1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Residue

  1. Jan 10, 2016 #1
    Hello everyone,
    I have a problem with finding a residue of a function:
    [itex]f(z)={\frac{z^3*exp(1/z)}{(1+z)}}[/itex] in infinity.
    I tried to present it in Laurent series:
    [itex]\frac{z^3}{1+z} sum_{n=0}^\infty\frac{1}{n!z^n}[/itex]

    I know that residue will be equal to coefficient [itex]a_{-1}[/itex], but i don't know how to find it.
     
  2. jcsd
  3. Jan 10, 2016 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Expand [itex]\frac{z^3}{1+z}=z^3-z^4+z^5-z^6...[/itex]. The multiply the two series together to find the coefficient you want (as an infinite series).
     
  4. Jan 10, 2016 #3
    Then i get:
    [itex]\frac{z^3}{1+z}=(-1)^n*z^{n+3}[/itex]

    and when i multiply I always get [itex]{z^3}[/itex] with some fraction
     
  5. Jan 11, 2016 #4

    Ssnow

    User Avatar
    Gold Member

    ## \left(z^{3}-z^{4}+z^{5}-\cdots \right)\left(1+\frac{1}{z}+\frac{1}{2z^{2}}+\frac{1}{6z^{3}}+\frac{1}{24z^{4}}+\cdots \right)##, the only interested terms are of the forms ##\frac{a_{-1}}{z}##, that are ##\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}-\cdots ## so it is ## e^{-1}-\frac{1}{2}+\frac{1}{3!} ## (if I did not make mistakes ...)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding Residue
  1. Residue of z/cos(z)? (Replies: 15)

  2. Quadratic residues (Replies: 3)

Loading...