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Residue of a complex function with essential singularity

  1. Jul 9, 2011 #1
    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.
     
  2. jcsd
  3. Jul 9, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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].
     
  4. Jul 9, 2011 #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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