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Questions tagged [laurent-series]

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.

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Classification of singularities of complex multivalued function

I have some problems dealing with multivalued functions when it comes to handling singularities. I'll give an example and try to ask questions based on it. I want to classify the singularities of $f(...
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262 views

Laurent expansion for $1/\cos(z)$

I have a quick question. How to find the Laurent expansion for $1/\cos(z)$ In the link above the person asks how to find the Laurent expansion for $\frac{1}{cos(z)}$. The accepted answer utilizes ...
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Showing that $\csc\left(\frac{1}{z}\right)$ does not have a Laurent series convergent in $B^{\circ}(0,r)$

I am trying to show that the function, $$f(z)=\csc\left(\frac{1}{z}\right),$$ does not have a Laurent series about $0$ that converges to $f$ in $B^{\circ}(0,r)$ for any $r>0$. I thought I could ...
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Laurent series $\sin(\frac{z}{1-z})$ at point $z_0 = \infty$

I need to find the Laurent series of $$\sin\left( \frac{z}{1-z} \right)$$ at the point $z_0 = \infty$. How I can decompose this? At point $z_0 = 1$ everything turned out well. There I tried 2 things. ...
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52 views

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}^...
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42 views

Laurent Series of $f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4}$ that converges at $z=1$ in powers of $z-2$

I am trying to find the Laurent series of, $$f(z)=\frac{2}{(z+2)^2}-\frac{5}{z-4},$$ in powers of $z-2$ that converges at $z=1$. My attempt: I think our radius for convergence is $|z-2|<2$. Now, \...
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34 views

Residue of $z_0=1$ for $f(z)=\frac{z^3+5}{z(z-1)^3}$

Consider $$f(z)=\frac{z^3+5}{z(z-1)^3}.$$ I am trying to find the residue of the pole of order $3$, $\ z_0=1$. I know from calculations that $$\text{Res}(f,1)=\frac{1}{2}\lim_{z\to 1}\frac{\partial^2}...
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59 views

Find the Residues of $f(z)=\frac{z(z-\pi)^2}{\sin^2(z)}$

I am trying to find the residues of the function, $$f(z)=\frac{z(z-\pi)^2}{\sin^2(z)}.$$ My attempt: I have considered three singularities: $z_0=0,\pi,k\pi \ (k\in\mathbb{Z}, \ k\neq 0,1).$ For $...
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Laurent series of $f(z)=\text{cosec}(z)$ that is valid at $\frac{1}{2}$

I am trying to find a Laurent series for $f(z)=\text{cosec}(z)$ up to the $z^3$ term about $z=0$ that is valid at $\frac{1}{2}$. Now $f$ has singularities at $z=k\pi, \ \ k\in\mathbb{Z}$. So if we ...
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39 views

Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ that converges at 3

I am trying to find the Laurent series for $f(z)=\frac{1}{z-2i}-\frac{1}{z+i}$ with centre $1$ that converges at $3$. I was thinking that if we want the series to converge at $3$, we are interested ...
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38 views

Laurent Series Question: Summation Starts from $-\infty$?

Consider $f(z)=\frac{2z-4}{z^2-4z+3}$. Find the Laurent series for $f(z)$ that converges when $|z-1|>2$. My working: \begin{align} f(z)&=\frac{1}{z-3}+\frac{1}{z-1} \\ &=\frac{1}{z-1}\...
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Help with understanding the contour integrals in Laurent series.

i have been reading about Laurent series. Their formal definition is given as $$ f(z) = \sum_{n=1}^\infty \frac {b_n} {(z-a)^n} + \sum_{n=0}^\infty a_n(z-a)^n $$ Where $ b_n = \frac { 1}{2i\pi}\...
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114 views

Laurent Expansion of $f(z)=\frac{1}{z^2(1-z)}$

I am trying to find two distinct Laurent expansions of $f(z)=\frac{1}{z^2(1-z)},$ in powers of $z$. My attempt: I believe I have found one Laurent expansion of the region $0<|z|<1$. \begin{...
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Laurent Series of the Log-derivative of the Xi Function

So I am curious about the product representation of the Xi function. Some operations applied to the Hadamard product can result in the identity: $$\frac{\xi'(x) }{\xi(x) }=-\sum_{n=0}^{\infty }x^n\...
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64 views

How can I evaluate this complex integral $\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$?

