Solution 3 (a).

See text and/or instructor's solution manual.

Answer.       valid for   .  

Solution.   Start by writing    .

Expand the expression in brackets by replacing z with z-1 in the geometric series (valid, therefore, for  ),  then multiply by the    term.  Get  

              

Therefore,       is valid for   .  

We are done.   

                    The images of the disk    under   ,  for  .  

                    The image of the disk    under the mapping   .  

Aside.  In Section 10.2 we will learn that the image of the disk    is the disk  

We are really done.   

Aside.  We can let Mathematica double check our work.

This solution is complements of the authors.

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