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.

8
votes
2answers
527 views

finding the expansion of $\arcsin(z)^2$

Is there a fast and nice way to find the expansion of $\arcsin(z)^2$ without squaring expansion of $\arcsin(z)$ ? For $|z|<1$ show that $$(\sin^{-1}(z))^2 = z^2 + \frac{2}{3}\cdot \frac{z^4}{...
2
votes
1answer
978 views

expansion of $\text{ cosh}(z+1/z)$

It is given that if $|z|>0$, $$\text{cosh}(z+1/z) = c_0 + c_1 (z + 1/z) + c_2(z^2 + 1/z^2) + \dots $$ where $$c_n = \frac{1}{2\pi}\int_0^{2\pi} \cos (n \phi) \text{cosh}(2\cos \phi ) d\phi$$ I ...
2
votes
1answer
86 views

complex analysis explanation

Can someone please explain why : "Residue at a finite point is zero if the function is analytic at that point". Some explanation going by the definition or Laurent's expansion will be helpful.
0
votes
1answer
348 views

Find the Laurent series and residue of $\frac{az+b}{cz+d}$ at $z_0 = -\frac{d}{c}$.

Find the Laurent series for the given function about the indicated point, and the residue of the function at the point: $$\frac{az+b}{cz+d}\quad \text{at}\quad z_0 = -\frac{d}{c}.$$ I'm not sure ...
1
vote
2answers
669 views

Find the Laurent series at the given point and its residue

a) $\displaystyle \frac{\sin(z)}{(z-\pi)^2}$ at $z_0=\pi$ b) $\displaystyle \frac{1}{1-\cos(z)}$ at $z_0=0$ I'm having trouble understanding Laurent series, please help!
2
votes
3answers
3k views

Difficulties performing Laurent Series expansions to determine Residues

The following problems are from Brown and Churchill's Complex Variables, 8ed. From §71 concerning Residues and Poles, problem #1d: Determine the residue at $z = 0$ of the function $$\frac{\cot(z)}{...
0
votes
1answer
210 views

Finding Laurent Series

This was an exam question and I was absolutely lost. Find positive constants $a,b$ such that the function $$ \frac{1}{z^2+a} + \frac{1}{z+b} $$ has three Laurent series, one for each of the domains ...
1
vote
2answers
213 views

product $abababab…$ in clifford algebra

Let $a,b$ are vectors in vector space $V \leq \mathcal{Cl}_n(V)$. I would like to know if product $ababab...ab=(ab)^r$ can be written in form $\sum_{\alpha \in A} F_\alpha(a) G_\alpha(b)$. For some $...
0
votes
1answer
303 views

Find the 3 laurent series of $f(z)=\frac{3}{(1-z)(z+2)}$.

Find the Laurent series of $$f(z)=\frac{3}{(1-z)(z+2)}$$ centered at $0$ with the three domains $|z|<1$, $1<|z|<2$, $2<|z|$. I know you use partial fractions and I got $\frac{1}{1-z} + \...
0
votes
1answer
1k views

Laurent Series for $\frac{\cos(z)}{z^2}$ centered at $0$

How would you prove that this is entire if f(z) is (cos(z)-1) if z doesn't equal zero and is -1/2 if z equals zero?
3
votes
1answer
7k views

How do I find the Laurent series for $\frac{1}{z^2 - 4}$ at $z = 2$?

How do I find the Laurent series for $\frac{1}{z^2 - 4}$ at $z = 2$? Can anyone help me out with this? I used partial fraction decomposition and got the $\frac{1}{4}$ part of it, just don't ...
1
vote
2answers
602 views

Help evaluating an integral complex analysis

Question is: Evaluate $\int_\gamma z^n e^{1/z} \, dz$ where $\gamma$ is the circle of radius $1$ centered at $0$ and traveled once in the counterclockwise direction. I know that $\int_\gamma f(z) \, ...
1
vote
2answers
692 views

Laurent series for function $(z-1)/z^2$ in domain $ |z - 1| > 1$

I'm having trouble with the following Laurent series question: Laurent series for function $$\frac{z-1}{z^2}$$ in domain $|z - 1| > 1$. Specifically, how to modify $z^2$ into powers of $(z-1)$ and ...
1
vote
1answer
585 views

The field of Laurent series over $\mathbb{C}$ is quasi-finite

How can I prove that the field of Laurent series over $\mathbb{C}$ is quasi-finite, which means that it has a unique extension of degree $n$ for all $n \geqslant 1$ ? The article http://en.wikipedia....
2
votes
2answers
4k views

Find principal part of Laurent expansion of $f(z) = \frac{1}{(z^2+1)^2}$ about $z=i$.