I'm trying to evaluate the following complex integral using the residue method. $$\int_{|z|=1}e^{\frac{1}{z}}\cos{\frac{1}{z}}dz$$ The point $z_0=0$ seems to be a singularity. I'm not sure but I ...
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272 views

How can I find the residue of this removable singularity?

We have the following function : $$f(z)=\frac{z^2}{1-cosz}$$ where $z_0=0$ is a removable singularity since the limit as z goes to 0 is 2. In such cases, in order to find the residue I proceed by ...
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Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$

in this exercise I am asked to provide the principal part of the Laurent series of $$\frac{1}{(1-\cosh(z))^2}$$ And i am kinda struggling with fonding a solution or even a pattern towards one Thanks ...
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$\frac {1}{(x^2+1)^2}$ as a Laurent series

How can I write this function $\frac {1}{(z^2+1)^2}$ as a Laurent series around the point $z =i$? I am struggling with the partial fractions part. It's unreducible. The only thing that came to my ...
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Laurent series of $f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)}$

I want to find the Laurent series of the following function around the point $z=2$, $z$ is a complex number. $f(z)= \frac {z^2-2z+5}{(z-2)(z^2+1)}$ After the fraction decomposition we received ...
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Poles of alternating series

Consider the function $f(\cdot)$ defined as follows, $$ f(x) = \sum_{k=0}^{\infty} a_k \left(\frac{-1}{x}\right)^k$$ where $a_0 = 1$ and $a_k > 0$ for all $i$. Assume the series converges in ...
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Integral of $e^\left(1/z^2\right)$ around $|z|=1$ in the complex plane [closed]

$e^\left(1/z^2\right)$ has an essential singularity at $0.$ Don't know how to do this integral.
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Laurent Expansion of $\frac{1}{f(z)^2}$, for $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$ and $f''(a) = 0$.

First, we are asked: For $f$ a holomorphic function on a nbd of $a \in C$ with $f(a) = 0$ $f'(a) \ne 0$, find the residue of $\frac{1}{f(z)}$. I used $f'(a) \ne 0$ to conclude the $f$ is not constant,...
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Show that the field of fractions of $\mathbb{Z}[[x]]$ is properly contained in $\Bbb Q((x))$

I've been working on this for a while but I don't know how to proceed. Here it's what I've done: Clearly, my goal is to show that every element of the field of fractions is in $\Bbb Q((x))$ and to ...
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80 views

Laurent series expansion of a function with removable singularities

Let $$f(z)=\frac{\sin^2z}{z}, \quad z\neq0$$ with $z_0=0$ being an removable singularity of $f$ since $$\lim_{z \to 0}\, zf(z)=\lim_{z \to 0}\,\sin^2z=0$$ To find its Laurent series expansion within ...
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Doubt on Laurent series of $f(z) = 1/z(z-2)^2$

I've a doubt about this exercise Find the Laurent series of the function $$f(z) = \frac{1}{z(z-2)^2}$$ addend $z=0$ in the annular regions of interest The problem it's not totally about the how ...
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Confusion with the intuitive understanding of Laurent's series and it's reduction to Taylor's

I know that Laurent's series is used for an annulus with the function $f(z)$ being analytic in the region R defined by the annulus ($r<|z|<R$, where $R > r$). Now my question is that ...
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Uniform convergence and series of expansion

Assume the following sum: $$ s(x) = \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\frac{1}{n \sqrt{n^2+x^2} \left(\sqrt{n^2+x^2}+n\right)}, $$ where $x \in \mathbb{R}$. Since $f_n(x) = f_n(-x)$, it ...
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Find the Laurent expansion $f(z)=\frac{1}{z(1-z)^2}$.

