Solution 15.

Answer.  The image of the horizontal strip     is the region that lies exterior to both of the circles  

    and    ,  

i.e. the image region is   .

Solution.   Method I.   Part I.  The boundary of the upper half plane    is the real axis   ,  

and we can give this boundary a positive orientation by using the points  .

The image points  ,  

all lie on the circle    and make the circle

a positively oriented boundary for the region  .  

Therefore, the image of the upper half plane    under the mapping    is the region  .  

Furthermore, as a double-check we can choose the point    in the upper half plane,  then    lies in the region  ,

which leads us to conclude that the image of    is the exterior of the circle  .  

Solution.   Method I.   Part II.  The boundary of the lower half plane    is the horizontal line   ,  

and we can give this boundary a positive orientation by using the points  .

The image points  ,  

all lie on the circle    and make the circle

a positively oriented boundary for the region  .  

Therefore, the image of the lower half plane    under the mapping    is the region  .  

Furthermore, as a double-check we can choose the point    in the upper half plane,  then    lies in the region  ,

which leads us to conclude that the image of    is the exterior of the circle  .  

Solution.   Method I.   Conclusion.  

Therefore, the image of the horizontal strip     is the region that lies exterior to both of the circles  

    and    ,  

i.e. the image region is   .

We are done.   

Aside.  We can let Mathematica double check our work.

The x-axis is given positive orientation by using the points ,  ,  and  .  

Check our work and by looking at the images of  .  

The image points  ,  all lie on the circle  .  

The horizontal line    is given positive orientation by using the points ,  ,  and  .  

Check our work and by looking at the images of  .  

The image points  ,  all lie on the circle  .  

We are really done.   

Solution.  Method II.   Start by finding the inverse transformation for   .

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

(10-13)             ,

(10-14)             .

Here we have    and   .  

Then  

                
            
Hence, the inverse transformation is   .  

Then get  

                       

Then    implies that    which implies that  ,  which in turn implies that  .  

Therefore, the image of the upper half plane    under    is  .  

Also,    implies that  

                       
                    

Therefore, the image of the lower half plane    under    is  .  

Solution.   Method II.   Conclusion.  

Therefore, the image of the horizontal strip     is the region that lies exterior to both of the circles  

    and    ,  

i.e. the image region is   .

We are really really done.   

Aside.  We can let Mathematica double check our work.

We are really really really done.   

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

                    The image of the horizontal strip    under    is the region

                    that lies exterior to both of the circles    and  .

This solution is complements of the authors.

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