Exercise 14.  Find the image of the first quadrant    under the mapping   .

Solution 14.

Answer.   The image of first quadrant    under the mapping   .
        
is the semi infinite vertical strip   .  

Hint.   Extend the results in Example 10.13.  You can use the information in Figure 10.17.

Short Solution.   The transformation    maps the first quadrant    onto the upper half-plane  .  

Then   maps the upper half-plane    onto the semi-infinite vertical strip .  

Therefore, the image of the first quadrant    under the composition mapping     

is the semi-infinite vertical strip  .  

We might be done.   

Solution.   The first quadrant     can be written as

                    .

The mapping    can be written as a composition  

                    ,    

where   

                    ,    and   

                    .

        The mapping   can be expressed as

                    ,  

                    or       and    .  

Then      implies      implies   .   Here we have   .  

Also,      implies      implies   .  

        Hence, the image of      under      is   ,

which is the upper half-plane   .  

        For the mapping      we will use known properties about the inverse mapping   .

        For the inverse mapping    we must determine the set in the w-plane so that it's image is  .

        Now we can use Equation (5-33) to write  

                    .  

If  ,  then the image of the vertical line    is the curve in the Z-plane given by the parametric equations  

                        and    ,    for    .  

The inequalities    imply that the image points lie

in the upper half-plane  .   

Eliminate v from the parametric equations and the result is  .  

When the image curve approaches the ray  .  

When the image curve approaches the ray  .  

        Hence, the image of the semi infinite vertical strip      

under the inverse mapping     is the upper half-plane  .   

        Hence, the image of the upper half-plane      under the mapping     

is the semi infinite vertical strip     

        Therefore, the image of first quadrant    under the mapping   .
        
is the semi infinite vertical strip   .  

We are done.   

Aside.  We can look at some graphs of the mapping  .

          The mapping   ,   followed by the mapping   .

          Observe the points and their images    and    

                                          

Aside.   In Section 5.5 we introduced the formula  

                      

which is correct at least for values of z in the upper half plane  .

The following compositions experiment with the composition  

                    .

where  

                    

                    ,    and  

                    .

            The mapping   ,   followed by   ,   followed by   .   

          The mapping   ,   followed by   ,   followed by   .   

          Observe the points and their images  ,     and    

                                

We are really really done.   

Remark.   For curiosity, consider the negative square root    in the form    and make this replacement

in the mapping  .  Then the result would be

                    .

Now focus your attention on the term    and notice that it is similar to  .  

In Section 11.8 we will see that the inverse of the mapping  

                        is    .

The Russian scientist Nikolai Egorovich Joukowsky studied the function

            ,

and how it is used to determine the so called "Joukowski airfoil."

We leave it for the reader to investigate the similarities between      and   .

This solution is complements of the authors.

(c) 2008 John H. Mathews, Russell W. Howell