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Help understanding Laurent series in complex analysis

  1. Jun 16, 2011 #1
    The part about Laurent series in my Complex Analysis book is somewhat vague and Wikipedia etc. didn't help me much.

    I am hoping someone would tell me the exact mathematical definition of a Laurent series (around a given point?) of a given function, perhaps providing an example. Also, how can i use this to find the residue of the contour integral around a single singularity, provided that i have already located the singularities?

    Any help is much appreciated!
     
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  3. Jun 16, 2011 #2

    micromass

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    Hi Waxbear! :smile:

    As you know, we can find the Taylor expansion of an analytic function around every point of the domain (which is assumed open).

    We want to expand the notion of Taylor series a bit such that it also holds for singularities. That is, we want to take a singularity (which is a point p not included in the domain, but there is a small ball around p such that every point in that ball, expect p, is included in the domain), and we want to write some sort of Taylor expansion of the function around this point. However, to do this, we need to allow negative powers.

    So, what's the idea? Take a singularity a, we want to express that if z is close to a, then

    [tex]f(z)=...+a_{-3}(z-a)^{-3}+a_{-2}(z-a)^{-2}+a_{-1}(z-a)^{-1}+a_{0}+a_{1}(z-a)^{1}+a_{2}(z-a)^{2}+a_{3}(z-a)^{3}+...[/tex]

    It can be shown that this is indeed possible and that the Laurent series is actually unique!!

    Now, how do we find the Laurent series? Let me give three examples:
    • Let [itex]f(z)=\frac{\sin(z)}{z}[/itex], this has a singularity at 0. So, we use the Taylor series of the sine to find the Laurent series:
      [tex]f(z)=\frac{\sin(z)}{z}=\frac{z-\frac{z^3}{3!}+\frac{z^5}{5!}-...}{z}=1-\frac{z^2}{3!}+\frac{z^4}{5!}+...[/tex]

      As you can see, there are no negative exponents of z in this expansion. This is because the singularity 0 was actually removable. That is, if we add 0 to the domain and define f(z)=1, then this extends f to an entire analytic function! So our Laurent series is in this case the Taylor series! (this is always the case with removable singularities)
    • Let [itex]f(z)=\frac{\cos(z)}{z}[/itex], this has a singularity at 0. So, we use the Taylor series of the cosine to find the Laurent series:
      [tex]f(z)=\frac{\cos(z)}{z}=\frac{1-\frac{z^2}{2!}+\frac{z^4}{4!}-...}{z}=z^{-1}-\frac{z}{2!}+\frac{z^3}{4!}+...[/tex]

      Now there is a negative exponent of z. But there is only one negative exponent. We say that the function f has a simple pole in 0. If there are only finitely many negative exponents, then we say that the function f has a pole in 0.
    • Let [itex]f(z)=e^{\frac{1}{z}}[/itex], this has a singularity at 0. So, we use the Taylor series of the exponent to find the Laurent series:

      [tex]e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+...[/tex]

      we just need to substitute 1/z for z:

      [tex]f(z)=e^{\frac{1}{z}}=1+\frac{1}{z}+\frac{1}{2z^2}+\frac{1}{6z^3}+...[/tex]

      Thus f has an infinite number of negative exponents of z! We say that 0 is an essential singularity of f.

    An important feature of the Laurent series is the residue. The residue around a is just the coefficient of [itex](z-a)^{-1}[/itex]. This residue is important because of the residue theorem, which allows us to easily calculate integrals.
     
  4. Jun 16, 2011 #3
    This is a very good explanation, thank you! I finally feel like i understand this.. This also gave me a better understanding of the difference between the various types of singularity. Your examples seem to be general Laurent series, they do not include the point a where the singularity is. If for example a function had 2 singularities(poles) and i wanted to find the value of an integral of a circle that enclosed both of these, i would have to find both of the residues, so i guess i would need 2 Laurent series, one for each point? I hate to ask for another example, because i feel that you have already supplied me with a very elaborate explanation, but if it isn't too much? :P
     
  5. Jun 16, 2011 #4
    i see now that all your examples were with singularities in 0, so i guess i can just substitute all your z's with (z-a) in the examples...
     
  6. Jun 16, 2011 #5

    micromass

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    Yes, if you have two singularities, then you'll need two Laurent series. Remember that the Laurent series is only defined on a ball around the singularity (but not containing the singularity) such that the function is defined in every point of the ball. Thus if we have two singularities, then the balls around each of the singularities will not contain the other singularity.

    As an example, let's look at the function

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

    this has two singularities: 0 and 2. Let's develop the Laurent series around 0 (the Laurent series around 2 will be analogous and will probably be a very good exercise for you!). First we take the partial fraction decomposition:

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

    Remember that (memorize this, it's very handy)

    [tex]\frac{1}{1-z}=1+z+z^2+z^3+z^4+...[/tex]

    under the condition that z<1. Thus

    [tex]\frac{1}{z-2}=-\frac{1}{2}\frac{1}{1-(z/2)}=-\frac{1}{2}(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+...)[/tex]

    if |z/2|<1, thus if |z|<2

    [tex]\frac{2}{z(z-2)}=-\frac{1}{2}(1+\frac{z}{2}+\frac{z^2}{4}+\frac{z^3}{8}+... ) -\frac{1}{z}[/tex]

    which can be simplified to


    [tex]\frac{2}{z(z-2)}=-\frac{1}{z}-\frac{1}{2}-\frac{1}{4}z-\frac{1}{8}z^2-\frac{1}{16}z^3+... [/tex]

    which is valid if 0<|z|<2. Thus the singularity around 2 is not included in the Laurent series. Note that the residue of this function is simply -1.

    If you do the calculation of the Laurent series around 2, then you'll get an expression for 0<|z-2|<2, thus 0 is not included in the Laurent series.
     
  7. Jun 16, 2011 #6

    micromass

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    I did all the singularities in 0 because that was easiest. For example if you want to know the Laurent series of

    [tex]f(z)=\frac{\sin(z)}{z-2}[/tex]

    in 2, then you will want to calculate the Taylor series of the sine in 2, and that's pretty tedious. So I did it in 0.
     
  8. Jun 16, 2011 #7
    Thank you very much! i have just found a practice sheet with this problem: Find the Laurent series for the function [itex]f(z)=\ufrac{1}{z(z+2)}[\itex] i'll try to solve this now, thank you again! :)
     
  9. Feb 25, 2013 #8
    Thanks a bunch, actually really helped! Cheers!
     
  10. Feb 27, 2013 #9

    mathwonk

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    some random general remarks, as footnotes to micromass' beautiful answer.

    a laurent series is what you get when you try to divide two power series. so they exist in order to flesh out and compute the fraction field of the ring of power series.

    the basic case of dividing power series is, as micromass said, 1/(1-z) = 1+z+z^2+....

    this then gives all the others as follows. to take the reciprocal of a+bz + cz^2+...

    where a≠0, just divide by a, to get a series starting with 1, then write it as 1 -(some power series P divisible by z), so 1/(1-P) = 1+P +P^2 +......

    Then to invert any power series at all, of form az^k + bz^(k+1)+.....

    first divide it by z^k, to get one beginning with a non zero constant, invert that as above, and finally multiply through by z^(-k). so basically all laurent series can be computed from the basic one 1/(1-z) = 1+z+z^2+.....
     
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