Exercise 10.  Find the image of the portion of the upper half-plane    that lies outside the circle    

under the transformation  .  

Solution 10.

Remark.  Compare this exercise with Exercise 9.

Answer.   The image of the portion of the upper half-plane    that lies outside the circle    

under the mapping      is the horizontal strip   .   

Short Solution.   The image of    under    is Quadrant II: ,  

then the image of    under    is the horizontal strip   .   

Solution.   The portion of the upper half-plane    that lies outside the circle   is the domain

                    .

The mapping    can be written as a composition  

                    ,    

where   

                    ,    and   

                    .

        Start by finding the inverse transformation for   .

Use equations (10-13)  and  (10-14).  

(10-13)             ,

(10-14)             .

Here we have    and   .  

Then  

                        

Thus, the inverse transformation is   .  

Then get  

                       

Then      implies    


                       

Thus,       implies    .    

Also,      implies        implies        implies    .    

        Hence, the image of the portion of the upper half-plane    that lies outside the circle    

under the mapping      is   Quadrant II:  .

        Notice that Quadrant II:    can be written in the form  .

Recall that      which can be written as  

                       and   .

Then,      implies that      in turn implies that   .  

Also,      implies that   .   

        Hence, the image of      under    is   .

        Therefore, the image of the portion of the upper half-plane    that lies outside the circle     

under the mapping      is the horizontal strip   .   

We are done.   

          The mapping   ,   followed by the mapping   .

          Observe the points and their images    and    

                                                            

We are really done.   

Aside.   If it is important then we can include the following details.

          Here the values    ,        and        are calculated as limits.

                    ,   

                    ,  

                    .  

          Similarly, the values    ,    and    ,    and        are calculated as limits.

                    ,  

                    ,  

                    .

This solution is complements of the authors.

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