Solution 2 (k).

See text and/or instructor's solution manual.

Answer.      has simple poles at      for   ,  

Solution.   Use a convenient method to determine where the denominator is zero.   

Consider    at the point     ,  here we have

                    ,  
                    
and     

                      

Applying Definition 7.6 we can conclude that    has simple zeros at the points    for  .  

Now apply  Corollary 7.5 to conclude that   has simple poles at      for   ,  

We are  done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

          The simple poles at      for   ,  

                              A plot for   .  

This solution is complements of the authors.

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