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Coefficients of a complex laurent series

  1. Oct 24, 2004 #1


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    My question is about the coefficients of a complex laurent series. As far as I know, there are three kinds of series: those which converge in a finite circular region around the expansion point z0,(aka taylor series), those that converge in a ring shaped region between two circles centered at z0, and those that converge for all points outside a circle centered at z0. For the first kind, I know that cn=0 for all n<0, because otherwise it would blow up at z0. My question is about the second and third cases.

    Do there have to be both negative and positive n coefficients in the second case? I think so, because if there werent any negative n terms, what would stop you from shrinking the inner circle? and the same goes for positive n terms and expanding the outer circle. But I'm not sure.

    For the third case, where the series converges for all z such that |z-z0| is bigger than some number, can there be any positve n terms? Again, it seems like no, because they would diverge if you got far enough out. But if the coefficients got small fast enough, say as 1/n!, then it might still work. For example, the e^x taylor series converges for all x, so it is sort of like this kind of a laurent series, and it has positive n terms.

    Basically, my question is this: When can you assume the cn's are only nonzero for n>0, only for n<0, and when must you have both? If you have a bunch of singularites inside the contour your integrating over to find the coefficients, but you know there won't be an upper limit on |z-z0| for convergence, can you safely assume these singularities will all cancel out and leave 0 for n>0? If not always, when can you do this?
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 26, 2004 #2
    There has little to do the number of therms of the Laurent and Taylor series with their convergence radii.

    The convergence of the series will hapen until it reaches a singularity. Check this example


    for [itex]|z|<1[/itex]


    for [itex]1<|z|<2[/itex]

    [tex]\frac{1}{(1-z)(2-z)}=\frac{1}{z(z-2)(\frac{1}{z}-1)}=-\frac{1}{2z(1-\frac{z}{2})} \sum_{n=0}^{\infty}\frac{1}{z^{n}}=-\frac{1}{2z}\left(\sum_{n=0}^{\infty}\frac{1}{z^{n}}\right)\left(\sum_{n=0}^{\infty}\frac{z^n}{2^n}\right)[/tex]

    for [itex]|z|>2[/itex]

    [tex]\frac{1}{(1-z)(2-z)}=\frac{1}{z^{2}(\frac{1}{z}-1)(\frac{2}{z}-1)}=\frac{1}{z^2} \left(\sum_{n=0}^{\infty}\frac{1}{z^{n}}\right) \left(\sum_{n=0}^{\infty}\frac{2^n}{z^n}\right)[/tex]

    These are the Laurent Expantions in the respective regions.
  4. Oct 26, 2004 #3
    Heh, i always tend to do it the complicated way...


    using this, we will have

    for [itex]|z|<1[/itex]


    for [itex]1<|z|<2[/itex]


    for [itex]|z|>2[/itex]


    50 posts woohoo!!!
  5. Oct 26, 2004 #4


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    yea, but thats what i expected. you have just positive powers of n for the taylor series, positive and negative for the laurent series that converges in a ring, and only negative powers for the laurent series that converges out to infinity. my question was whether you could always assume this will happen.
  6. Oct 27, 2004 #5
    Well, the exponential function only has positive (and infinte) terms and it converges for all z. An you can do a number of composite functions using this fact.
    Last edited: Oct 27, 2004
  7. Apr 11, 2010 #6
    I find it difficult to understand why we have to make a transition from Taylor series to Laurent series....why does the taylor series not work when we have an annulus?(I could put it this way:What is the special importance of these negative terms needed to describe a function around a singularity?)

    I found an explanation of Taylor series on https://www.physicsforums.com/blog.php?b=1758...which [Broken] intuitively describes how the Taylor series comes into being and why it works...is there any similar reasoning for Laurent series? What is it about the negative powers in the Laurent series that allow it to define the function around a singularity?
    (This is the vital difference between Taylor series and Laurent series,right?)