Find the Laurent Series expansion of $$f(z)=\frac{1}{z(1-z)^2}$$ at $z=1$. How do I do this when I am not given any region?
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Expand $\frac{e^{5z}}{(z-5)^3}$

Expand $\frac{e^{5z}}{(z-5)^3}$ around $z=5$ $$\frac{e^{5z}}{(z-5)^3}=\frac{e^{5z-25+25}}{(z-5)^3}=\frac{e^{25}e^{5(z-5)}}{(z-5)^3}=\frac{e^{25}\sum_{n=0}^{\infty}\frac{5^n}{n!}(z-5)^n}{(z-5)^3}={e^{...
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How to multiply in the formal Laurent series ring

I'm working on a problem that asks me to show that the ring of formal Laurent series $F((x))$ is actually a field when $F$ is a field. My problem is that the problem doesn't define the product between ...
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530 views

Laurent series expansion (example)

$$ \text{Let} \quad f(z)=\frac{1}{z}+\frac{2}{3-z},\quad z\in \mathbb{C}\smallsetminus \{0,3\}.$$ To find the Laurent series expansion of $f$ about $z_0=1$ within the annulus $1<|z-1|<2,$ one ...
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Laurent series of matrix function

I need to find the Laurent series of a matrix function around $z=\infty$, but I lack the math knowledge. I can't find a way to get it with WolframAlpha or ...
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Laurent series of an integral with parameter

To find the Laurent series of function $f(a)$ at point $a=0$ $$ f(a)=\int^1_0 \frac{d x}{x^2+a^2} $$ one can first do the integral $$ f(a)=\frac{1}{a}\arctan(1/a) $$ then expand $\arctan(1/a)$ and ...
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161 views

Laurent Series $e^{\frac1{1-z}}$, $|z|>1$.

I tried to find Laurent expansion for: $e^{\frac1{1-z}}$, $|z|>1$. I tried next: $\frac1{1-z}=-\sum_{n=1}^{\infty}\frac1{z^n}$, then using $e^{\frac1{1-z}}=1+\frac1{1-z}+\frac{1}{{2!(1-z)}^2}+\...
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40 views

Finding Laurent series of a function by changing variable

I'm trying to find Laurent series of the following function at $$\mid z\mid<1$$ My function is as follows: $$f(z)=\dfrac{1}{z(z-1)(z-2)}=\dfrac{1}{z}\left(\dfrac{1}{(z-2)}+\dfrac{1}{(1-z)}\right)$$ ...
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Laurent series of a fractional cosine and sine

i was wondering how i can write the laurent series of $$\cos(\dfrac{1}{z-2})$$ in z=2 ? and if i want the laurent series of $$\sin(\dfrac{1}{z})$$ ( for Z=2 too , how it is because here 2 is ...
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characterizing complex functions whose Laurent series coefficients have bounded 1-norm

Suppose $h:\mathbb{T}\to\mathbb{C}$ is a complex function on the unit circle with a valid Laurent series about zero: $$ h(z) = \sum_{k=-\infty}^\infty c_k z^k\qquad\text{for }|z|=1 $$ The coefficients ...
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61 views

Laurent Series Of $\frac{z}{(z+1)(z+2)}$ Around $z=-1$

Expand $\frac{z}{(z+1)(z+2)}$ Around $z=-1$ $$\frac{z}{(z+1)(z+2)}=-\frac{1}{z+1}+\frac{2}{z+2}=-(z+1)^{-1}+\frac{2}{z+2}=\\=-(z+1)^{-1}+\frac{2}{1-(-z-1)}=-(z+1)^{-1}+2\sum_{n=0}^{\infty}(-1)^n(z+1)^...
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Expand $\frac{z-\sin z}{z^3}$ Around $z=0$

