Revisit and Explore Solution 10.7.

Example 10.7.  Find the bilinear transformation that maps the crescent-shaped region that lies inside the disk   and outside the circle   onto a horizontal strip.

Solution.  For convenience we choose    and the image values  ,  respectively.  The ordered triple    gives the circle    a positive orientation and the disk    has a left orientation.  The image points    all lie on the extended u axis, and they determine a left orientation for the upper half-plane  .  Therefore we can use the second implicit formula (Equation (10-21)) to write  

            ,  

which determines a mapping of the disk    onto the upper half-plane  .  Use the fact that   to simplify the preceding equation and get

            

which can be written in the form  

            .

A straightforward calculation shows that the points    are mapped onto the points  

              

respectively.  The points    lie on the horizontal line    in the upper half-plane.  Therefore the crescent-shaped region is mapped onto the horizontal strip  ,  as shown in Figure 10.7.

            Figure 10.7  The mapping  .  

Enter three points and their images and solve for  w = S(z).  For convenience we choose    which are mapped onto the points  ,  respectively.  In this case we remove the terms involving  ,  in the implicit formula, because this implies that  .  

Check our work and look at the images of  .  

And we need to look at the images of three more points;     and  .  

To illustrate the mapping, consider the inverse image of the horizontal strip.  We will need to use the inverse function.

Now plot the graphs using T[w] then the identity map.

We have constructed the transformation   and see that the image of the crescent-shaped region that lies inside the disk   and outside the circle   is the horizontal strip  .  

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