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Laurent Series Around a Given Point

  1. Jul 10, 2013 #1
    "[F]ind the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

    8. ##\frac{z^{2}}{z^{2} -1}##; ##z_{o} = 1##" (Complex Variables, 2nd edition; Stephen D. Fisher, pg. 150)

    I'm not very comfortable with Laurent series (or series in general) and this is a problem I've been assigned. I have found two approaches to this problem - one of them makes sense to me but disagrees with the residue that I would expect, while the other solution I do not entirely understand.

    Firstly, the residue should be given by:

    ##Res(f;z_{o}) = c_{m-1} = \frac{H^{m-1}(z_{o})}{(m-1)!}##

    I would write the following:

    $$\frac{z^{2}}{z^{2} -1} = \frac{z^2}{z+1} \frac{1}{z-1} = H(z) \cdot \frac{1}{(z-z_{o})^{m}}$$

    Then finding the residue is easy; we have ##m=1## and so we find:

    $$Res(f;z_{o}) = \frac{H^{m-1}(z_{o})}{(m-1)!} = \frac{H(z_{o})}{0!} = H(z_{o}) = \frac{1}{2}$$


    My difficulty lies in writing the Laurent series.

    Here was my first attempt, which seemed promising:

    $$f(z) = \frac{z^{2}}{z^{2}-1} = 1 + \frac{1}{z^{2} -1} = 1 - \frac{1}{1 - z^2} = 1 - \sum (z^{2})^{n}$$

    This last equivalency is by the geometric series (or am I only allowed to use this for ##|z| < 1##?). However, this representation suggests a residue of 0 and I'm pretty gosh-darned certain my residue above is correct.


    After browsing the internet I found the solution (as this is a textbook problem and was assigned at another university) - but I just don't understand it. The solution is found on the 4th page of the following link and it is question 2.5.8: http://www.math.rutgers.edu/~alexandra/Math403_hw8slns.pdf

    If anyone could kindly explain this solution to me it would be greatly appreciated! I don't want to simply copy out a solution because it agrees with what I expect the residue to be; I would much rather understand the method of achieving this solution. Thank you very much in advance!
     
  2. jcsd
  3. Jul 11, 2013 #2

    Dick

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    After you found the correct residue at z=1 you expanded the Laurent series around z=0 and found a residue around z=0 to be 0. Which it clearly is since the function is analytic at z=0. You want the residue at z=1. To use laurent series, find the power series in u where u=z-1. So you want the u^(-1) coefficient of the power series expansion of (u+1)^2/(u(u+2)).
     
    Last edited: Jul 11, 2013
  4. Jul 11, 2013 #3
    That is correct.

    Also, the 2nd solution in the OP takes a different approach: Instead of finding the Laurent series for the original function, it breaks the function up to pieces and finds the Laurent series for one piece, then multiplies that with the other parts it left out.
     
  5. Jul 12, 2013 #4
    Thank you for the responses. I have attempted this problem again in the following manner:

    Consider the function ##f(z) = \frac{z^{2}}{z^{2}-1} = \frac{z^{2}}{(z+1)(z-1)} = \frac{z^{2}}{z+1} \cdot \frac{1}{z-1}## and consider this a function of the form ##f(z) = \frac{H(z)}{(z-z_{o})^{m}}## where the function ##f## has a pole at ##z_{o}##. Then we write a power series expansion for the function ##H(z)## centered at the pole.

    In general it is possible to write a power series in the form

    $$H(z) = \sum^{\infty}_{n = 0} a_{n} (z-z_{o})^{n}$$

    where ##a_{n}## is the coefficient of the ##n^{th}## term and ##z_{o}## is the center of the series. Furthermore, we can determine ##a_{n}## from the following formula:

    $$a_{n} = \frac{H^{(n)}(z_{o})}{n!}$$


    From this it is possible to expand the above function ##H(z)## about the point ##z_{o} = 1## by computing ##a_{n}## for several terms of the series (and try to establish a pattern; that is, a representation for all ##a_{n}##).


    From 'brute computation' it is easy to see:

    $$H(z) = \frac{1}{2} + \frac{3}{4} \cdot (z - 1) + a_{2}(z - 1)^{2} + ...$$

    $$f(z) = \frac{H(z)}{(z-z_{o})^{m}} = \frac{\frac{1}{2} + \frac{3}{4} \cdot (z - 1) + a_{n}(z - 1)^{2} + ...}{z -1} = \frac{1}{2(z-1)} + \frac{3(z-1)}{4(z-1)} + \frac{a_{2}(z-1)^{2}}{z-1} + ... = \frac{1}{2(z-1)} + \frac{3}{4} + {a_{2}(z-1)} + ...$$

    And from this it is easy to see that the residue of the function ##f(z) = \frac{z^{2}}{z^{2} -1}## is ##\frac{1}{2}##.


    I'm feeling much more confident about my approach now. Not only does this yield the correct residue, but it also makes sense to me, which is a good sign, I hope. My only concern, then, is how best to represent ##a_{n}## in general. Should I continue computing derivatives of ##H(z)## and evaluating for ##a_{n}## until I find a pattern? Is there an easier way to do this - for example, if the pattern were very unorthodox?
     
  6. Jul 12, 2013 #5
    Sadly, no, there is not. Finding the power series for a function is always hard when there is no obvious pattern. However, one can resort to your method (or the 2nd method in the OP) and break the function up to pieces that have easier power series to find.
     
  7. Jul 12, 2013 #6
    Ahh...that's a shame. Thanks again for your time and all your help!
     
  8. Jul 12, 2013 #7
    Here is another approach which seems simpler to me. Just apply partial fractions:
    [tex]\frac{z^2}{z^2-1}=1 - \frac{1/2}{z+1} + \frac{1/2}{z-1}[/tex]
    from which it is obvious that the residue at 1 is 1/2.
     
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