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.

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Laurent series of $e^z$

find the Laurent series centered at $z=1$ $$ f(z)=\frac{e^z}{(z-1)^2} $$ I thought that the denominator part is safe by our center and the expansion is just about the exponential which is a Taylor ...
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Laurent series of this function

Find the Laurent series of the function$$ f(z) = \frac{z+1}{z(z-4)} $$ in the annulus $0<|z-4|<4$. My approach: $$ \begin{aligned} f(z) &= \frac{z+1}{z(z-4)} =\left(\frac{-1}{4}\frac{1}{z}+\...
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What is the value of the sum of $e^{i(-ak^2 + bk)}$ when $k$ goes through every integer? [closed]

I'm doing a physics question sheet and I stumbled upon a series of the kind $$ \sum_{k \in \mathbb{Z}} e^{i(-ak^{2}+bk)} $$ Where $a$ and $b$ are real numbers with $a>0$. I have no idea how to do ...
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Laurent series $\ \ \ \ \ $

Trying to compute first few terms of the Laurent series for: $$ \frac{e^z}{z^2(z^2+1)} =\sum_{n=-2}^\infty c_n z^n = \quad ?$$ I know the expansion of $$e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$$ and ...
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Why does $\frac{z+2}{z-1}$ not have a series expansion around $|z-1|>1$?

Determine the Laurent series of $$z \mapsto \frac{z+2}{z-1}$$ over the region $$C := \left\{ z \in \mathbb{C} \mid |z-1| > 1 \right\} $$ With the region being simplified to $$ \frac{1}{|z-1|} < ...
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The Laurent series of $\frac{1}{(z-i)}+\frac{1}{(z-1)}$ in the annulus $0<|z-1|<√2$

I have no idea how to figure out how exactly I'm supposed to get these functions into the form $$\frac{1}{1-w}=\sum_{n=0}^\infty w^n$$ in this question: Find the Laurent Series of the function $$f(z)=...
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Laurent Expansion of $\frac{1}{z(z-1)}$

Find the Laurent expansion of $$\frac{1}{z(z-1)}=\frac{1}{z-1}-\frac{1}{z}$$ in $0<|z|<1$ and $|z|>1$ I think I'm getting confused about which expansion corresponds to which domain. I've ...
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Finding Laurent Series Expansion for $f(z) = \frac{z^3}{z^2-3z+2}$ for different domains

So I wanted to find the Laurent series expansion for $f(z) = \frac{z^3}{z^2-3z+2}$. So I guess more specifically, my question pertains how the expansion may change for different domains: $|z| < 1; ...
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2 answers
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How do I find the Laurent series expansion for this complex function? [duplicate]

I have given $f(z)=\frac{1}{z^2-1}$ and I want to find the Laurent series expansion for $0<|z-1|<2$. I have just remarked that $$f(z)=\frac{\frac{1}{2}}{z-1}-\frac{\frac{1}{2}}{z+1}$$I know ...
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name for the "last" nonzero coefficient of a polynomial or laurent series [duplicate]

the leading coefficient of a polynomial $f$ is its first non-zero coefficient and is sometimes denoted by “$l(f)$”. is there a similar name and notation for the last non-zero coefficient? for $f = X^2 ...
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1 answer
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Laurent series of $\frac{1}{(z+2)}\frac{1}{(e^{i\pi z /2}+1)}$

I'm re-learning this stuff. It is a little confusing to me how to obtain the Laurent expansion of exponential functions in denominator. I have to determine the Laurent series of $$\frac{1}{(z+2)}\frac{...
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Finding the Laurent series of $\frac{-z^2-2z+3i}{z-2i}$

Finding the Laurent series of $$f(z) = \frac{-z^2-2z+3i}{z-2i}$$ on the annulus of infinite external radius $|z-i| > 1$ This function looks fairly messy. We clearly have a simple pole at $z=2i$. ...
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2 answers
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Laurent Series for $\frac{1}{1 - z}$ and $\frac{1}{(1 - z)^2}$ around $|z| > 1$.

