Whats a Laurent series? And how do I use one to represent a function?

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SUMMARY

The discussion focuses on deriving two Laurent series for the function f(z) = 1/(z(1+z^2)) in specified domains. The first series is given by the expression ∑(−1)^(n+1) * z^(2n+1) + 1/z for the domain 0 < |z| < 1, while the second series is ∑[(-1)^(n+1)] / z^(2n+1) for the domain 1 < |z| < ∞. Key steps include identifying poles, using the geometric series for expansion, and ensuring convergence conditions are met based on the location of the poles.

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  • Understanding of Laurent series and their applications in complex analysis.
  • Familiarity with geometric series and Taylor series expansions.
  • Knowledge of poles and convergence criteria in complex functions.
  • Ability to manipulate rational functions and perform partial fraction decomposition.
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  • Study the properties and applications of Laurent series in complex analysis.
  • Learn about identifying and classifying poles of complex functions.
  • Explore the use of partial fraction decomposition in deriving series expansions.
  • Investigate convergence criteria for series expansions in complex analysis.
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Students and professionals in mathematics, particularly those studying complex analysis, as well as educators looking for examples of Laurent series applications in function representation.

laura_a
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Homework Statement


Write TWO laurent series in powers of z that represent the function

f(z)= \frac{1}{z(1+z^2)}

In certain domains, and specify the domains


Homework Equations



Well that's my prob, not sure what the terms in the Laurent series are

The formula I'm looking at is

\sum^{infty}_{n=0} a_n * (z-c)^n

for a complex function f(z) about a point c and a_n is a constant.

I sort of understand that, but do I use it to represent a function? Thats the part I'm not sure about

The Attempt at a Solution




My lame guess is that sub in the f(z) I'm given into the equation. I've seen one example where the fraction is split into two and then the Laurent expansion is applied. I have the answers (because it's a textbook question)

They are

\sum^{\infty}_{n=0} (-1)^{n+1} * z^{2n+1} + 1/z (0 < |z| < 1)

and

\sum^{\infty}_{n=0} [(-1)^{n+1}] / z^{2n+1} (1 < |z| < \infty)

All I need to know is what to I plug into where and I'll work on the rest :) any suggestions will make my day (night actually, but who's counting)

Thanks
Laura
 
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laura_a said:

Homework Statement


Write TWO laurent series in powers of z that represent the function

f(z)= \frac{1}{z(1+z^2)}

In certain domains, and specify the domains


Homework Equations



Well that's my prob, not sure what the terms in the Laurent series are

The formula I'm looking at is

\sum^{infty}_{n=0} a_n * (z-c)^n

for a complex function f(z) about a point c and a_n is a constant.

I sort of understand that, but do I use it to represent a function? Thats the part I'm not sure about

The Attempt at a Solution




My lame guess is that sub in the f(z) I'm given into the equation. I've seen one example where the fraction is split into two and then the Laurent expansion is applied. I have the answers (because it's a textbook question)

They are

\sum^{\infty}_{n=0} (-1)^{n+1} * z^{2n+1} + 1/z (0 < |z| < 1)

and

\sum^{\infty}_{n=0} [(-1)^{n+1}] / z^{2n+1} (1 < |z| < \infty)

All I need to know is what to I plug into where and I'll work on the rest :) any suggestions will make my day (night actually, but who's counting)

Thanks
Laura
Btw, put your TeX code between [ itex ] and [ /itex ] (with no spaces...I left spaces there just so that they would show up in my post.)

First you must identify the poles of your function.

Once you have the poles, pick one and use that point as the center of your Laurent series. Then you must expand everything in powers of (z-c)^n as you said, where c is the location of the pole.
The formula to use is the geometric series
[tex]\frac{1}{1-q} = \sum_{n=0}^\infty q^n[/tex]
at the condition that q < 1. (in your case, it will be a condition on the magnitude of a complex quantity).


For the domain, you will have to check if you have any restriction from using the geometric series and you must laos make sure that you don't go far enough from your pole that you run into one of the other poles.
 
Last edited:
To answer the question in your title, a Laurent series is a power series that includes negative powers. When your original function is a rational function, like the one you give, [tex]f(x)= \frac{1}{z(1+z^2)}[/tex], a good method is to take out one of the factors in the denominator, say 1/z, and expand the remaining function, 1/(1+z2), in a Taylor's series. Here we can write [tex]\frac{1}{1+z^2}= \frac{1}{1- (-z^2)}[/tex] and recognize that as a geometric series with "common factor" -z2 (I just noticed that is what nrqed said): [itex]1- z^2+ z^4- z^6+ \cdot\cdot\cdot[/itex]. Finally, put the 1/z back in: [itex]z^{-1}- z+ z^3- z^5+ \cdot\cdot\cdot[/itex]. Obviously, that cannot converge at z= 0. Further more the original power series only converges for the absolute value "common factor" less than 1. Here that is z2< 1 so the Laurent series converges for 0< |z|< 1.

For a different Laurent series, notice that 1/(1+ z^2) is discontinuous at z= i. Factor out 1/(z- i), leaving [tex]\frac{1}{z(z+ i)}[/tex] and find a Taylor's series for that, about z= i. A good method is to separate into two different fractions using "partial fractions" and write each as a geometric series.
 

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