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Laurent Seireis

  1. Jul 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Laurent series at z0=i, which is convergent in the annulus A ={z:0<|z-i|<51/2 } of

    1/[(z-i)(z-2)]


    2. Relevant equations



    3. The attempt at a solution
    |z-i|/51/2 <1

    i make
    1/[(z-i)(z-2)] = 1/[51/2 (z-i)((i-2)/51/2 + (z-i)/51/2 )
    now how do i make it so that i have 1-(z-i)/51/2 to use the binomial series?
    Is this a right approach to this type of questions? I could use partial factions to get

    1/(2-i)*[1/(z-2)-1/(z-i)] but again i don't know how i can take advantage of |z-i|/51/2 <1 to use in the binomial series.

    Thanks in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 17, 2009 #2

    HallsofIvy

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    [tex]\frac{1}{(z-i)(z-2)}= \frac{1}{z-i}\left(\frac{1}{z-2}\right)[/tex]

    [tex]\frac{1}{z-2}[/tex]
    is analytic in the annulus given so it has a Taylor's series about z= i. Find that Taylor's series and divide each term by z-i.

    (Hint: write 1/(z-2) as
    [tex]\frac{1}{z-2}= \frac{1}{z- i+ i- 2}= -\frac{1}{(i-2)+ (z-i)}= -\frac{\frac{1}{i-2}}{1- \frac{z-i}{i-2}}[/tex]
    and think of geometric series.)
     
    Last edited: Jul 18, 2009
  4. Jul 17, 2009 #3
    Thanks a lot . I was doing the same thing, except I've forgotten about geometric series :) . Got it now :) .
     
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