I am supposed to find the Laurent series for $f(z) = \frac{1}{1 - z}$ and $ f(z) = \frac{1}{(1 - z)^2}$ around $|z| > 1$. I think I can find the first one, but the second one to me seems a little ...
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Why is $z\mapsto\frac{1}{z}$ the universal example of a holomorphic function that fails to have a primitive?

This Wikipedia article (on Morera's theorem) mentions the following in the introduction. In a certain sense, the $\frac{1}{z}$ counterexample is universal: for every analytic function that has no ...
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Find Laurent expansion of $\frac{1}{\sin z^3}$ at $0 < ∣z-\sqrt[3]{2 \pi}∣ < ∣\sqrt[3]{4 \pi} - \sqrt[3]{2 \pi} ∣$

I need to find the principal part of the Laurent expansion of the function in the annulus described in the title. In my attempt, I have replaced $z$ by $w:=z - \sqrt[3]{2 \pi}$. I know the Taylor ...
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How do we find the power series of $f(z) = (1+2z)/(1-2z)$ centered at the origin?

$$f(z) = (1+2z)/(1-2z)$$ Centered at 0. I was struggling to find its power expansion for f(z) All I got is $$z<1/2$$ being the only radius of convergence? $$f(z) = 1/(1-2z) + 2z/(1-2z) = 1/(1-2z) +(...
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1 vote
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Conway's Proof of Laurent Series Development

I was learning Laurent development (Conway's text, page 107), where Conway uses a two-page (a little tedious) proof to show the existence of Laurent Series, and uniqueness is left as exercise. Then I ...
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Half of a Mean-Value Theorem

For an entire function $$f(z)=\sum_{n=0}^\infty c_n z^n \, ,$$ we simply have that $$\frac{1}{2\pi} \int_0^{2\pi} f(Re^{it}) \, {\rm d}t = c_0$$ $\forall R>0$. The same is true if $R\rightarrow 0$, ...
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How can I find the limit of the complex function: $f\left(z\right)=\frac{z\left(e^{\frac{1}{z}}-1\right)}{e^z-1}$?

I was asked to determine the type of the isolated singularity of the function $f(z)$ at $z=0$: $$f\left(z\right)=\frac{z\left(e^{\frac{1}{z}}-1\right)}{e^z-1}$$ I know that if the function's limit ...
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Where does Laurent series for zeta function converge?

I am wondering in what region of the complex plane the Laurent series $$ \zeta(s)=\frac{1}{s-1} + \sum_{k=0}^{\infty} \frac{(-1)^k \gamma_k}{k!} (s-1)^k $$ converges. It is straight forward to derive ...
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Find laurent series for function $\frac{z + i}{z^2}$ for $|z - i| > 1$

So, I have function $$f = \frac{z + i}{z^2}$$ Since $f \in H(\{|z - i| > 1\})$ I was hoping to find laurent expansion. It is pretty easy to do it in case of writing down expansion for $|z - i| < ...
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1 answer
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Why is $F(x)\cong F[x]\otimes F(x^n)$?

Let $F$ be field. Let $F[x]$ denote the polynomial ring over $F$, and $F(x)$ its field of fractions. Let $F(x^n)$ be the field of fractions of $F[x^n]$, the subring of polynomials in $x^n$. Consider ...
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1 vote
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Find the coefficients of $\prod_i\frac{x_i+y}{1-e^{-(x_i+y)}}$

I would like to compute the coefficients (in $y$) of the following series $$\prod_{i=1}^n\frac{x_i+y}{1-e^{-(x_i+y)}}$$ Is there a way to extract them using residue calculation?
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What can we conclude about the coefficients of Laurent expansion for two functions being equal on the real axis?

Consider two functions $f(z)$ and $g(\bar{z})$ where $z=x+iy$ and $\bar{z} = x-iy$. It is given that these functions admit a Laurent expansion around $z=0$ as $$f(z) = \sum_{n=-\infty}^{\infty} f_n z^...
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1 vote
1 answer
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How to determine which fraction to use in Laurent series

I wish to find the Laurent series for the function $$f(z) = \frac{1}{z(z-1)}$$ on the annulus $2 < |z+2| < 3$. From the bounds of the domain I deduce that we would like something of the form $$\...
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What is the Laurent series of $e^{z+z^{-1}}$? (Detailed derivation using Cauchy product of series)

We know that: $$e^{z+z^{-1}}=\underbrace{\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=0}^{\infty}\frac{z^{-m}}{m!}\right)}_{=:\alpha}.$$ Now I am having trouble, applying the ...
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5 votes
2 answers
120 views

Can the field of Laurent series be made into an Archimedean ordered field?

