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    Can anyone help me out with this question please?

    Find the residue of:

    f(z)=1/(e^(z)-1) at z=0

    I can't seem to expand this through a Laurent expansion. Has anyone got a simple solution to working it out step by step please.
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    (Original post by coheed94)
    Can anyone help me out with this question please?

    Find the residue of:

    f(z)=1/(e^(z)-1) at z=0

    I can't seem to expand this through a Laurent expansion. Has anyone got a simple solution to working it out step by step please.
    Until somebody more experienced comes by:

    \displaystyle \frac{1}{e^z - 1} = \frac{1}{z\left(1 + \left(z/2! + z^2/3! + \cdots\right)\right)}, now let g(z) = z/2! + z^2/3! + \cdots, so we get:

    \displaystyle \frac{1}{e^z - 1} = \frac{1}{z} - \frac{g(z)}{z} + \frac{g(z)^2}{z} - \frac{g(z)^3}{z} + \cdots

    I'll let you do the simplification.

    You should get
    Spoiler:
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    1/z - 1/2 + z/12 + \cdots
    and hopefully Greg/atsruser/firegarden/somebody will correct me if I'm wrong.
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    (Original post by Zacken)
    Until somebody more experienced comes by:

    \displaystyle \frac{1}{e^z - 1} = \frac{1}{z\left(1 + \left(z/2! + z^2/3! + \cdots\right)\right)}, now let g(z) = z/2! + z^2/3! + \cdots, so we get:

    \displaystyle \frac{1}{e^z - 1} = \frac{1}{z} - \frac{g(z)}{z} + \frac{g(z)^2}{z} - \frac{g(z)^3}{z} + \cdots

    I'll let you do the simplification.

    You should get
    Spoiler:
    Show
    1/z - 1/2 + z/12 + \cdots
    and hopefully Greg/atsruser/firegarden/somebody will correct me if I'm wrong.
    (Original post by coheed94)
    Can anyone help me out with this question please?

    Find the residue of:

    f(z)=1/(e^(z)-1) at z=0

    I can't seem to expand this through a Laurent expansion. Has anyone got a simple solution to working it out step by step please.
    too late ...
    I would do the same
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    (Original post by Zacken)
    and hopefully Greg/atsruser/firegarden/somebody will correct me if I'm wrong.
    Looks right to me. Getting at residues often involves hacking at it with the mathematical equivalent of a machete.
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    (Original post by Gregorius)
    Looks right to me. Getting at residues often involves hacking at it with the mathematical equivalent of a machete.
    Thanks. They have their occasional elegance once in a blue moon. :lol:
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    (Original post by Gregorius)
    Looks right to me. Getting at residues often involves hacking at it with the mathematical equivalent of a machete.
    Can't we pierce it with a rapier by saying that since:

    \displaystyle \lim_{z \to 0} \frac{z}{e^z-1} = \lim_{z \to 0} \frac{z-0}{e^z-e^0} = \frac{1}{\frac{d (e^z)}{dz} |_{z=0} } = \frac{1}{e^0}=1

    then f(z) has a pole of order 1 at 0, so that limit also gives the residue?
 
 
 
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