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Complex Laurent Series

  1. May 1, 2012 #1

    gbu

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    1. The problem statement, all variables and given/known data
    Find the Laurent series of
    [itex]
    \frac{(z+2)}{(z-1)}
    [/itex]

    on

    [itex]C_1: 1 < |z|[/itex]
    and
    [itex]C_2: 0 < |z| < 1[/itex]


    2. Relevant equations

    I have a formula for computing Laurent series, but it includes an integral that is impossible to solve. For everything that I've read, no one actually solves them this way. I don't think there's any relevant equations that I know that can help.

    3. The attempt at a solution
    I've been stumped on this for four days now. I've been trying to self-work through a complex analysis textbook over the summer, and I'm really close to giving up as I completely can't wrap my head around the annulus type Laurent series stuff.

    Any advice helps. I've read lots and Googled lots. I know Taylor series and such...
     
  2. jcsd
  3. May 1, 2012 #2

    jbunniii

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    Note that

    [tex]\frac{z+2}{z-1} = \frac{z - 1 + 3}{z - 1} = 1 + \frac{3}{z - 1}[/tex]

    Can you express the second term as a power series?
     
  4. May 2, 2012 #3

    gbu

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    OK, yes. Thank you!

    I recognize that... but a power series around what value? What do I do with the C_0 and C_1 annuli?

    A power series around 0 makes it pretty easy (though the answer doesn't appear sensical), but the C_0 says 1 < |z| and the |z| doesn't meet that condition at 0. Although, I really don't understand what we're doing with these rings at all...
     
  5. May 2, 2012 #4
    Zero. They're both expansions about the point zero but Laurent series are convergent in annular domains separated by singular points. z=1 is a singular point of that function so any laurent series about the point zero will converge in one ring with |z|<1 and outside that ring another different Laurent series represents the function again between singular points but this time it's 1 and infinity. To find out that second one, write:

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

    Now, what is the region of convergence of a power series representation for that expression?
     
  6. May 2, 2012 #5

    gbu

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    How did I know to write it like that? It makes sense (and thanks a TON for your explanation of the annulus stuff)... I guess I just don't see how you go from the z+2/z-1 to the proper representation for the right series "on demand" so to speak.
     
  7. May 2, 2012 #6

    jbunniii

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    Some general guidelines:

    You know that if your regions of convergence are centered at the origin, you will need a series that is centered at the origin, so you will have powers of z. More generally, for a region of convergence centered at a point w, you will need powers of (z-w).

    If the series has positive powers, i.e. z^n where n is positive, then z can't be too large, so you know the region of convergence will be inside a circle, not outside of it.

    Conversely, if the series has negative powers, i.e. z^(-n) for positive n, then z can't be too small, so the region of convergence will be outside a circle.

    Then you just need to "pattern match" your given function to standard functions whose series expansions you know.

    Taking the simplest example, f(z) = 1/(1-z), we know that

    [tex]\frac{1}{1-z} = 1 + z + z^2 + \ldots = \sum_{n=0}^{\infty}z^n[/tex]

    which converges inside the unit circle. If I need convergence outside the unit circle, I know I'll need powers of z^(-1), so I need to manipulate the function accordingly:

    [tex]\frac{1}{1-z} = \frac{z^{-1}}{z^{-1} - 1} = -z^{-1} \frac{1}{1 - z^{-1}}[/tex]

    which equals

    [tex]-z^{-1}(1 + z^{-1} + z^{-2} + \ldots) = -z^{-1}\sum_{n=0}^{\infty}z^{-n} = -\sum_{n=1}^{\infty}z^{-n}[/tex]

    which converges outside the unit circle.
     
  8. May 2, 2012 #7

    jbunniii

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    P.S. If everything I wrote in the previous post makes sense, the next step would be to try a slightly different function and see if you can manipulate it into a suitable form for expressing as a series. I suggest trying something like

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

    If you get stuck or want to check your answer, post your work here.
     
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