Solution 2 (g).

See text and/or instructor's solution manual.

Answer.      has simple poles at    for  ,  and a removable singularity at the origin.  

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

Write  .

We know that    has simple zeros at     for  ,  

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

Also, we know that    has a removable singularity at  ,  and that    is analytic at   .  

Thus    has a removable singularity at  .  

Therefore,     has simple poles at    for  ,  and a removable singularity at the origin.  

We are  done.   

Aside.  We can let Mathematica double check our work.

We are really done.   

          The simple poles at    for  ,  and a removable singularity at the origin.  

                              A plot for   .  

This solution is complements of the authors.

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