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

1827 questions
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How to find Laurent Series for $z/(z-1)(z+4)$

How do I find the Laurent series for $$\frac{z}{(z-1)(z-4)}?$$ on: i) $0<|z-1|<5$; ii) $5<|z-1|$. I broke it up into $$\frac{1}{5}\left(\frac{4}{z+4}+\frac{1}{z-1}\right)$$ but now I am stuck.
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Problem with Laurent series

I am trying to find a Laurent series for $\cos(\frac{1}{z})z$. I know that $$ \cos(1/z) = \frac{1}{2} e^{(-i/z)}z + \frac{1}{2} e^{(i/z)}z = z \sum_{n=0}^\infty \frac{(-1)^n (\frac{1}{n})^{(2n)}}{(2n)!}. $$ I just don't know what I should do next
Jezzie
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Laurent series of $z/(z-6i)^9$

I'm really confused by the Laurent series, and need a little help with this one: $ f(z) = \frac z{(z-6i)^9} $ around $z = 6i$ The partial fraction method doesn't seems adapted for this problem. I wrote $$ f(z) = \sum_{n=-\infty}^{\infty} a_{n}…
Sihtam
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Laurent series solver

Given a fractional function containing polynomials in both numerator and denominator; find its Laurent series in all convergence domains. The polynomials are given by its zeros. For example, the function…
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Laurent Series 1/(z-3i)

Apologies if this is a simple question. I've just read a chapter on Laurent series which seems to indicate that in practice we don't calculate the coefficients of a Laurent series using the integral, instead we often manipulate Taylor series, adjust…
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Laurent Series for this function

$f(z) = \frac{1}{(z^2 + 1)^2}$ in the domain $\{z: 0<|z-i|<2\}$ but I'm struggling to get to the result, thanks in advance for your time
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laurent series of a function defined by an integral

let be the 2 functions defined by an integral a) $$ F(x)= \int_{0}^{x}g(t) dt $$ b) $$ H(x)= \int_{0}^{\infty} dt \frac{g(t)}{1+tx} $$ assuming that i only can evaluate $ F(x)$ and $ H(x)$ by numerical methods , how could i expand this functions…
Jose Garcia
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Suppose that I'm trying to construct Laurent series for the following complex function: $$f(z)=(z^2+4)^{\frac{1}{3}}$$ Since I have a 1/3 exponent, this means that my branch points must extend to infinity when I define my branch cut (to make f…
Incognito
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Laurent Series Region

So, I'm trying to find the regions in which the function may be represented as a Laurent series (expanded about the origin). Then I want to find those expansions: $$f(z)=(z^2+1)^{-1/2}$$ Now, since there's a singularity at $-i \space \space and…
Incognito
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Odd Laurent Series

So, I've been studying Laurent series, and I'm fine with series such as $ \frac {1}{(z-1)(z+1)} $ for example. For these, we can just use partial fraction decomposition and then geometric series. However, I'm not even sure how to get started with…
Incognito
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Laurent series and residue.

Why is $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, using the identity ($m=-n$)? Why shouldn't the exponent be equal to $2m-1$? Also, why is the following correct - $\sum_{n=0}^\infty (1/n!)(1/z)^n = \sum_{m=-\infty}^0…
Grtv
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Need Hint on Laurent series

I stuck at part (a) after 1 have done the partial fraction, can anyone give me some hints? for part (b) I express $$f_1(z)=(-1)^n-1(n-1)!f_n(z)$$ What I suppose to do next?
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Determine the analytic and principal part of Laurent series

Let $$f(z) = \frac{1}{(z+1)(z-1)(z-3)} = \frac{1}{8(z+1)}-\frac{1}{4(z-1)}+\frac{1}{8(z-3)}$$ If we want to find all possible Laurent series at center $z=0$, there will be three domains. $|z|<1$ $1<|z|<3$ $|z|>3$ The first and last domain can be…
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Using long division with negative exponents

I have the following function $$f(x)=\frac{\sum_{n=0}^{+\infty}a_n \, x^{-n}}{\sum_{n=0}^{+\infty}\left(b_n \, x^n + c_n \, x^{-n}\right)}$$ Is there a way, maybe using long division to find the following equivalence: $$\sum_{n=0}^{+\infty}d_n \,…
Balfar
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what is the condition for the laurent series coefficient become zero?

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Tong Su
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