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!

Laurent series(a really hard one!)

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data

    ((e^{iz})/((1+z²)))
    Find a Laurent expansion valid for 0<|z-i|<1and use it to show that the derivative will leave you with the residue a₋₁.
    So we have an annulus with its center at one of the singularities. Since the other singularity at z=-i is outside of the annulus it is not a problem, because the function is analytic. .
    The problem is i have never found a laurent expansion for a function this complicated! with double poles!


    2. Relevant equations



    3. The attempt at a solution

    ((e^{iz})/((1+z²)²))=((e^{iz})/((1+z²)(1+z²)))=((e^{iz})/((i+z)(z-i)(i+z)(z-i)))=((e^{iz})/((i+z)²(z-i)²))
    let w=z-i
    z+i=w+2i
    z=w+i
    ((e^{iz})/((i+z)²(z-i)²))=((e^{i(w+i)})/((w+2i)²(w)²))=((e^{iw})/(e(w+2i)²(w)²))=((e^{iw})/e)((1/(w²)))((1/(4[i-(-w)]²))

    =((e^{iz})/4)*∑{n=0 to ∞}(-w)²ⁿ⁻²

    I cant figure this question out! so if anyone can help
     
  2. jcsd
  3. Oct 14, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Write the function as
    [tex]f(z)= \frac{e^{iz}}{z+i}\frac{1}{z- i}[/itex]
    As you say, [itex]e^{iz}/(z+ 1)[/itex] is analytic at z= i and so can be written as a Taylor series there. Multiplying that by 1/(z-i) only decreases each power of (z- i) in the series by 1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Laurent series(a really hard one!)
  1. Laurent series (Replies: 2)

  2. Laurent series (Replies: 6)

Loading...