Exercise 13.  Find the image of the first quadrant    under the transformation   .  

Solution 13.

Answer.   The image of      under      is   .  

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

Short Solution.   For the mapping    determine the set in the w-plane so that it's image is  .

The image of the semi infinite vertical strip      under the mapping     

is known to be the first quadrant  .   Therefore, the image of the first quadrant    

under the mapping      is the semi infinite vertical strip   .  

We might be done.   

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

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    and    imply that the image points lie

in the first quadrant  .  

Eliminate v from the parametric equations and the result is  .  

When the image curve approaches the y-axis where  .  

When the image curve approaches the ray  .  

        Hence, the image of the semi infinite vertical strip    

under the inverse mapping    is the first quadrant  .  

        Therefore,  the image of the 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   .

                              Observe the points and their images    

                                                    

We are really done.   

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