Questions tagged [laurent-series]

This tag is for questions about finding a Laurent series of functions and their convergence. The Laurent series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

Laurent series: Suppose that $~f(z)~$ is analytic on the annulus $~A : r_1 <|z − z_0| < r_2~$. Then $~f(z)~$ can be expressed as a series $$f(z) = \sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}+\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ The coefficients have the formulas $$a_n=\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\, dw$$and$$b_n=\frac{1}{2\pi i}\int_\gamma {f(w)}{(w-z_0)^{n-1}}\, dw$$where $~\gamma~$ is any circle $~|w − z_0| = r~$ inside the annulus, i.e. $~r_1 < r < r_2~.~$

The entire series is called the Laurent series for $~f~$ around $~z_0~$.

Notes: $~(a)~~~$ The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ is called the analytic or regular part of the Laurent series.

$(b)~~~$ The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$ is called the singular or principal part of the Laurent series.

$(c)~~~$ Since $~f(z)~$ may not be analytic (or even defined) at $~z_0~$ we don’t have any formulas for the coefficients using derivatives.

Remarks:

  • The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$converges to an analytic function for $~|z − z_0| < r_2~$.
  • The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$converges to an analytic function for $~ |z − z_0| > r_1~$.
  • Together, the series both converge on the annulus $~A~$ where $~f~$ is analytic.

The Laurent series is calculated over contour integrals of counterclockwise self-avoiding rectifiable paths of the function. For holomorphic functions the Taylor series and Laurent series are identical.

The Laurent series has a principal part, which consists entirely of negative-degree terms. When the principal part vanishes (there are no negative indices) the function is holomorphic; when it is an infinite sum the function has an essential pole.

Reference:

https://en.wikipedia.org/wiki/Laurent_series

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What's with the lower bound of Laurent series?

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Luthier415Hz
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Finding the Laurent series when a residue at $z=0$ is given

I have $a_n=\frac{1}{n^4+1}$, where n is an integer. I want to determine the Laurent series for a function f(z) such that the residue for $\frac{f(z)}{z^n}$ in $z=0$ is $a_n \forall \ n$. Based on this introduction, I start with …
Luthier415Hz
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Generic method for determining the order of a pole?

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HelloGoodbye
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Expand using Laurent series

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I've been having a bad time with the Laurent Series, so I would really appreciate if someone can help me with this problem: Consider $f(z)$ $$ f(z) = \frac{z^2+1}{2z-1} $$ Then: i) Classify the zeros and poles of $f(z)$, consider the extended…
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Silence
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Calculation of the Jacobi triple product near 1

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In order to find the Laurent series of $\frac{1}{\sin{x}}$, I used a well-known formula: $\sum_{n=0}^{\infty}{x^n} = \frac{1}{1-x} \quad (|x| < 1)$ by $\frac{1}{\sin{x}} =…
S tomio
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Laurent Expansion $\frac{1}{z(1-z)^2}$

Hi I was wondering if anyone could help me with this Laurent expansion $ f(z)=\frac{1}{z(1-z)^2} $ about $z=1$ I don't think I have done it correctly but this is what I did: $f(z)=\frac{1}{z} \frac{1}{1-z} \frac{1}{1-z} =\frac{1}{z}…
H99
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ogirkar
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Laurent expansion of $\frac{1}{z^2+i}$ at $z=i$

I want to find the Laurent series of $1\over(z^2+i)$at $z=i$. How do I start and go on from there? Do I start with either of these? $$1 \over{(z+i\sqrt i)(z-i\sqrt i)}$$ $$\frac{1}{i} \frac{1}{1+(z^2/i)}$$ answer for this question
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I have the following question : Find a Laurent expansion of $f(z) = \dfrac{z}{(z^2 + 1)}$, valid for $|z-3| > 2$. I have learnt finding the Laurent expansion of functions where the denominator is of the form $(z+a)^n$, but this form in the given…
Jasmine
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Laurent series $e^{\frac{1}{1-z}}$ at $|z| > 1$

Firstly, I use this: $e^{\frac{1}{1-z}} = \sum_{n=0}^{\infty} \frac{(-1)^n \frac{1}{z}}{n!\big(1 - \frac{1}{z}\big)^n}$ Next I use the binomial expansion for $\big(1 - \frac{1}{z}\big)^n = \sum_{k=0}^{n} \frac{n! (-1)^k z^{-k}}{k!(n-k)!}$ Combining…