In today's analysis class, my professor introduced the field of formal Laurent series $\Bbb R((x))$. He also talked about the dictionary order on $\Bbb R((x))$ and why it is not an Archimedean ordered ...
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  • 353
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0 answers
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Holomorphic function satisfying an integral property

Find a holomorphic function $f : \Bbb C^* \to \Bbb C$ such that $$\int_0^{2 \pi}f(e^{it})e^{int} dt = \frac{2 \pi}{\lvert n\rvert !} \quad \forall n \in \Bbb Z $$ I said that $$\int_0^{2 \pi}f(e^{it})...
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4 votes
2 answers
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Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$

I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$. My work so far By partial fraction decomposition we can get that: $$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(...
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  • 1,781
1 vote
0 answers
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Laurent series of the function $f(z)=e^{z+\frac{1}{z}}$ around zero [duplicate]

let define : Laurent series of the function $f(z)=e^{z+\frac{1}{z}}$ around zero: $$\sum^{\infty}_{n=-\infty}a_n z^n$$ prove that for all $n\in\mathbb{N}$ : $$a_n=a_{-n}=\sum^{\infty}_{k=0}\frac{1}{k!(...
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1 answer
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Laurent series of $\frac{1}{(z^2 + 1)(z^2 - 9)}$

I want to derive Laurent series of $$f(z) = \frac{1}{(z^2 + 1)(z^2 - 9)}$$ for two sets: $1 < |z| < 3$ and $3 < |z|$. My work so far First thing to do is to apply partial fraction ...
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  • 1,781
0 votes
1 answer
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Laurent Series and Residue for $f(z) = z^4 \sin(1/z)$ at $z=0$

I had to find the Laurent series and residue of the function. We know that the $res_{z_0} = a_{-1}$, the coefficient of $\frac{1}{z-z_0}$ of the Laurent series. For $f(z) = z^4 \sin(1/z)$ on the ...
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0 votes
1 answer
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How can I find the residue of this function at $z=0$?

How can I find the residue of $$\frac{\text{sin}(\frac{1}{z})}{z^2+1}\quad ?$$ at $z=0$? In order to find this, I tried to calculate $a_{-1}$ in the Laurent series, and then arrived at $$\frac{\text{...
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0 votes
2 answers
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Calculate the integral $\int_{\partial D} \frac{z+4}{z-4}\frac{e^z}{\sin z} dz$.

Take the rectangle $D = \{ z\in\mathbb{C};|x|\leq 2,|y|\leq 1\}$. I need to calculate the integral $$\int_{\partial D} \frac{z+4}{z-4}\frac{e^z}{\sin z} dz.$$ The only singularity in this case is in $...
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1 answer
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Expand the Laurent Series for $f(z) = \frac{1}{z(z-i)^2}$ for $0 < |z - i| < 1$

My attempt As $|z - i| < 1$, this means, that $$\sum_{n=1}^{\infty} (z - i)^n = \frac{1}{1 - (z - i)}$$ But more than that, I don't really know what to do. The problem here is, that what we get is: ...
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  • 459
1 vote
0 answers
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Find the Laurent Series for $f(z) = \frac{1}{z(z-i)}$ for $1 < |z| < \infty$

My attempt We know, that $$\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$$ for $|z| < 1$ As we have $1 < |z| < \infty$, this means that we have such $z$ that $\left|\frac{1}{z}\right| < 1$, thus ...
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  • 319
1 vote
1 answer
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Why is it that functions with nonisolated singularities at a point do not have Laurent series at that point?

