In mathematics, the Laurent series of a complex function f(z) is a representation of that function as a power series which includes terms of negative degree. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied. The Laurent series was named after and first published by Pierre Alphonse Laurent in 1843. Karl Weierstrass may have discovered it first in a paper written in 1841, but it was not published until after his death.The Laurent series for a complex function f(z) about a point c is given by
f
(
z
)
=
∑
n
=
−
∞
∞
a
n
(
z
−
c
)
n
,
{\displaystyle f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-c)^{n},}
where an and c are constants, with an defined by a line integral that generalizes Cauchy's integral formula:
a
n
=
1
2
π
i
∮
γ
f
(
z
)
(
z
−
c
)
n
+
1
d
z
.
{\displaystyle a_{n}={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{(z-c)^{n+1}}}\,dz.}
The path of integration
γ
{\displaystyle \gamma }
is counterclockwise around a Jordan curve enclosing c and lying in an annulus A in which
f
(
z
)
{\displaystyle f(z)}
is holomorphic (analytic). The expansion for
f
(
z
)
{\displaystyle f(z)}
will then be valid anywhere inside the annulus. The annulus is shown in red in the figure on the right, along with an example of a suitable path of integration labeled
γ
{\displaystyle \gamma }
. If we take
γ
{\displaystyle \gamma }
to be a circle
|
z
−
c
|
=
ϱ
{\displaystyle |z-c|=\varrho }
, where
r
<
ϱ
<
R
{\displaystyle r<\varrho <R}
, this just amounts
to computing the complex Fourier coefficients of the restriction of
f
{\displaystyle f}
to
γ
{\displaystyle \gamma }
. The fact that these
integrals are unchanged by a deformation of the contour
γ
{\displaystyle \gamma }
is an immediate consequence of Green's theorem.
One may also obtain the Laurent series for a complex function f(z) at
z
=
∞
{\displaystyle z=\infty }
. However, this is the same as when
R
→
∞
{\displaystyle R\rightarrow \infty }
(see the example below).
In practice, the above integral formula may not offer the most practical method for computing the coefficients
a
n
{\displaystyle a_{n}}
for a given function
f
(
z
)
{\displaystyle f(z)}
; instead, one often pieces together the Laurent series by combining known Taylor expansions. Because the Laurent expansion of a function is unique whenever
it exists, any expression of this form that actually equals the given function
f
(
z
)
{\displaystyle f(z)}
in some annulus must actually be the Laurent expansion of
In complex analysis, what exactly is the purpuse of the luarent series, i mean, i know that it apporximates the function like a taylor series, an if the function is analytic in the whole domain it simplifies into a taylor series. But i fail to see its purpose - what does it do that the taylor...
Hi everyone!
I'm studying Laurent Series, and I thought I understood it but after solving one exercises I get confused. The problem was to get the Laurent series for the following function:
\frac{1}{\left(z-1\right)^2\left(z+3\right)}
I do it this way:
\frac{1}{\left(z-1\right)^2}\cdot...
Homework Statement
Could anyone help me with Laurent series? I do not understand it at all even though the book has several examples.
And here is one with my comments
Find the Laurent series of
\frac{1}{(z-1)(z-2)}
a in the region abs(z) < 1
b in the region 1< abs(z) < 2
c in the...
Homework Statement
Find the Laurent series representation for f(z)=(z-1)/(z-2) at z=i.
Homework Equations
NA
The Attempt at a Solution
I have taken multiple derivatives but I keep getting stuck at what to do after I find my representation of my nth derrivative. PLEASE HELP
Homework Statement
Find the Laurent series about 0 of 1/sinh up to (and including 0) the z5 term
Homework Equations
The Attempt at a Solution
Since 1/sinh is equal to
(1/z) * (1/(1+(z^2/3!)+(z^4/5!)+(z^6/7!)+...))
So if we work on the second term by dividing 1 by denominator and multiply...
Homework Statement
1) Find the Laurent series for (z^2)*cos(1/3z) in the region \left|z\right|
2) Find the Laurent series expansion of (z^2 - 1)^(-2) valid in the following region
a) 0 < \left|z - 1\right| < 2
b) \left|z + 1\right| > 2
Homework Equations
The Attempt at a Solution
I...
Laurent series "throwing away" terms
Homework Statement
Veeeery similar to https://www.physicsforums.com/showthread.php?p=1868354#post1868354":
Determine the Laurent series and residue for f(z) = \frac{1}{(e^{z} - 1)^{z}}
Homework Equations
We know that the Taylor series expansion...
How does one form a laurent series about the point z0 = 2i
for the function:
4cos(z*pi) / (z-z0).
Could one take advantage of the power series
1 / (z - z0)
1 / (z - 2i)
SUMMATION q^n
= SUMMATION (2i)^n
= 1 + 2i -4 -8i . . . . .
and somehow integrate the rest of the...
We were discussing them in my math methods class today however I'm not really sure how the idea works.
