Example 10.11.  Consider the function  [Graphics:Images/MapElementaryFunMod_gr_107.gif],  which is the composition of the functions  [Graphics:Images/MapElementaryFunMod_gr_108.gif]  and  [Graphics:Images/MapElementaryFunMod_gr_109.gif]  where the branch of the square root is  [Graphics:Images/MapElementaryFunMod_gr_110.gif],  where  [Graphics:Images/MapElementaryFunMod_gr_111.gif] ,  [Graphics:Images/MapElementaryFunMod_gr_112.gif],  and  [Graphics:Images/MapElementaryFunMod_gr_113.gif].  Then the transformation  [Graphics:Images/MapElementaryFunMod_gr_114.gif]  maps the upper half-plane  [Graphics:Images/MapElementaryFunMod_gr_115.gif]  one-to-one and onto the upper-half plane  [Graphics:Images/MapElementaryFunMod_gr_116.gif]  slit along the segment  [Graphics:Images/MapElementaryFunMod_gr_117.gif].  

Figure 10.13  The composite transformation  [Graphics:Images/MapElementaryFunMod_gr_123.gif]  and the intermediate steps  [Graphics:Images/MapElementaryFunMod_gr_124.gif]  and  [Graphics:Images/MapElementaryFunMod_gr_125.gif].

Explore Solution 10.11.

Enter the function  [Graphics:../Images/MapElementaryFunMod_gr_126.gif].  

[Graphics:../Images/MapElementaryFunMod_gr_127.gif]


[Graphics:../Images/MapElementaryFunMod_gr_128.gif]

 

 

To show  w = f(z) = [Graphics:../Images/MapElementaryFunMod_gr_129.gif]  is one-to-one conformal we need to find the inverse function.

[Graphics:../Images/MapElementaryFunMod_gr_130.gif]



[Graphics:../Images/MapElementaryFunMod_gr_131.gif]

 

 

 

However, in Mathematica  the branch cut for square root is along the negative x-axis.  For graphing purposes, we need the version of the function f(z) with a branch cut along the positive x-axis  [Graphics:../Images/MapElementaryFunMod_gr_132.gif].  

[Graphics:../Images/MapElementaryFunMod_gr_133.gif]



[Graphics:../Images/MapElementaryFunMod_gr_134.gif]

 

 

The image is traced using a graph.

[Graphics:../Images/MapElementaryFunMod_gr_135.gif]





[Graphics:../Images/MapElementaryFunMod_gr_136.gif]

[Graphics:../Images/MapElementaryFunMod_gr_137.gif]

[Graphics:../Images/MapElementaryFunMod_gr_138.gif]

[Graphics:../Images/MapElementaryFunMod_gr_139.gif]


We see that the transformation  [Graphics:../Images/MapElementaryFunMod_gr_140.gif]  maps the upper half-plane  [Graphics:../Images/MapElementaryFunMod_gr_141.gif]  one-to-one and onto the upper-half plane  [Graphics:../Images/MapElementaryFunMod_gr_142.gif]  slit along the segment  [Graphics:../Images/MapElementaryFunMod_gr_143.gif].  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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