Learning complex analysis, I've been taught that a function like csc(1/z) cannot have a Laurent series at 0, because there is a nonisolated singularity there. If I recall correctly, one needs to not ...
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1 vote
1 answer
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What is the Laurent series of $f(z)=1/z^2$

I have just looked at about 10 different Laurent series in the past days, but I cannot solve $f(z)=\frac{1}{z^2}$. I attempt, $|z|>0$, with u=$z^2+1$: \begin{equation} \frac{1}{z^2}=\frac{1}{z^2-1+...
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1 answer
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What's with the lower bound of Laurent series?

I have the simple example of finding the Laurent series of: $f(z)=\frac{1}{z-2}$ on $|z|\leq2$ and $|z|\geq2$ Procedure to find it is rather straight forward, develop it as a Power series (in the form ...
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0 votes
1 answer
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Developing Laurent series on three types of regions: centered at the origin, and non-centered at the origin

So, in lieu of this post, I want to deepen the discussion of the Laurent series. We have the Laurent theorem: If f(z) is analytic on $R_1\leq |z-z_o|\leq 2$, then $f(z)=\sum_{n=-\infty}^\infty c_n(z-...
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2 votes
1 answer
44 views

How to grasp the relationship between Laurent series depending on the region they are developed at

in trying to understand how to set up the Laurent series for a fractional expression, I have the given function \begin{equation} \frac{1}{(z-2)}-\frac{1}{(z-1)} \end{equation} The Laurent series of ...
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1 answer
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Finding a Laurent series for a complex function with two poles where one is outside of the region

when finding the Laurent series for $\frac{1}{(z-1)(z-2)}$ in the region $|z| <1$, I thought that one should first evaluate where the poles are, inside or outside the given region. z=2 is outside ...
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1 answer
60 views

Laurent Series of $f(z) = \frac{(z+1)^2}{z(z^3+1)}$ about $z = 0$?

This is supposed to be a non-calculator question, so I managed to get this far; $ z^3 + 1 = (z + 1)(z^2 - 1 +1)$ , by polynomial division. Therefore, $ f(z) = \frac{z+1}{z(z^2-z+1)} = (1 + \frac{1}{...
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-2 votes
1 answer
45 views

Challenging Laurent series

I want to find the Laurent series expansion of $f(z)=\frac{e^{2z}}{z^2}$, and apprently, the pole is at 0, so this series would be for $|z|>0$. I am no sure this is correct, but if I consider the ...
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2 votes
1 answer
84 views

Manipulation of Infinite Series Example

I'm struggling to understand the manipulation of an infinite series shown in the text below. We begin with the series $\sum_{n=0}^{\infty} u^n$ which we know converges since $|u|<1$. We then ...
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1 answer
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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 ...
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0 answers
33 views

Meromorphic function can be represented by a rational function

Let $z_1,...,z_n\in \mathbb{C}$ and $f: \mathbb{C}\setminus \{z_1,...,z_n\} \to \mathbb{C}$ be holomorphic. Furthermore, $\operatorname{ord}(f,z_i)$ is finite for all $i$ and $\lim\limits_{|z|\to\...
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  • 1,887
0 votes
1 answer
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Laurent series outside region of convergence of $\frac{1}{z^2+1}$

I have already computed the Laurent series of $\frac{1}{z^2+1}$ at $z_0=i$ in $R=\{z\in \mathbb{C} : 0<|z-i|<2\}$. I have to compute the Laurent series in $|z-i|>2$ right now, how can I do ...
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  • 1,887
1 vote
1 answer
50 views

How does one arrive at the basic limit formulation for the Stieltjes constants?

The starting definition that I am using is: $$\zeta(s)=\frac{1}{s-1}+\sum_{n=0}^\infty\gamma_n\cdot\frac{(-1)^n}{n!}(s-1)$$ If I naively differentiate, I find: $$\zeta'(s)=\sum_{m=1}^\infty-\frac{\ln(...
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  • 8,219
1 vote
3 answers
66 views

How to find the order of a pole

How can I find the order of the pole $z = \frac{\pi}{2}$ for $f(z)=\frac{1}{(2\log(z))(1-\sin(z))}$? I know the answer should be 2, but I can't solve it, mostly due to poor understanding of the pole ...
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