Solution 1 (g).

See text and/or instructor's solution manual.

Answer.      has simple zeros at   ,  ,   and   .  

Solution Method I.   Factorization reveals that  

                    ,

and it is evident that there are six simple zeros at   ,  ,   and   .  

We are done.   

Solution Method II.   Consider    and    at the point  .  

Then      and    .

Thus  ,  and  .

Applying Definition 7.6 we conclude that    has a zero of order 1  at the point  .  

Similarly, it can be shown that    has a zeros of order 1  at the points  .  

We are really done.   

Solution Method III.   Consider    at the point  .

Expand  f(z) in its Taylor series for centered at    and get

                    ,

and it is evident that    has a zero of order  1  at the point  .  

Remark.  Horner's method could be used to divide    by    and obtain this result.

Similarly,     has zeros of order  1  at the points  ,  ,   and   .  

We are really really done.   

Another Solution using Method III.   Consider    at the point  .

Expand  f(z) in its Taylor series for centered at    and get

                      

and it is evident that    has a zero of order  1  at the point  .  

We are really really really done.   

Aside.  We can let Mathematica double check our work.

We are really really really really done.   

          The zeros    
          of order  , respectively.  

                              A contour plot for   .  

                                   A plot for   .  

This solution is complements of the authors.

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