Exercise 1.  Find   [Graphics:Images/ResidueCalcModHome_gr_1.gif]   for the following functions:

1 (c).   [Graphics:Images/ResidueCalcModHome_gr_65.gif].

Solution 1 (c).

See text and/or instructor's solution manual.

Answer.   [Graphics:../Images/ResidueCalcModHome_gr_66.gif]

Solution.  Use Definition 7.6 or Theorem 7.10 in Section 7.4 to determine that  [Graphics:../Images/ResidueCalcModHome_gr_67.gif]  has a simple zero at the point [Graphics:../Images/ResidueCalcModHome_gr_68.gif].

Applying Definition 7.6 with  [Graphics:../Images/ResidueCalcModHome_gr_69.gif]  and  [Graphics:../Images/ResidueCalcModHome_gr_70.gif]  we compute

                    [Graphics:../Images/ResidueCalcModHome_gr_71.gif]   and   [Graphics:../Images/ResidueCalcModHome_gr_72.gif].   

Hence, we conclude that so that  [Graphics:../Images/ResidueCalcModHome_gr_73.gif]  has a simple zero at  [Graphics:../Images/ResidueCalcModHome_gr_74.gif].  

Or you can apply Theorem 7.10 and expand  [Graphics:../Images/ResidueCalcModHome_gr_75.gif] in a Taylor series about the point [Graphics:../Images/ResidueCalcModHome_gr_76.gif].

                    [Graphics:../Images/ResidueCalcModHome_gr_77.gif].  

The first few coefficients of the Taylor series are  [Graphics:../Images/ResidueCalcModHome_gr_78.gif].

Hence, we can conclude that  [Graphics:../Images/ResidueCalcModHome_gr_79.gif]  has a simple zero at  [Graphics:../Images/ResidueCalcModHome_gr_80.gif].  

        Here we have  [Graphics:../Images/ResidueCalcModHome_gr_81.gif]  and the denominator  [Graphics:../Images/ResidueCalcModHome_gr_82.gif]  has a simple zero at the point  [Graphics:../Images/ResidueCalcModHome_gr_83.gif].

Now apply  Corollary 7.5 to conclude that [Graphics:../Images/ResidueCalcModHome_gr_84.gif]  has a simple pole at  [Graphics:../Images/ResidueCalcModHome_gr_85.gif],  and by Theorem 8.2 the residue is calculated as follows:

                     [Graphics:../Images/ResidueCalcModHome_gr_86.gif]  

We are done.   

Aside.  We can let Mathematica double check our work.

[Graphics:../Images/ResidueCalcModHome_gr_87.gif]

[Graphics:../Images/ResidueCalcModHome_gr_88.gif]

Maple can check our work too!

     > limit( (z-0)*1/sin(z), z=0 );

               1

We are really done.   

Alternate Solution.   If you prefer, you can expand  [Graphics:../Images/ResidueCalcModHome_gr_89.gif]  in a Laurent series about  [Graphics:../Images/ResidueCalcModHome_gr_90.gif].

                    [Graphics:../Images/ResidueCalcModHome_gr_91.gif]
                    
This might be quite tedious, so we will show how to do it with Mathematica and Maple.

[Graphics:../Images/ResidueCalcModHome_gr_92.gif]

[Graphics:../Images/ResidueCalcModHome_gr_93.gif]


     > series( 1/sin(z), z=0,10 );

               [Graphics:../Images/ResidueCalcModHome_gr_94.gif]

Here we see that  [Graphics:../Images/ResidueCalcModHome_gr_95.gif][Graphics:../Images/ResidueCalcModHome_gr_96.gif]  

has a simple pole at  [Graphics:../Images/ResidueCalcModHome_gr_97.gif],  and according to Definition 8.1 the residue is  [Graphics:../Images/ResidueCalcModHome_gr_98.gif].  

We are really really done.   

Aside.  Both [Graphics:../Images/ResidueCalcModHome_gr_99.gif] and [Graphics:../Images/ResidueCalcModHome_gr_100.gif] are capable of finding residues.

[Graphics:../Images/ResidueCalcModHome_gr_101.gif]

[Graphics:../Images/ResidueCalcModHome_gr_102.gif]


     > residue( 1/sin(z), z=0 );

               1

We are really really really done.   

Extra Stuff.

      Applying Corollary 7.5 we know that  [Graphics:../Images/ResidueCalcModHome_gr_103.gif] has simple zeros at the points  [Graphics:../Images/ResidueCalcModHome_gr_104.gif]  where  n  is an integer.

Hence,  [Graphics:../Images/ResidueCalcModHome_gr_105.gif]  has a simple poles at the points  [Graphics:../Images/ResidueCalcModHome_gr_106.gif]  where  n  is an integer,  and the residues are

                     [Graphics:../Images/ResidueCalcModHome_gr_107.gif]

[Graphics:../Images/ResidueCalcModHome_gr_108.gif]

[Graphics:../Images/ResidueCalcModHome_gr_109.gif]


[Graphics:../Images/ResidueCalcModHome_gr_110.gif]

[Graphics:../Images/ResidueCalcModHome_gr_111.gif] 

 

 

 

 

 

 

 

 

 

 

 

 

 

 



This solution is complements of the authors.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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