1. Not finding help here? Sign up for a free 30min 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

  1. Jul 9, 2008 #1
    i get the idea of the laurent expansion but i get confused with the constraints and how they change the way you work with the expansion.
    by now you can prob. tell im trying to get to grasp with complex analysis as a whole.

    for example i have this :

    find laurent series for :[itex]\[
    f(z) = \frac{z}
    {{z^2 - 1}}
    \][/itex]

    given the constraints:
    (a) [itex]\[
    0 < \left| {z - 1} \right| < 2
    \][/itex] ............... (b) [itex]\[
    \left| {z + 1} \right| > 2
    \][/itex] ............... (c)[itex]\[
    \left| z \right| > 1
    \][/itex]

    ------------------------------------------------------------
    My Attempt: for part (a)

    first I break up the function using partial fractions:
    [itex]\[
    \frac{z}
    {{z^2 - 1}} = \frac{z}
    {{(z - 1)(z + 1)}} = \frac{A}
    {{(z - 1)}} + \frac{B}
    {{(z + 1)}}
    \][/itex]

    [itex]\[
    z = A(z + 1) + B(z - 1)
    \][/itex]

    setting: z=1:
    [itex]\[
    1 = 2A,A = \frac{1}
    {2}
    \][/itex]

    setting: z=-1:
    [itex]\[
    - 1 = B( - 2),B = \frac{1}
    {2}
    \]
    [/itex]

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



    so for: 0 < |z-1| < 2:
    [itex]\[
    \begin{gathered}
    \frac{1}
    {{2(z - 1)}} + \frac{1}
    {{2(z + 1)}} \hfill \\
    = \frac{1}
    {{2(z - 1)}} + \frac{1}
    {2}\left[ {\frac{1}
    {{2 - ( - (z - 1)}}} \right] \hfill \\
    = \frac{1}
    {{2(z - 1)}} + \frac{1}
    {2}\frac{1}
    {2}\left[ {\frac{1}
    {{1 - \left[ { - (\frac{{z - 1}}
    {2})} \right]}}} \right] \hfill \\
    = \frac{1}
    {{2(z - 1)}} + \frac{1}
    {4}\left[ {\frac{1}
    {{1 - \left[ { - (\frac{{z - 1}}
    {2})} \right]}}} \right] \hfill \\
    = \frac{1}
    {{2(z - 1)}} + \frac{1}
    {4}\sum\limits_{n = 0}^\infty {\left[ {( - 1)^n \frac{{(z - 1)^n }}
    {{2^n }}} \right]} \hfill \\
    \frac{1}
    {{2(z - 1)}} + \sum\limits_{n = 0}^\infty {\left[ {( - 1)^n \frac{{(z - 1)^n }}
    {{2^{n + 2} }}} \right]} \hfill \\
    \end{gathered}
    \]
    [/itex]

    is this correct? i'm predicting that i may have missed out two parts:
    1) the first part of the final equation : 1/2(z-1) : can this be simplified and integrated into the sum formula?

    2) the constraint for part (a) was 0 < |z-1| < 2. I didn't know how to interpret |z-1| being between 0 and 2.



    -------------------------------------------------------------------------------

    for the other two parts - the constraints are (part (b) |z+1|>2 , part (c) |z|>1) - please could you advise me on how the constraints are meant to be used and how the final answer changes? is there a trick to this?
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Laurent Series
  1. Major laurent help (Replies: 0)

  2. Fourier series (Replies: 0)

  3. Fourier series (Replies: 0)

  4. Fourier series (Replies: 0)

  5. Fourier series (Replies: 0)

Loading...