Exercise 3.  Evaluate  

3 (a).  [Graphics:Images/ResidueCalcModHome_gr_443.gif].

Solution 3 (a).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/ResidueCalcModHome_gr_444.gif].  

Solution.  Factor the denominator and get

                    [Graphics:../Images/ResidueCalcModHome_gr_445.gif]

and conclude that  [Graphics:../Images/ResidueCalcModHome_gr_446.gif]  has simple zeros at  [Graphics:../Images/ResidueCalcModHome_gr_447.gif],  [Graphics:../Images/ResidueCalcModHome_gr_448.gif],  [Graphics:../Images/ResidueCalcModHome_gr_449.gif]  and  [Graphics:../Images/ResidueCalcModHome_gr_450.gif].

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  [Graphics:../Images/ResidueCalcModHome_gr_451.gif]  has simple poles at [Graphics:../Images/ResidueCalcModHome_gr_452.gif],  [Graphics:../Images/ResidueCalcModHome_gr_453.gif],  [Graphics:../Images/ResidueCalcModHome_gr_454.gif]  and  [Graphics:../Images/ResidueCalcModHome_gr_455.gif].

                         [Graphics:../Images/ResidueCalcModHome_gr_456.gif]

                    The point  [Graphics:../Images/ResidueCalcModHome_gr_457.gif]  that lies inside the contour  [Graphics:../Images/ResidueCalcModHome_gr_458.gif].  

        Now apply Theorem 8.1 the Cauchy's Residue Theorem.

The point  [Graphics:../Images/ResidueCalcModHome_gr_459.gif]  is the only singularity that lies inside the contour  [Graphics:../Images/ResidueCalcModHome_gr_460.gif].  

Since  [Graphics:../Images/ResidueCalcModHome_gr_461.gif] is a simple pole, by Theorem 8.2 the residue is calculated as follows:

                    [Graphics:../Images/ResidueCalcModHome_gr_462.gif]   

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

                              [Graphics:../Images/ResidueCalcModHome_gr_463.gif]  
        

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ResidueCalcModHome_gr_464.gif]

[Graphics:../Images/ResidueCalcModHome_gr_465.gif]


[Graphics:../Images/ResidueCalcModHome_gr_466.gif]

[Graphics:../Images/ResidueCalcModHome_gr_467.gif]

Maple can check our work too!

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

               [Graphics:../Images/ResidueCalcModHome_gr_468.gif]

We are really done.   

Aside.  Both [Graphics:../Images/ResidueCalcModHome_gr_469.gif] and [Graphics:../Images/ResidueCalcModHome_gr_470.gif] are capable of finding residues.

[Graphics:../Images/ResidueCalcModHome_gr_471.gif]

[Graphics:../Images/ResidueCalcModHome_gr_472.gif]


[Graphics:../Images/ResidueCalcModHome_gr_473.gif]

[Graphics:../Images/ResidueCalcModHome_gr_474.gif]


     > residue( 1/(z^4+4), z=-1+I );

               [Graphics:../Images/ResidueCalcModHome_gr_475.gif]

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

               [Graphics:../Images/ResidueCalcModHome_gr_476.gif]

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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