Exercise 1.  Find      for the following functions:

1 (e).   .

Solution 1 (e).

See text and/or instructor's solution manual.

Answer.   

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

,  "this is a zero of order  ."

The denominator    has simple zero at , "this is a zero of order  ."

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)*cos(z)/sin(z), z=0 );

               1

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( cos(z)/sin(z), z=0,10 );

               

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( cos(z)/sin(z), z=0 );

               1

Extra Stuff.

          Applying Theorem 7.12 and Corollaries 7.5 and 7.8 we know that   has simple zeros at the points    where  n  is an integer,

and     is analytic at the points    and    where  n  is an integer.  

Hence,    has a simple poles at the points    where  n  is an integer,  and the residues are

                      

This solution is complements of the authors. 

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