Exercise 20.  Show that  [Graphics:Images/ComplexFunReciprocalModHome_gr_856.gif]  is mapped onto the point  [Graphics:Images/ComplexFunReciprocalModHome_gr_857.gif]  on the Riemann sphere.  

Solution 20.

See text and/or instructor's solution manual.

Solution.  The cartesian form of the point  [Graphics:../Images/ComplexFunReciprocalModHome_gr_858.gif]  in three dimensional space is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_859.gif].  
The Riemann sphere sphere is centered at  [Graphics:../Images/ComplexFunReciprocalModHome_gr_860.gif]  and has radius  [Graphics:../Images/ComplexFunReciprocalModHome_gr_861.gif].
If we use the notation  [Graphics:../Images/ComplexFunReciprocalModHome_gr_862.gif]  for an arbitrary point in three dimensional space [Graphics:../Images/ComplexFunReciprocalModHome_gr_863.gif] then the equation of the Riemann sphere is  [Graphics:../Images/ComplexFunReciprocalModHome_gr_864.gif].  

      The line segment L from the north pole [Graphics:../Images/ComplexFunReciprocalModHome_gr_865.gif] to the point   [Graphics:../Images/ComplexFunReciprocalModHome_gr_866.gif].  is given by the vector function

            [Graphics:../Images/ComplexFunReciprocalModHome_gr_867.gif]

and we must find the point   [Graphics:../Images/ComplexFunReciprocalModHome_gr_868.gif]  where L intersects the Riemann sphere.  The coordinate functions for [Graphics:../Images/ComplexFunReciprocalModHome_gr_869.gif]  are  

            [Graphics:../Images/ComplexFunReciprocalModHome_gr_870.gif].

      Substitute these values into [Graphics:../Images/ComplexFunReciprocalModHome_gr_871.gif]  and get:

            [Graphics:../Images/ComplexFunReciprocalModHome_gr_872.gif]   

    Now substitute [Graphics:../Images/ComplexFunReciprocalModHome_gr_873.gif] into  [Graphics:../Images/ComplexFunReciprocalModHome_gr_874.gif]  and get  [Graphics:../Images/ComplexFunReciprocalModHome_gr_875.gif].

Therefore,   [Graphics:../Images/ComplexFunReciprocalModHome_gr_876.gif].

We are done.   

Aside.  The lower half of the Riemann sphere corresponds to the unit disk   [Graphics:../Images/ComplexFunReciprocalModHome_gr_877.gif]  and the upper half of the Riemann sphere corresponds to the exterior of the unit circle [Graphics:../Images/ComplexFunReciprocalModHome_gr_878.gif], i. e. the region  [Graphics:../Images/ComplexFunReciprocalModHome_gr_879.gif].   

          

              [Graphics:../Images/ComplexFunReciprocalModHome_gr_880.gif]            [Graphics:../Images/ComplexFunReciprocalModHome_gr_881.gif]

  

                    The Riemann hemispheres corresponding to the unit disk  [Graphics:../Images/ComplexFunReciprocalModHome_gr_882.gif]  and the region  [Graphics:../Images/ComplexFunReciprocalModHome_gr_883.gif].  

Aside.  Consider the inverse transformation  [Graphics:../Images/ComplexFunReciprocalModHome_gr_884.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_885.gif],  and the equations

[Graphics:../Images/ComplexFunReciprocalModHome_gr_886.gif]  and  [Graphics:../Images/ComplexFunReciprocalModHome_gr_887.gif].  When these values are substituted into [Graphics:../Images/ComplexFunReciprocalModHome_gr_888.gif]  we get  [Graphics:../Images/ComplexFunReciprocalModHome_gr_889.gif].

 

[Graphics:../Images/ComplexFunReciprocalModHome_gr_890.gif]

                    [Graphics:../Images/ComplexFunReciprocalModHome_gr_891.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_892.gif][Graphics:../Images/ComplexFunReciprocalModHome_gr_893.gif]

 

 

[Graphics:../Images/ComplexFunReciprocalModHome_gr_894.gif]

                    [Graphics:../Images/ComplexFunReciprocalModHome_gr_895.gif]

 

 

[Graphics:../Images/ComplexFunReciprocalModHome_gr_896.gif]

                    [Graphics:../Images/ComplexFunReciprocalModHome_gr_897.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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