Exercise 1.  Find      for the following functions:

1 (a).   .

Solution 1 (a).

See text and/or instructor's solution manual.

Answer.   .  

Solution.   Here we have    and the numerator    is non-zero at  ,

i. e.  ,  and the denominator    has a simple zero at the point  .

Applying Theorem 7.12 we have    is analytic at the point    and

  where  .  Hence    has a simple pole at  ,  and by Theorem 8.2 the residue is

                       

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > limit( (z-0)*exp(z)/z, z=0 );

               1

We are really done.   

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

Use the fact that  ,  and then get  

                      

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(z)/z, z=0 );

               1

This solution is complements of the authors.

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