Exercise 5.  Find  ,   when  

5 (a).  .  

Solution 5 (a).

See text and/or instructor's solution manual.

Answer.   

Solution.  Factor the denominator and get

                    ,  

and conclude that    has a zero of order  2  at  ,  and simple zeros at    and  .  

Use the techniques for determining the order of poles: apply  Corollaries 7.5 and 7.8  in Section 7.4 to conclude that

the integrand    

has a pole of order  2  at  ,  and simple poles at    and  .  

                         The point    that lies inside the contour  .  

        Now apply Theorem 8.1 the Cauchy's Residue Theorem.

The point    is the only singularity that lies inside the contour  .  

Since   is pole of order  2,  by Theorem 8.2 the residue is calculated as follows:

                       

The "Residue Calculus" gives the value of the integral:

                                

We are done.   

Aside.  We can let Mathematica double check our work.

Maple can check our work too!

     > limit( diff((z-1)^2*1/((z-1)^2*(z^2+4)),z), z=1 );

               

We are really done.   

Aside.  Both and are capable of finding residues.

     > residue( 1/((z-1)^2*(z^2+4)), z=1 );

               

     > 2*Pi*I*residue(1/((z-1)^2*(z^2+4)),z=1);

               

This solution is complements of the authors.

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