Does anyone know of any online references that might be of some help? Google wasn't much help for me =/
why NO multiple Laurent series ??
why are ther Taylor series in several variables (x_{1} , x_{2} ,..., x_{n} but there are NO Laurent series in several variables ? why nobody has defined this series , or why they do not appear anywhere ?
i think there are PADE APPROXIMANTS in serveral...
Homework Statement
Question is= Find all Laurent series expansion of f(z)=z^4/(3+z^2) around 1. I will be very very thankful if someone can help me to do this question.
Homework Equations
The Attempt at a Solution
can I assume (z-1=u) here and change the function in terms of...
f(x)=\frac{-2}{z-1}+\frac{3}{z+2}
our distance is from -2 till 1
we develop around 1
so our distances are 3 and zeo
so our areas are
0<|z-1|<3
0<|z-1|
3>|z-1|
but i was told to develop around
0<|z-1|<1
there is no such area
?
find the laurent series of f(x)=\frac{-2}{z-1}+\frac{3}{z+2}
for
1<|z|<2
i was by my teacher that the radius of convergence
is what smaller then the number which makes the denominator 0.
if
f(x)=\frac{1}{1-z}
then
the radius is 1 and
because 1-1=0
so
it is analitical on
|z|<1
so...
Homework Statement
Calculate a Laurent series about z = 0 for 1/(z2(z+i)) in the region D = {z: |z| < 1}
Homework Equations
The Attempt at a Solution
I used partial fractions to get 1/(z2(z+i)) = 1/z -1/z2 - 1/(z+i) but where do I go from here.
Homework Statement
Find the Laurent series for f(z)=1/(z(z-1)) valid on 1<|z-1|<infinity
Homework Equations
1/(1+a)=1-a+a^2-a^3... where |a|<1
we are not supposed to use integrals for this problem
The Attempt at a Solution
I want 1/(z-1) to be in my final answer, so I have...
Homework Statement
find the Laurent series for
\frac{z+2}{z^{5}-8z^{2}} in 2<|z|<\infty
Homework Equations
The Attempt at a Solution
Well, I factored out z^{5} in the denominator, which left me with a geometric sum (since |z|>2). I've come up with...
Homework Statement
f(z) = (z + 2)/(z - 2)
a) Find the Maclaurin Series for f on the doman |z| < 2.
b) Find the Laurent Series for f centered at z0 = 0 on domain 2 < |Z| < inf.
Homework Equations
The Attempt at a Solution
I'm having a hard time figuring out how (z + 2)/(z -...
[b]1. I am trying to calculate the laurent series expansion of the function 1/(1-z)² in the region 1<|z|
[b]2. None
[b]3. I can get an answer informally by doing the polynomial division like in high school, but I don't know if this is the right answer and in case it is I cannot prove...
Homework Statement
I'm asked to find the Laurent series of some rational function and using partial fractions I encounter something like 1/(c-z)^2 with c > 0.
Homework Equations
The Attempt at a Solution
I've tried several 'algebraic tricks' like multiplying for z^2 or just...
can we define a multivariable power series (laurent series)
\sum_{i,j,k,l,...=-\infty}^{\infty}a_{i,j,k,l,...}(X-a)^{i}(Y-b)^{j}(Z-c)^{k}(W-k)^{l}...
indices i,j k and l run over ALL the integers positive and negatives
how could i calculate the coefficients ?? a_{i,j,k,l} ?
Homework Statement
Hi all,
I've just calculated the first three nonzero terms of the Laurent series of 1/(cos(z)-1) in the region |z|<2pi, and now I've been asked to 'find the three non-zero central terms of the Laurent expansion valid for 2pi<|z|<4pi' - firstly, what does it mean by...
I don't quite understand a few details here. First, What is the difference between geometric series and laurent series? Than, how do I multiply/divide 2 series with each other? Finally, I have this problem, and I'm really clueless as of what to do.
Turn 1/(1-cos(z)) into a laurent series.
The problem:
find the laurent series for 1/(z^2-1)^2 valid in 0 < |z-1| < 2 and |z + 1| > 2
we know that f(z) has poles of order 2 at 1 and -1...
In the first region, there are no poles (since z=-1 isn't a part of it). We can write the equation as a product of 1/(z-1)^2 and 1/(z+1)^2...
Homework Statement
Find the Laurent series expansion of f(z) = (e^z - 1) / (sinz)^3 at z = 0.The Attempt at a Solution
Ok, so I'm confused on a number of fronts here. For e^z - 1, I assume you just use the standard power series expansion of e^z and then tack on a -1 at the end, which would...
Homework Statement
I am trying to understand Laurent series from the few and limited examples given by my texts.
I understand the basic idea, which is to expand the series about a function's pole.
So how would one go about finding the Laurent expansion for \frac{z + 1}{z^{2} - z - 6}?
Or...
Homework Statement
Hello, I'm having massive troubles with Laurent series'. I'm pretty shocking with series', so I'm probably making some fundamental mistake that you'll want to slap me for, but everytime I try one of these questions I'm wrong ever so slightly. I've attached a couple of...
Hello -
I have a problem in general finding the region in which the Laurent series converges...
