Exercise 1.  Find      for the following functions:

1 (k).   .

Solution 1 (k).

See text and/or instructor's solution manual.

Answer.   

Solution.   Use Definition 7.6 or Theorem 7.10 in Section 7.4 to determine that    has a simple zero at the point .

Applying Definition 7.6 with    and    we compute

                       and   .   

Hence, we conclude that so that    has a zero of order  2  at  .  

Or you can apply Theorem 7.10 and expand   in a Taylor series about the point .

                    .  

The first few coefficients of the Taylor series are  .

Hence, we can conclude that    has a zero of order  2  at  .

        Here we have    and the denominator    has a zero of order  2  at  .

Now apply  Corollary 7.5 to conclude that   has a pole of order  2  at the point  ,  and by Theorem 8.2 the residue is calculated as follows:

                       

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > limit( diff((z-0)^2*1/(z*sin(z)),z), z=0 );

               0

We are really done.   

Alternate Solution.   If you prefer, you can expand    in a Laurent series about  .

                    

This might be quite tedious, so we will show how to do it with Mathematica and Maple.

     > series( 1/(z*sin(z)), z=0,10 );

               

We are really really done.   

Aside.  Both and are capable of finding residues.

     > residue( 1/(z*sin(z)), z=0 );

               0

This solution is complements of the authors.

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