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Laurent Series problem

  1. May 21, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Laurent series for the function that converges at [itex]0 < \left|z-z_0\right| < R[/itex]

    2. Relevant equations

    [itex]\frac{z^2-4}{z-1}[/itex]
    [itex]z_0 = 1[/itex]


    3. The attempt at a solution

    I think this is going to be a finite sum. I think it could be wrong, though, because it certainly differs with the answer in the book.

    [itex]\frac{z^2-4}{z-1} = \frac{z^2}{z-1}-\frac{4}{z-1}[/itex]

    I tried to put the first part in terms of [itex]z-1[/itex]...

    [itex]\frac{z^2}{z-1} = \frac{z-1+1}{z-1}\frac{z-1+1}{z-1} =[/itex]

    [itex]\left( \frac{z-1}{z-1}+\frac{1}{z-1}\right)\left(\frac{z-1}{z-1}+\frac{1}{z-1}\right) [/itex]

    [itex] = \left(1+\frac{1}{z-1}\right)\left(1+\frac{1}{z-1}\right) = 1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}[/itex]

    I recombined with the last term...

    [itex]1+\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}-\frac{4}{z-1} = 1-\frac{2}{z-1}+\frac{1}{\left(z-1\right)^2}[/itex]

    This is the final answer that I get. However, the book says that it is

    [itex]-\frac{3}{z-1}+2+(z-1)[/itex]

    I am absolutely stumped and I've been racking my brain over it for hours.
     
    Last edited: May 21, 2013
  2. jcsd
  3. May 21, 2013 #2

    micromass

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    You found the Laurent series at ##1##. I think they want you to find the Laurent series at ##0##.
     
  4. May 21, 2013 #3
    They may have wanted me to find the Laurent series at 1. Sorry, I forgot to add to the problem...

    [itex]z_0 = 1[/itex]
     
  5. May 21, 2013 #4

    micromass

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    OK, so I'll take a closer look.

    I see right away that this is not a correct thing to do:

     
  6. May 21, 2013 #5
    I figured it was the algebra that was holding me back here. Is it because it creates a singularity at 1 or something? Why is this wrong? I do not see it.
     
  7. May 21, 2013 #6

    micromass

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    [tex]\frac{z-1+1}{z-1}\frac{z-1+1}{z-1} = \frac{z}{z-1}\frac{z}{z-1} = \frac{z^2}{(z-1)^2}[/tex]

    This does not equal ##\frac{z^2}{z-1}##.
     
  8. May 21, 2013 #7
    Holy crap, oh geeze. That's embarrassing. Thanks for pointing that out. Let me try this again.
     
  9. May 21, 2013 #8
    Alright, well, realizing that that technique is wrong has completely erased the paths that I see to solving this problem. Any hints as to an appropriate next step to take? I've gone the method of splitting [itex]z^2-4[/itex] into [itex](z-2)(z+2)[/itex] but I do not know how to go from there.
     
  10. May 21, 2013 #9

    micromass

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    Try to write

    [tex]\frac{z^2 - 4}{z-1} = \frac{(z-1)^2 + P(z)}{z-1}[/tex]

    and try to find the polynomial ##P(z)##.
     
  11. May 21, 2013 #10
    Oh, that's clever. Thanks for suggesting this method. This immediately gave me the path I needed to do.

    [itex]\frac{z^2-4}{z-1} = \frac{\left(z-1\right)^2+P(z)}{z-1}=\frac{{(z-1)}^2+2z-5}{z-1} = \frac{{(z-1)}^2+2(z-1)-3}{z-1} = (z-1)+2-\frac{3}{z-1}[/itex] which is exactly what I was looking for. I will keep this technique in my toolbox. Thanks for the suggestion ;)

    Oh, and it converges for all R > 0.
     
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