Expand $\frac{z-\sin z}{z^3}$ Around $z=0$ $$\sin z=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}$$ So $$\frac{z-\sin z}{z^3}=\frac{z-\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}}{z^3}=\...
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Laurent series $\frac{e^{2z}}{(z-1)^3}$ around $z=1$

Expand $$\frac{e^{2z}}{(z-1)^3}$$ around $z=1$ $$\frac{e^{2z}}{(z-1)^3}=\frac{e^{2z-2+2}}{(z-1)^3}=\frac{e^{2(z-1)+2}}{(z-1)^3}=\frac{e^{2(z-1)}e^2}{(z-1)^3}$$ Now can we say that $e^{2(z-1)}=(e^{(z-...
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Laurent Series for singularities and poles

Hi guys I was wondering how I can understand if the sin and the cos has essential singularities. for instance if I want to understand if 0 which singularity is i, can write the Laurent series only of ...
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69 views

Laurent series expansion of $f(z)=\frac 1{z^2+z+1}$ for $|z| > 1$ centered at $z=0$

Since $|z| > 1$, with finding a convergent series in mind I wanted to write an equivalent closed-form expression of $f$ in terms of $\frac 1z$ because $|\frac 1z| < 1$. So I tried to factor out $...
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1answer
47 views

Laurent Expansion $\frac{e^z}{z-1}$

Expand $\frac{e^z}{z-1}$ at $|z-1|>1$ $t=z-1\iff t+1=z$ $e^z=\sum_{n=0}^\infty \frac{z^n}{n!}=\sum_{n=0}^\infty \frac{(t+1)^n}{n!}$ $\frac{1}{t}=\frac{1}{1-1+t}$ Is it the way?
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59 views

Laurent Expansion Of $\frac{1}{z^2+1}$

Expand $\frac{1}{z^2+1}$ around $0<|z-i|<2$ $$\frac{1}{z^2+1}=\frac{1}{2i(z-i)}-\frac{1}{2i(z+i)}=\frac{1}{2i}\cdot\frac{1}{(z-i)}-\frac{1}{2i}\cdot\frac{1}{(z+i)}$$ How can I expand it to ...
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131 views

Explicit calculation of Laurent series coefficients by evaluating integral

I want the Laurent expansion \begin{equation} f(z) = \sum_{n=-\infty}^{\infty}c_n (z-a)^n \end{equation} of $f(z) = \frac{1}{(z+1)(z+3)}$ around the isolated pole $a =-1$ (for $|z+1|<2$). From a ...
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1answer
80 views

complex analysis Laurent series around no singularity point?

hi Guys I am trying to determine a series for the following function: $$f(x)=\frac{2z}{(z+i)(z-1)}$$ about $|z-1|<\frac{1}{2}$ I already developed partial fractions: $$\frac{1 + i}{z + i} + \...
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1answer
70 views

Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence.

I'm working on the following problem: Find the two Laurent series for $f(z)=\frac{1}{z^3-z^4}$ that involve powers of z, and state their domains of convergence. I'm not sure how to go about it, any ...
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1answer
48 views

An apparent contradiction based on considering the series of $f(z)$ and relating it to $f(1/z)$

I have a function with a relationship between $f(z)$ and $f(1/z)$ that, when interpreted naively using the series for $f$, gives wrong results. I'm going to give this naive argument in the hopes that ...
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1answer
81 views

Laurent series for $f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$ at $-3i$ (not leaving as a product of series)

I got this problem in my complex analysis class: Find the Laurent series of $$f(z) = \frac{\sinh(z + 3i)}{z(z + 3i)^3}$$ to calculate the residue at $z=-3i$. Is there an easy way to calculate the ...
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1answer
165 views

Finding Residues Using Partial Fractions

I'm calculating some countour integrals of complex functions by residues, but I can't understand an certain way to calculate these residues in some cases. We have a theorem that states that the ...