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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

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    Science Advisor

    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

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    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 ...)
     
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