Exercise 15.  Find the image of the quadrant    under the mapping  .  

Make sketches of the domain set and range set.

Hint.  Consider the points  ,  ,  ,  ,    in the z-plane
and their images  ,  ,  ,  ,    in the w-plane.

Solution 15.

See text and/or instructor's solution manual.

Answer.  The intersection of    
and  ,  
i. e.  the image region is  .  

Solution using algebra.

      The inverse mapping is  .   

Now use the substitution    and get:

          

Thus, the image of the right half-plane    is the disk  .

And get:  

          

Thus, the image of the right half-plane    is the disk  .

Therefore, the image of the quadrant    under the mapping   is the intersection of these two disks, i.e.  

.  

We are done.   

                    The quadrant    and the image region  .

                    The points  ,   ,   ,   ,   ,   and
                    the images   ,   ,   ,   ,   .

Solution using point images.  

       We need to show that the image of the line    in the z-plane is the circle    in the w-plane, and the latter will be a boundary curve of the image region. Then show that the image of the line    in the z-plane is the circle    in the w-plane, and the latter will be a boundary curve of the image region.  The image is the region that is bounded by the two circles and .

First consider the points  ,  ,  ,  ,    in the z-plane

and their images  ,  ,    in the w-plane.

First consider the points  ,  ,    lie on the vertical line   in the z-plane.  

The image points  ,  ,  ,  lie on the the circle    in the w-plane,

and this is shown by observing that the center of the circle    is  ,  and then making the computations:  

        ,  

        ,  and  

        .  

and we have verified this observation.   

This verifies our claim that the vertical line    is the circle  .  

Second consider the points  ,  ,    in the z-plane

and their images  ,  ,    in the w-plane.

The points  ,  ,    lie on the horizontal line   in the z-plane.  

The image points  ,  ,    lie on the circle    in the w-plane,  

and this is shown by observing that the center of the circle    is  ,  and then making the computations:  

,  

,  and  

.  

This verifies our claim that the image of the horizontal line   is the circle  .  

Furthermore, we observe that the point    is mapped onto the point     so the image region lies inside  .

Therefore, the image of the quadrant    is the region  .  

Aside.  We can let Mathematica double check our work.

                    The quadrant    and the image region  .

                    The points  ,   ,   ,   ,   ,   and
                    the images   ,   ,   ,   ,   .

This solution is complements of the authors.

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