Solution 2 (i).

See text and/or instructor's solution manual.

Answer.      has simple poles at    for  ,  and a pole of order  3  at the origin.

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

Here we have    and the denominator    has a zero of order  3  at the point  .  

Now apply  Corollary 7.5 to conclude that   has a pole of order  3  at the point  .

Also, for     the denominator    has a simple zeros at    for  .

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

Therefore,     has simple poles at    for  ,  and a pole of order  3  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 the pole of order 3 at the origin.

                              A plot for   .  

This solution is complements of the authors.

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