Exercise 3. Evaluate
3 (a).
.
Solution 3 (a).
See text and/or instructor's solution manual.
Answer.
.
Solution. Factor
the denominator and get
![[Graphics:../Images/ResidueCalcModHome_gr_445.gif]](../Images/ResidueCalcModHome_gr_445.gif)
and conclude that
has
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
simple poles at
,
,
and
.
![[Graphics:../Images/ResidueCalcModHome_gr_456.gif]](../Images/ResidueCalcModHome_gr_456.gif)
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 a simple pole, 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( (z-(-1+I))*1/(z^4+4), z=-1+I );
We are really done.
Aside. Both
and
are capable of finding residues.
>
residue( 1/(z^4+4), z=-1+I );
>
2*Pi*I*residue(1/(z^4+4),z=-1+I);
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell