Solution 13.

Answer.   The portion of the disk    that lies in the upper half plane  .  

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

and we can give the upper half plane    a left orientation by using the points  .

The image points  ,  

lie on the real axis  ,  and give the upper half plane    a left orientation.  

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

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

then    lies in the upper half plane   ,

which leads us to conclude that the image of the upper half plane    is the the upper half plane  .

        Now consider the boundary of the right half plane    is the imaginary axis  ,  

and we can give the right half plane   a left orientation by using the points  .

The image points  ,  give the unit circle  

  a positive orientation and the disk    a left orientation.

Therefore, the image of right half plane    under    is  .  

Furthermore, as a double-check we can choose the point    in the right half plane  ,  

then    lies in the unit disk  ,

which leads us to conclude that the image region lies inside the unit circle  .

We are done.   

Aside.  We can let Mathematica double check our work.

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

Check our work and by looking at the images of  .  

The image points  ,  ,  and    give the upper half plane a positive orientation.

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

Check our work and by looking at the images of  .  

The image points  ,  ,  and    give the unit disk a positive orientation.

Furthermore, as a double-check we can choose the point    in first quadrant   ,

then    lies in the portion of the disk    that lies in the upper half plane  .  

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        implies        implies    ,   

and      implies        implies        implies    .

Therefore, the image of the first quadrant    under  ,

is the portion of the disk    that lies in the upper half plane  .  

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 first quadrant    under    is the region  .  

This solution is complements of the authors.

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