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Residue of a complex function with essential singularity

by benygh2002
Tags: complex analysis, residue theorem
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benygh2002
#1
Jul9-11, 04:23 AM
P: 3
1. The problem statement, all variables and given/known data

Hello friends,
I'm a student of mechanical engineering and I have a problem with computing residues of a complex function. I've read some useful comments. Now I ve got some ideas about essential singularity and series expansion in computing the residue. However, I still can't find the solution to my problem.
I arrived at a complex function in the process of finding a solution to a mechanical problem.
Then I have to obtain the residues to proceed to the next steps.



2. Relevant equations

The function has the following form:

f(z)=exp(A*Z^N+B*Z^-N)/Z

where A, B and N are real constants (N>=3).

3. The attempt at a solution

I want to compute the resiude at z=0. I wrote the Laurent serie of f but got an infinite sum. I do not even know if I am at the right direction.
I would be really thankful if someone could give me a hint on this and put me back in the right direction.
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HallsofIvy
#2
Jul9-11, 07:34 AM
Math
Emeritus
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Thanks
PF Gold
P: 39,339
An essential singularity does not have a residue.

However, your Laurent series should not have an infinite number of negative powers and so this is NOT an essential singularity.

[tex]e^{Az^n+ Bz^{-n}}= e^{Az^n}e^{Bz^{-n}}[/tex]
is analytic and so has a Taylor's series- with no negative power terms. Dividing by z gives a series have "[itex]z^{-1}[/itex]" as its only negative power. The residue is the coefficient of [itex]z^{-1}[/itex].
benygh2002
#3
Jul9-11, 10:49 AM
P: 3
Thanks for your reply.

1) You mentioned that "an essential singularity does not have a residue". I'm a bit confused about this. For example:

[itex]f\left(z\right)=e^{z+\frac{1}{z}}[/itex]

has an essential singularity at z=0. However. after writing Laurent expansion of f(z) the residue at z=0 can be obtained as a serie:

[itex]Res\left(f,0\right)=1+\frac{1}{2!}+\frac{1}{2!3!}+\frac{1}{3!4!}+...[/itex]


2) Laurent expnasion of my desired function f(z) is:

[itex]f\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z}[/itex]

[itex]=\frac{\sum{(Az^n+Bz^{-n})}^k}{z k!}[/itex]

(k=0 to infinity)

in which the coefficients of 1/z terms give the residue as a serie (but the serie does not converge).

3) I would be thankful if you could tell me what I should do to find the residue of the following function which is similar to the previous function f(z):

[itex]g\left(z\right)=\frac{e^{Az^n+Bz^{-n}}}{z-a}[/itex]

where a is a constant.
What should I do to find the poles? At what point I should write the Taylor expansion?

Thanks for your help.


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