Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    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.
    Offline

    22
    ReputationRep:
    (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:
    Show
    1/z - 1/2 + z/12 + \cdots
    and hopefully Greg/atsruser/firegarden/somebody will correct me if I'm wrong.
    Offline

    19
    ReputationRep:
    (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
    Offline

    13
    ReputationRep:
    (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.
    Offline

    22
    ReputationRep:
    (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:
    Offline

    11
    ReputationRep:
    (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?
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Brexit voters: Do you stand by your vote?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.