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Complex Analysis. Laurent Series Expansion in region(22C).

  1. Mar 31, 2017 #1
    <Moderator's note: moved from a technical forum, so homework template missing>

    Hi. I have solved the others but I am really struggling on 22c. I need it to converge for |z|>2. This is the part I am really struggling with. I am trying to get both fractions into a geometric series with 1/(1-(1/(z/2)) so that it converges for 1/(z/2) <1; which becomes, (z>2). I however cannot manipulate either series into this form that I am looking for. I have tried many times. I have also tried combining both fractions into one fraction over z^2-1, but am not getting anywhere. Any pointers, or tips so that I can get it to converge for |z|>2 by manipulating the fractions (just simply cannot get in right form). Thank you.
    media%2F49b%2F49b7fdac-72e2-4e80-b4b1-48fc16fb8003%2Fphpdyb4Ql.png
     
    Last edited by a moderator: Apr 3, 2017
  2. jcsd
  3. Apr 1, 2017 #2

    BvU

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    Worries me. You get ##\displaystyle {2\over z^2 - 1}##, right ?
    Perhaps a simple transform to ##y = z-1## can shed some light ?
     
  4. Apr 1, 2017 #3

    FactChecker

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    Two approaches:
    1) You can do long division of polynomials the same way the you do it with integers.
    2) Use 1/(1-r) = 1+r+r2+r3+... and 1/(1+r) = 1-r+r2-r3+...
     
  5. Apr 1, 2017 #4

    BvU

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    Yet another ? From some searching in PF I really like the suggestion ##
    \displaystyle {1\over z^2 - 1} = - {1\over z^2 \left ( 1 - {1\over z^2}\right ) } ##

    And please forget about my ##
    y = z-1
    ## :nb)
     
  6. Apr 1, 2017 #5

    FactChecker

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    Ha! Of course! That is the easiest way. (but the leading minus sign on the right is not needed)
     
  7. Apr 1, 2017 #6

    BvU

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    Oops
     
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