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Confused about Laurent series of analytic functions

  1. Dec 4, 2011 #1
    I am a bit confused about laurent series. I know the definitions where the coefficients are expressed as integrals.

    However, I am confused about how to actually find the laurent series in practice, for analytic functions.
    The information I can find online is just terrible, some of them do solve them but they don't use integration at all, they all seem to have some different but similar tricks to find the series and none of them actually explain the steps.

    For example a problem I am looking at says to find all taylor and Laurent expansion about the origin, of the function:
    [tex]\frac{1}{z^{2}+1}[/tex]
    There are two poles at [itex]z=\pm i[/itex]

    So by my limited understanding of how to expand this function, I think the idea is to just do a taylor series about z=0, which is only valid in the region |z|<1

    Then I think the next step is to do a Laurent series which is valid in the region |z|>1
    but I am unsure how to do this.
    Some sources I found seem to suggest expanding about the poles but which one do I pick?


    Just to be clear I am not really asking for a solution to the function I posed above, what I would like is an understanding of how to find laurent expansions of functions such as the one I wrote above. Nothing too complicated, this doesn't seem like too complex of a subject, but I just can't find any decent coherent source explaining how to perform the expansion online.

    So thank you for any help or general advice on performing laurent expansions you can give me.
     
  2. jcsd
  3. Dec 5, 2011 #2
    Ok, so you understand how to compute the series about a region where the function is analytic right? The series is just a Taylor series and an easy way to do that as well a lot of other ones is to manipulate the expression so that it looks like a geometric series which you already know what the form is right? So:

    [tex]\frac{1}{1+z^2}=\frac{1}{1-(-z^2)}=\sum_{n=0}^{\infty}(-1)^n z^{2n},\quad |z|<1[/tex]

    Now when you're expanding within an annular region bordered by poles, you will always obtain a series which has an essential singularity at the borders. That means of course, you'll get an expression with z's always on the bottom but again, manipulate the expression so that it looks like a geometric series so:

    [tex]\frac{1}{1+z^2}=\frac{a}{z+i}+\frac{b}{z+i}[/tex]

    I'll do one and see if you can do the other:

    [tex]\frac{a}{z+i}=\frac{a}{z(1+i/z)}=\frac{a}{z}\frac{1}{1-(-i/z)}=a/z\sum_{n=0}^{\infty} (-1)^n\left(\frac{i}{z}\right)^n,\quad |z|>1[/tex]

    Also, get a good book on Complex Analysis like "Basic Complex Analysis" by Marsden and Hoffman then work towards deriving a general expression for expressing the Laurent series in the respective annular regions between poles of the expression:

    [tex]\frac{1}{P_n(z)}[/tex]

    where P_n(z) is a polynomial, that is derive the expression:

    [tex]\frac{1}{P_n(z)}=\text{my series in my annular region} [/tex]

    for any polynomial what so ever.
     
    Last edited: Dec 5, 2011
  4. Dec 5, 2011 #3
    Did you mean the following?
    [tex]\frac{1}{1+z^{2}}=\frac{a}{z+i}+\frac{b}{z-i}[/tex]
    You had a + instead of a - in the denominator for b.

    Then the other one would be:
    [tex]\frac{b}{z-i}=\frac{b}{z(1-i/z)}=\frac{b}{z}\frac{1}{1-(i/z)}=\frac{a}{z}\sum\limits_{n=0}^{\infty}\left( \frac{i}{z}\right)^{n}, |z|>1[/tex]

    Also I solved for:
    [tex]a=\frac{i}{2}, b=\frac{-i}{2}[/tex]

    Thanks it's starting to make sense now. So I guess I just plug in the a and b values and add the two series together and that is equal to the Laurent expansion in the annular region then?


    Also, one other question, sorry if this is an obvious question, but why is it that in the |z|<1 region we do a geometric series for:
    [tex]\frac{1}{1-(-z^{2})}[/tex]
    and in the annular region we do a geometric series for:
    [tex]\frac{a}{z+i}+\frac{b}{z-i}[/tex]

    how do you know using the second one will give you the answer you want in the annular region and not the first one?
     
  5. Dec 5, 2011 #4
    Yes. Very good then.

    Need b in that one.

    Yes.


    Because the geometric series converges for |z|<1 so in the first case if |z|<1, we can just use it as is and represent it in terms of a geometric series for the variable z^2. Howeer, for the second case, we need to force-fit the quotient expressions to contain terms usually z/k or k/z such that when |z| is in the annular region, then the absolute value of z/k or k/z are less than one and then we can express the quotient in terms of the geometric series for the variable z/k or k/z..
     
  6. Dec 5, 2011 #5
    Perfect, that is all very clear now.

    I just have one last question if you don't mind.

    Say I did a similar problem but instead of 1 in the numerator there was some function f(z) which has no poles.
    So:
    [tex]\frac{f(x)}{1+z^{2}}[/tex]

    Could I just take off the f(z) and expand the same function we just did with the 1 in the numerator,
    [tex]\frac{1}{1+z^{2}}[/tex]

    and then separately expand f(z) (which has no poles) in a taylor expansion for example?
    and then just multiply the two series? So I would end up with the product of two sums?

    Does that sound right, or do I have to completely modify the method you have outlined here if there exists an f(z) instead of a 1 in the numerator?

    I mean, I know that doing the product of two series would be equal to the original function, but what I'm asking is would it be considered as a "laurent expansion" of the function?
     
  7. Dec 5, 2011 #6
    That is correct although there may be easier ways of computing the Laurent series than computing the product of two series.
     
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