Example 2.22.  Show that the image of the right half plane  [Graphics:Images/ComplexFunReciprocalMod_gr_71.gif]  under the mapping  [Graphics:Images/ComplexFunReciprocalMod_gr_72.gif]  is the closed disk  [Graphics:Images/ComplexFunReciprocalMod_gr_73.gif]  in the w-plane.  

Explore Solution 2.22.

Enter the function f[z] and find its inverse.

[Graphics:../Images/ComplexFunReciprocalMod_gr_83.gif]



[Graphics:../Images/ComplexFunReciprocalMod_gr_84.gif]

 

 

 

Solve for the inverse function and the real and imaginary parts.

[Graphics:../Images/ComplexFunReciprocalMod_gr_85.gif]



[Graphics:../Images/ComplexFunReciprocalMod_gr_86.gif]

[Graphics:../Images/ComplexFunReciprocalMod_gr_87.gif]

 

 

Now find the image of the right half plane  [Graphics:../Images/ComplexFunReciprocalMod_gr_88.gif].  

[Graphics:../Images/ComplexFunReciprocalMod_gr_89.gif]



[Graphics:../Images/ComplexFunReciprocalMod_gr_90.gif]

 

 

This last inequality  [Graphics:../Images/ComplexFunReciprocalMod_gr_91.gif]  is the same as  [Graphics:../Images/ComplexFunReciprocalMod_gr_92.gif]  and is the disk of radius 1 centered at  w = 1  in the w-plane.

[Graphics:../Images/ComplexFunReciprocalMod_gr_93.gif]



[Graphics:../Images/ComplexFunReciprocalMod_gr_95.gif]

[Graphics:../Images/ComplexFunReciprocalMod_gr_97.gif]


We see that the image of the right half plane  [Graphics:../Images/ComplexFunReciprocalMod_gr_98.gif] under the reciprocal transformation  [Graphics:../Images/ComplexFunReciprocalMod_gr_99.gif]   is the disk  [Graphics:../Images/ComplexFunReciprocalMod_gr_100.gif]  in the w-plane.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c) 2006 John H. Mathews, Russell W. Howell