Exercise
1. Find
for
the following functions:
1
(c).
.
Solution 1 (c).
See text and/or instructor's solution manual.
Answer. ![]()
Solution. Use
Definition 7.6 or Theorem 7.10 in Section
7.4 to determine that
has
a simple zero at the point
.
Applying Definition 7.6 with
and
we
compute
and
.
Hence, we conclude that so that
has
a simple zero at
.
Or you can apply Theorem 7.10 and expand
in a Taylor series about the point
.
.
The first few coefficients of the Taylor series
are
.
Hence, we can conclude that
has
a simple zero at
.
Here we
have
and
the denominator
has
a simple zero at the point
.
Now apply Corollary
7.5 to conclude that
has
a simple pole 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)*1/sin(z), z=0 );
1
We are really done.
Alternate
Solution. If you prefer, you can
expand
in
a Laurent series about
.
![[Graphics:../Images/ResidueCalcModHome_gr_91.gif]](../Images/ResidueCalcModHome_gr_91.gif)
This might be quite tedious, so we will show how to do it with
Mathematica and Maple.
>
series( 1/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( 1/sin(z), z=0 );
1
We are really really really done.
Extra Stuff.
Applying Corollary
7.5 we know that
has simple zeros at the points
where n is
an integer.
Hence,
has
a simple poles at the points
where n is
an integer, and the residues are
![[Graphics:../Images/ResidueCalcModHome_gr_107.gif]](../Images/ResidueCalcModHome_gr_107.gif)
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell