Exercise 1.  Find      for the following functions:

1 (m).   .

Solution 1 (m).

See text and/or instructor's solution manual.

Answer.   

Solution.   Here    and the numerator    has zero of order    at  ,  

The denominator    has a zero of order    at the point  .

Applying Corollary 7.8 we can conclude that    has a pole of order    at , 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( (z-0)*(exp(4*z)-1)/(sin(z))^2, z=0 );

               4

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( (exp(4*z)-1)/(sin(z)^2), z=0,8 );

               

Here we see that    

has a simple pole at  ,  and according to Definition 8.1 the residue is  .  

We are really really done.   

Aside.  Both and are capable of finding residues.

     > residue( (exp(4*z)-1)/(sin(z)^2), z=0 );

               4

This solution is complements of the authors.

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