Find principal part of Laurent expansion of $f(z) = \frac{1}{(z^2+1)^2}$ about $z=i$. My attempt at a solution: First, I noticed that if I plug in $z=i$, I get a zero in the denominator. This leads ...
9
votes
1answer
18k views

Difference between the Laurent and Taylor Series.

I have never taken complex analysis, but I am preparing for a GRE for this week end and I am trying to learn a bit about Laurent Series. So far what I get is that the Laurent Series are of form $$\...
5
votes
1answer
3k views

Multiplying Laurent series

I was solving the following problem from Bak & Newmans Complex Analysis (Chp. 9 # 12b): Find the Laurent series (in powers of z) for $$\frac{1}{z(z-1)(z-2)}$$ in the open annulus $1 < |z| < ...
3
votes
2answers
1k views

Laurent series $z=i$

To find the Laurent series expansion for $\frac{1}{1+z^2}$, centered at $z=i$ would using partial fraction decomposition be the right idea? So, $\frac{1}{1+z^2}$=$\frac{1}{(z+i)(z-i)}$=$\frac{\frac{-1}...
1
vote
1answer
174 views

Domain for Laurent Series

I need to evaluate $f(z)=\frac{e^{-z}}{(z-1)^2}$ around the circle $|z|=3$. I am trying to figure out which domain the Laurent series will be valid. Is it $-1<|z-1|<3$? Or would it be analytic ...
1
vote
2answers
412 views

Determining Laurent Series

This question is from Complex Variables and Applications by Brown & Churchill, 8ed. Section 62, #2. Determine the Laurent Expansion of: $$\frac{\exp(z)}{(z+1)^2}$$ for the interval, $ 0 < |...
2
votes
2answers
434 views

Find the Laurent expansion in powers of $\,z\,$ and $\,1/z\,$

$f(z)=\large\frac{z+2}{z^2-z-2}$ in $1<|z|<2$ and then in $2<|z|<\infty$ I am unsure how the two different regions of $z$ affect the series expansion. Any help would be appreciated.
1
vote
1answer
260 views

Behavior at infinity.

Classify the behavior at $\infty$ for $$f(z)=\frac{\sin z}{z^2},\,g(z)=\frac{1}{\sin z},\,h(z)=\exp\left(\tan\frac{1}{z}\right).$$ So I just considered $f(1/z),g(1/z),h(1/z)$ at $z=0$. For $f$ I get $...
2
votes
1answer
972 views

Radius of convergence of Maclaurin series for $\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$

What is the radius of convergence of the Taylor series about $z=0$ for $h(z)=\frac1{\sin z}-1/z+\frac{2z}{z^2-\pi^2}$? Here's a plot http://www.wolframalpha.com/input/?i=2z%2F%28z^2-pi^2%29%2B1%...
2
votes
2answers
1k views

Removable singularity and laurent series

How to show $z=\pm\pi$ is a removable singularity for $\frac1{\sin z}+\frac{2z}{z^2-\pi^2}$? I tried to compute the Laurent series, specifically the coefficients for $1/z,1/z^2,...$ since if we can ...
0
votes
1answer
110 views

Laurent series expansion of $f(z)=\frac {2(1-i)z} {z^2-2(1+i)z+4i}$ in $|z|>2$

Can anybody guide me how to find Laurent series expansion $f(z)=\frac {2(1-i)z} {z^2-2(1+i)z+4i}$ in $|z|>2$, I tried to complete the square in the denominator, but couldn't finish
0
votes
2answers
318 views

Finding the sum of a given Laurent series

A Laurent series around $0$ has $c_k=(-2)^k \frac {1} {(2k)!},k>0$; $c_0=0$; $c_k=(-1)^k\frac {1} {k},k<0$, I found the annulus of convergence to be $|z|>1$, How to find the sum function of ...
2
votes
1answer
851 views

Laurent expansion of $\csc^2(\frac{\pi}{z})$ about $\frac{1}{3}$ for $|z-\frac{1}{3}| \lt \frac{1}{12}$

The question is: Find the Laurent expansion of $\csc^2(\frac{\pi}{z})$ about $\frac{1}{3}$ for $|z-\frac{1}{3}| \lt \frac{1}{12}$. In particular what is the coefficient of $(z-\frac{1}{3})^{-2}$. I ...
2
votes
1answer
94 views

Evaluate $\int_C\frac{dw}{e^w-1}$ over some loop C contained in the annulus $0<|z|<2\pi$.

