Exercise 11.  Show that the function      maps the portion of the disk     

that lies in the first quadrant onto the portion of the upper half plane    that lies outside the unit circle.

Solution 11.

Remark.  Compare this exercise with Exercises 9 and 10.

Answer.   The image of      under      is     

which is the portion of the upper half plane      that lies in the region   .     

Short Solution.   The image of      under      is   ,

then the image of      under      is   .  

Solution.   The portion of the disk    that lies in the first quadrant can be  expressed as  

                    .  

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    .    

Also,      implies        implies        implies    .    

        Hence, the image of      under the mapping    

is  the portion of the first quadrant    that lies outside the unit circle  ,  i. e.

                    .  

The portion of the first quadrant    that lies outside the unit circle     can be expressed as

                    .  

The mapping    can be written as  

                    ,    where  

                        and    .  

Then      implies      which in turn implies that   .  

Also,      implies      which in turn implies that   .  

        Hence, the image of      under      is    

                    ,   

which is the portion of the upper half plane    that lies in the region  .    

        Therefore, the image of     under the mapping     is

                    .  

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        are calculated as limits.

                    ,  

                    .  

This solution is complements of the authors.

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