

          

      [Graphics:../Images/ComplexFunReciprocalModHome_gr_14.gif]    [Graphics:../Images/ComplexFunReciprocalModHome_gr_15.gif]

  

Graph of the horizontal line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_16.gif],  using the parametric equations  [Graphics:../Images/ComplexFunReciprocalModHome_gr_17.gif];  and the image circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_18.gif]  using the parametric
equations  [Graphics:../Images/ComplexFunReciprocalModHome_gr_19.gif],   and   [Graphics:../Images/ComplexFunReciprocalModHome_gr_20.gif].  

Solution using algebra.

      The inverse mapping is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_21.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_22.gif].   

Now use the substitution  [Graphics:../Images/ComplexFunReciprocalModHome_gr_23.gif]  and get:

        [Graphics:../Images/ComplexFunReciprocalModHome_gr_24.gif]  

This is an equation of the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_25.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_26.gif].  

We are done.   

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_27.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_28.gif]

  

                    The horizontal line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_29.gif] and the image circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_30.gif],  divide the z-plane and w-plane into two pieces.

Solution using point images.  

      The points [Graphics:../Images/ComplexFunReciprocalModHome_gr_31.gif]  lie on the horizontal line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_32.gif] in the z-plane.  

Their image points are  [Graphics:../Images/ComplexFunReciprocalModHome_gr_33.gif],  and these three points  [Graphics:../Images/ComplexFunReciprocalModHome_gr_34.gif]  determine the circle

[Graphics:../Images/ComplexFunReciprocalModHome_gr_35.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_36.gif]   in the w-plane.  

If we observe that the center of the circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_37.gif]  is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_38.gif],  then an easy calculation shows that

[Graphics:../Images/ComplexFunReciprocalModHome_gr_39.gif],  

[Graphics:../Images/ComplexFunReciprocalModHome_gr_40.gif],   and  

[Graphics:../Images/ComplexFunReciprocalModHome_gr_41.gif],  

and we have verified this observation .  

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ComplexFunReciprocalModHome_gr_42.gif]

          

     [Graphics:../Images/ComplexFunReciprocalModHome_gr_43.gif]     [Graphics:../Images/ComplexFunReciprocalModHome_gr_44.gif]

  

                    The horizontal line  [Graphics:../Images/ComplexFunReciprocalModHome_gr_45.gif] and the image circle  [Graphics:../Images/ComplexFunReciprocalModHome_gr_46.gif].

                    The points [Graphics:../Images/ComplexFunReciprocalModHome_gr_47.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_48.gif].

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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