Evaluate $\int_C\frac{dw}{e^w-1}$ (counterclockwise) over some loop C contained in the annulus $0<|z|<2\pi$. Considering the coefficient of $1/z$ in the Laurent series for $\frac{1}{e^z-1}$ by ...
2
votes
1answer
437 views

Laurent series expansions, complex variables

Given $f(z)=\frac{1}{z^2(1-z)}$ I am to find two Laurent series expansions. There are two singularities, $z=0$ and $z=1$. So for the first expansion, I used the region $0<|z|<1$ and I got $\sum_{...
3
votes
2answers
599 views

Laurent series, complex analysis.

I am working on a complex analysis exercice. I need to find the Laurent series of the function: $$f(z) = \frac{e^z}{z + 1}$$ about $z = −1$. I know that the result is $$\sum_{n=-1}^{\infty}\frac{(...
1
vote
1answer
223 views

Complex analysis Laurent series evaluated on unit circle

Let $f(z)$ be a function analytic on an annulus that includes the unit circle $z=e^{i\theta}$. By taking that circle as the path of integration for the coefficients in the Laurent series, show that $$ ...
1
vote
1answer
1k views

Laurent expansion with essential singularity

I am doing a multiple choice test for complex analysis, and I am stuck a bit at the following one. Let $f$ be holomorphic with an essential singularity at $0$. Then for every $z_0\in \mathbb{C}$ the ...
1
vote
1answer
1k views

Laurent Series and Taylor Series

I am trying to find the Laurent series of $\dfrac{1}{(1+x)^3}$; would this be the same as finding the Maclaurin series for the same function?
4
votes
2answers
2k views

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

Let $f(z)=\frac{1}{(z-1)^2(z+1)^2}$. While trying to expand this function into the Laurent series, convergent in $P(0,1,2):=\lbrace z\in\mathbb{C}:1<|z|<2\rbrace$, a few questions popped into ...
2
votes
1answer
775 views

Problem with the Laurent series

Find The Laurent series for the following function on the annulus $1<|z|<2$ : $\displaystyle f(z)=\frac{2z}{z^2+z-2}$ My work : $\displaystyle f(z)=\frac{2}{3} \left( \frac{1}{z-1}+\frac{2}{...
1
vote
1answer
154 views

Laurent expansion of an analytic function in a neighborhood of 0

Let $f$ be an analytic function in the annulus $ \{ 0<|z|<R \}$ Is it true that the Laurent series expansion of $f$ about $0$ has a finite number of negative coefficients? It came to my mind ...
1
vote
1answer
159 views

Residues and Laurent expansion

How can I prove that the coefficient of $(z-z_0)^{-1}$ when $f(z)$ has a pole of order $m$ at $z=z_0$ is given by: $$ a_{-1}=\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}\left[(z-z_0)^mf(z)\right]_{z=z_0} $...
6
votes
3answers
15k views

The degree of a polynomial which also has negative exponents.

In theory, we define the degree of a polynomial as the highest exponent it holds. However when there are negative and positive exponents are present in the function, I want to know the basis that we ...
10
votes
2answers
23k views

$e^{1/z}$ and Laurent expansion

$e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as $$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$ The coefficients of this Laurent expansion are computed ...
1
vote
1answer
140 views

Innermost Laurent Expansion

I've been really struggling with how to do this question. I can expand it by partial fractions but then have no idea what to do next. Find the principal part of the innermost laurent expansion for: $ ...
1
vote
1answer
2k views

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

How can we compute the Laurent expansion of $$f(z)=\frac{1}{z(z-i)^2}$$ about $0$ in $\Omega=B_1(0)\setminus\{0 \}$? Thoughts: By partial fractions we have that $$f(z)=\frac{1}{z-i}-\frac{i}{(z-...