Exercise 14.  Show that the transformation    maps the portion of

the first quadrant  ,  that lies outside the circle    onto the first quadrant  .

Solution 14.

Answer.   The image of the domain    

under the mapping      is the first quadrant   .  

Short Solution.   The image of      under      is   ,

then the image of      under      is the first quadrant   .  

Solution.   The portion of the first quadrant     that lies outside the circle     can be written as

                    .

The mapping    can be written as a composition  

                    ,    

where   

                    ,    and   

                    .

The mapping   can be expressed as

                    ,  

                    or       and    .  

Then      implies      implies   .   Here we have   .  

Also,      implies      implies   .  

        Hence, the image of      under      is   ,

which is the portion of the upper half-plane   ,   that lies outside the circle   ,    i. e.

                    .

(i).   First, 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   .

We now illustrate how to use the point      and its image      in the w-plane.  

Here, we can interpret      as the point at infinity along the negative u-axis.  

Thus, the image points    

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

         So that, 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  .  

(ii).   Second, the region    can be given a left orientation by using the points   .

Again, we now illustrate how to use the point      and its image      in the w-plane.  

Here, we can interpret      as the point at infinity along the positive v-axis.  

Thus, the image points    

give the right half plane      a left orientation.  
        
        So that, the image of the upper half plane    under    is  .  

        Now intersect the sets in  (i)  nd  (ii)  and get the conclusion.

        Hence, the image of the region     

under    is the first quadrant   .  

        Therefore, the image of the region    

under the mapping      is the first quadrant   .  

Remark.  The point      at infinity on the Riemann sphere joins all infinity points   

                       in the extended complex plane.

Thus, we are allowed to use      as our first choice for   ,  

and      as our second choice for    .

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 actually computed with directional limits

                    ,    and  

                    .  

          Similarly, the values        and        are computed with limits

                    ,    and  

                    .  

This solution is complements of the authors.

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