Solution 3 (e).
See text and/or instructor's solution manual.
Answer.
has
removable a singularity at the origin, and a simple pole
at -1.
Solution. Write ![]()
We know that
has
a removable singularity at
,
which implies that that
has
a removable singularity at ![]()
Now apply Corollary
7.5 to conclude that
has
a simple pole at -1.
Therefore,
has
removable a singularity at the origin, and a simple pole
at -1.
We are done.
Aside. We can let Mathematica double check our work.
We are really done.
![[Graphics:../Images/SingularityZeroPoleModHome_gr_991.gif]](../Images/SingularityZeroPoleModHome_gr_991.gif)
The removable a singularity at the origin, and a simple pole at -1.
![[Graphics:../Images/SingularityZeroPoleModHome_gr_992.gif]](../Images/SingularityZeroPoleModHome_gr_992.gif)
A
plot for
.
This solution is complements of the authors.
(c) 2008 John H. Mathews, Russell W. Howell