Solution 11 (a).

See text and/or instructor's solution manual.

Answer.     has simple poles at    for  ,  and a non-isolated essential singularity at the origin.

Solution.   We know that    has simple zeros at    for  .  

Hence,    has simple zeros at     for  .  

We have shown in Exercise 3 (b) that   has a non-isolated essential singularity at the origin,

i. e. every neighborhood contains points in the sequence  .

Then apply  Corollary 7.6 to conclude that    has simple poles at    for  ,  and a non-isolated essential singularity at the origin.

We are done.   

Aside.  We can look at the Laurent series expansion for    that was developed in Exercise 3 (b).

                      

It is possible to use a division algorithm to obtain the Laurent series for  ,  

but this is tedious work and we will not show the details.  

                      

It is interesting, that Mathematica can easily obtain some of the terms.

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