10.3  Mapping Involving Elementary Functions

    In Section 5.1 we saw that the function    is a one-to-one mapping of the fundamental period strip    in the z-plane onto the w-plane with the point    deleted.  Because  ,  the mapping    is a conformal mapping at each point z in the complex plane.  The family of horizontal lines  ,  and segments    form an orthogonal grid in the fundamental period strip.  Their images under the mapping    are the rays    and the circles  ,  respectively.  These images form an orthogonal curvilinear grid in the w-plane, as shown in Figure 10.9. If  ,  then the rectangle    is mapped one-to-one and onto the region  .   The inverse mapping is the principal branch of the logarithm  .  

Figure 10.9  The conformal mapping  .

    In this section we show how compositions of conformal transformations are used to construct mappings with specified characteristics.

Example 10.8.  Show that the transformation    is a one-to-one conformal mapping of the horizontal strip    onto the disk  .  Furthermore, the x-axis is mapped onto the lower semicircle bounding the disk, and the line    is mapped onto the upper semicircle.

Solution.  The function   is the composition of    followed by  .  The transformation maps the horizontal strip    onto the upper half plane  ;  the x axis is mapped on to the positive X axis;  and the line    is mapped onto the negative X axis.  Then the bilinear transformation    maps the upper half plane      onto the disk  ;  the positive X axis is mapped onto the lower semicircle;  and the negative X axis onto the upper semicircle.  Figure 10.10 illustrates the composite mapping.

Figure 10.10  The composite transformation  .

Explore Solution 10.8.

Example10.9.  Show that the transformation    is a one-to-one conformal mapping of the unit disk    onto the horizontal strip  .  Furthermore, the upper semicircle of the disk is mapped onto the line    and the lower semicircle onto  .  

Solution.  The function    is the composition of the bilinear transformation    followed by the logarithmic mapping  .  The image of the disk    under the bilinear mapping    is the right half-plane  ;  the upper semicircle is mapped onto the positive Y axis;  and the lower semicircle is mapped onto the negative Y axis.  The logarithmic function    then maps the right half-plane onto the horizontal strip;  the image of the positive Y axis is the line  ;  and the image of the negative Y axis is the line  .  Figure 10.11 shows the composite mapping.

Figure 10.11  The composite transformation  .

Explore Solution 10.9.

Example 10.10.  Show that the transformation    is a one-to-one conformal mapping of the portion of the unit disk    that lies in the upper half-plane    onto the upper half-plane  .  Furthermore, the upper semicircular portion of the boundary is mapped onto the line negative u-axis, and the segment    is mapped onto the positive u-axis.

Solution.  The function    is the composition of the bilinear transformation    followed by the mapping  .  The image of the half-disk under the bilinear mapping    is the first quadrant  ;  the image of the segment  ,  is the positive X axis;  and the image of the semicircle is the positive Y axis.  The mapping    then maps the first quadrant in the Z plane onto the upper half-plane  ,  as shown in Figure 10.12.

Figure 10.12  The composite transformation  .

Explore Solution 10.10.

Example 10.11.  Consider the function  ,  which is the composition of the functions    and    where the branch of the square root is  ,  where   ,  ,  and  .  Then the transformation    maps the upper half-plane    one-to-one and onto the upper-half plane    slit along the segment  .  

Solution.  The function    maps the upper half-plane    one-to-one and onto the Z-plane slit along the ray  .  Then the function    maps the slit plane onto the slit half-plane, as shown in Figure 10.13.

Figure 10.13  The composite transformation    and the intermediate steps    and  .

Explore Solution 10.11.

Remark 10.1.  The images of the horizontal lines are curves in the w plane that bend around the segment from  .  The curves represent the streamlines of a fluid flowing across the w plane. We discuss fluid flows in more detail in Section 11.7.  

10.3.1  The Mapping  

    The double-valued function    has a branch that is continuous for values of z distant from the origin.  This feature is motivated by our desire for the approximation    to hold for values of z distant from the origin.  We begin by expressing as  

(10-22)        ,  

where the principal branch of the square root function is used in both factors.  We claim that the mapping    is a one-to-one conformal mapping from the domain set , consisting of the z plane slit along the segment  ,  onto the range set , consisting of the w plane
slit along the segment  .  

    To verify this claim, we investigate the two formulas on the right side of Equation (10-22) and express them in the form  

              

    where    and  ,  and  
    
                  

    where    and  .

    The discontinuities of    are points on the real axis such that  ,  respectively.  We now show that is continuous on the ray  .

    We let    denote a point on the ray    and then obtain the following limit as z approaches from the upper half-plane:   

                

    We let    denote a point on the ray    and then obtain the following limits as z approaches from the lower half-plane:  

                  

    We can easily find the inverse mapping and express it similarly:  

            ,  

    where the branches of the square root function are given by  
    
                  

    where  ,  ,  and  ,  
    
                  

    where  ,  ,  and  .

    A similar argument shows that    is continuous for all w except those points that lie on the segment  .  Verification that  

               and   

hold for z in and w in , respectively, is straightforward.  Therefore we conclude that    is a one-to-one mapping from onto .  Verifying that is also analytic on the ray  ,  is tedious.  We leave it as a challenging exercise.  

10.3.2  The Riemann Surface for  

    Using the other branch of the square root, we find that , is a one-to-one conformal mapping from the domain set consisting of the z-plane slit along the segment , onto the range set consisting of the w-plane slit along the segment .  The sets and for and and form the Riemann surface for the mapping, as shown in Figure 10.14.  

Figure 10.14  The mappings    and  .  

    We obtain the Riemann surface for    by gluing the edges of together and the edges of together.  In the domain set, we glue edges  ,  ,  ,  and  .  In the image set, we glue edges  ,  ,  ,  and   .  The result is a Riemann domain surface and Riemann image surface for the mapping, as illustrated in Figures 10.15(a) and 10.15(b), respectively.

Figure 10.15  The Riemann surfaces for the mapping  .