    Also,one thing about a (complex) power series is that it contains powers of (z-z*)..where z* is the point at which the function is evaluated(a constant point)...but if we vary the variable z,wouldn't we get different values of the function?
    Moreover,when we write out the series at the point,we get an expression with the vaiable z still there ie we don't get a caonstant value for the function,,,then how indeed do we get the value of the function there?

    Lastly,why is it that only analytic functions can be expressed as power seires? Why not any other complex function?
    Last edited by a moderator: May 4, 2017
  8. Apr 11, 2010 #7
    but, could it be the function

    [tex] \frac{x+y}{1+x^{2}-y^{2}} =f(x,y) [/tex]

    be expanded into a convergent laurent series ??)(in powers of x and y), that is the question i have never heard about multi variable laurent series, although multivariable Taylor series exist

    also would it be possible to expand ONLY in one variable for example

    [tex] \frac{x+y}{1+x^{2}-y^{2}} =f(x,y) = \sum_{n=-\infty}^{\infty}a(y)(x+c)^{-n}[/tex] for some real 'c'
    Last edited: Apr 11, 2010
  9. Apr 12, 2010 #8
    I didn't understand how zetafunction's reply is related to my questions!
    Last edited: Apr 12, 2010
  10. Apr 12, 2010 #9
    Discussion of Laurent series should be in any textbook on complex analysis.
    For example: L. Ahlfors, COMPLEX VARIABLES.
  11. Apr 12, 2010 #10
    Actually,my exams are starting on Friday,and I don't really have time to hunt down books in the library...besides,our college library contains essentially books written by our own proffessors,and it's highy improbable that that they del with Laurent seroes in such detail.
    I would really appreciate if someone gave atleast some hints to my questions.
  12. Apr 12, 2010 #11
    Note that analytic means that the Taylor expansion converges to the function. This is not the case for all infinitely often real differentiable functions. But for complex differentiable functions one can prove that they are analytic.

    Now, the whole theory about Taylor and Laurent expansions cannot be explained in a few lines here. However, in complex analysis you have some powerful tools availble allowing you to understand certain results in some simple way without going through the standard textbook treatment.

    E.g., to address your point about Laurent expansion, I can set up the following reasing. I assume that you accept that a function that is complex differentiable has a converging Taylor expansion everywhere. Now, suppose that f(z) is a complex differentiable function except at some point z*, where it has a singularity. Then clearly, no series exansion around some arbitrary point p can converge at z*.

    Let's now look at what happens if we perform the following conformal transform from the z-plane to the w-plane:

    w(z) = 1/(z-z*)

    This sends the singular point to infinity in the w-plane. The inverse transform is:

    z(w) = z* + 1/w

    Let's now define a function g(w) on the w-plane that simply evaluates the function f at z(w) (so g is the transform of f under the above conformal transform as we are transfering the values taken by f on the z-pklane to the w-plane according to the conformal transform):

    g(w) = f[z(w)] = f(z* + 1/w)

    Then g(w) for all finite |w| will be complex differentiable and thus has a Taylor exansion that converges everywhere on the w-plane, as the point z* lies at infinity in the w-plane. If we now expand around w = 0 we get a Taylor expansion of the form:

    g(w) = a0 + a1 w + a2 w^2 + a3 w^3 + ....

    Since f(z) = g[w(z)] = g[1/(z - z*)], this means that:

    f(z) = a0 + a1 (z-z*)^(-1) + a2 (z-z*)^(-2)+ a3 (z-z*)^(-3) + ....

    So, we see that f can be expanded around z* in a Laurent expansion.
  13. Apr 12, 2010 #12
    Note that in the above we assume that the limit f(z) for z to infinity exists, so it is only a specific example.
  14. Apr 13, 2010 #13
    Wow! Thanks Count Iblis..this is all I really wanted....a rather intuitive approach to the topic.

    Thanks again.
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