Example 10.6.  Show that  the mapping    maps the disk    one-to-one and onto the upper half plane  .  

            Figure 10.6  The bilinear mapping  .

Explore Solution 10.6.

Enter the function  .  

To show  S(z)  is one-to-one, find the inverse function.

To show  S(z)  is onto, use the method of oriented points on the boundary curve.

Therefore, the image of the disk    is the upper half plane  .  In order to graph the image of the disk    under  w = S(z)  we use the change of variable and find the image of    under  .  

To plot the graph we use the shifted disk and the functions d[z] = z-1  and  g[z] = S[z-1].

We see that the transformation    maps the disk    one-to-one and onto the upper half plane  .  

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