Could someone please help me with this question - I know that this is is meant to be easy (as there is no fully worked solution to this) but I don't understand it:
f(z) = 1/ [(z^3)*cosz]...
Find the Laurent Series around z=0 for f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}}), so that the series absolutely converges in z = -i
The Attempt at a Solution
The singularities of f are z = 0, z = 1/2 and z=2. The Laurent series will converge absolutely in z = -i...
Laurent series (COMPLICATED, URGENT)
Homework Statement
Find the Laurent series for f(z) = \frac{{2{z^2}}}{{2{z^2} - 5z + 2}} + \sin (\frac{3}{{{z^2}}}) around z = 0, in which the series absolutely converges for z = -i.
The Attempt at a Solution
I have several problem with this. I...
Homework Statement
Find the expression of the Laurent series for f(z) = \frac{z}{{(2z - 1)(\frac{2}{z} - 1)}} so that \sum\limits_{ - \infty }^\infty {{a_n}} \ converges.
The Attempt at a Solution
First, I find that z = 1/2 and z = 2 (and infinite) are poles of the function f. Then, I...
Homework Statement
describe the laurent series for the function
f(z) = z^3 cos(\frac {1}{z^2})
b) use your answer to part a to compute the contour integral
\int z^3 cos(\frac {1}{z^2}) dz
where C is the unit counter-clockwise circle around the origin.Homework Equations
The Attempt at a...
Homework Statement
http://img243.imageshack.us/img243/4339/69855059.jpg
Homework Equations
i've heard that the solution requires the use of the exponential taylor series:
http://img31.imageshack.us/img31/6163/37267605.jpg
The Attempt at a Solution
i know that the first step is...
Hi,
can u pls help me on this?
Homework Statement
Find Laurent series that converges for
\, 0 < |z - z_0| < R }
and determine precise region of convergance
\, \frac {1}{z^2 + 1} \,\,
Homework Equations
The Attempt at a Solution
I tried to spilt this into...
Would anyone be willing to check and comment on my work for finding the Laurent series of
f(z)=\frac{1+2z}{z^2+z^3} ?
Page 1 - http://img23.imageshack.us/img23/7172/i0001.jpg"
Page 2 - http://img5.imageshack.us/img5/2140/i0002.jpg"
Page 3 -...
I uploaded a scanned page from Schaum's Outline of Complex Variables. I have some questions on how they found the Laurent series in Example 27.
http://img19.imageshack.us/img19/7172/i0001.jpg
If the image doesn't load, go to - http://img19.imageshack.us/img19/7172/i0001.jpg"
In...
Homework Statement
find the Laurent series for f(z) = 1/(z(z-1)(z-2)) on the annulus between 1 and 2. with the origin as center.
Homework Equations
The Attempt at a Solution
so i found the partial fraction decomposition of this function and it turns out to be f(z) = 1/2z +...
Homework Statement
Hey guys.
I need to develop this function into Laurent series.
I used the Sin Taylor series and got what I got.
Now, is there a trick or something to get the z-2 inside series or is this enough?
Thanks.
Homework Equations
The Attempt at a Solution
Homework Statement
Find the Principle part of the Laurent Expansion of f(z) about z=0 in the region
0 < mod z < 1, where f(z) = exp(z) / [(z^2)*(z+1)]
Homework Equations
1/(1-z) = Summation (n = 0 to n = infinity) { z^n}
The Attempt at a Solution
First, by using partial...
Hi,
My mathematics professor said that it is possible to construct a Laurent series of sqrt(z) about zero by integrating over a keyhole contour and then taking the limit R --> 0 where R radius of the inner circle. But I think he is mistaken. I don't understand how it is possible to have a...
Homework Statement
Find the Laurent series of the function f(z) = Sin(1/(z^2-z)) in the region 0<|z|<infinity.
The Attempt at a Solution
Now sin(z) = [e^(iz) - e^(-iz)]/(2i)
Shall we replace z by 1/(z^2-z) to obtain the Laurent series for f(z)?
I tried this but it gets messy. Is...
I'm puzzled by one thing concerning Laurent series.
If I have a series, for example f(z) =(z*sinz)/(2z-1) and I'm supposed to make a laurent series of f about the point z=1/2.
Now, what would the inner and outer radius of convergence be?
I would say that since z=1/2 is a pole, the...
Hi. I am having trouble getting started on this problem.
I need to find the Laurent series for: f(z) = exp[(a/2)*(z - 1/z)] in |z|>0.
I know that the coefficients are: (1/2pi)*integral[cos(kx) - a*sin(x)]dx |(0 to 2pi)
But I am having trouble seeing how to get started on showing that...
Homework Statement
We have
f(z) = \frac{1}{z^2 - 2z - 3}
For this function, we want to find the Laurent Series around z=0, that converges when z=2 and we want to find the area of convergence.
Homework Equations
\frac{1}{z-3} = -\frac{1}{3}\bigg( \frac{1}{1 - \frac{z}{3}}\bigg) =...
Homework Statement
I know the sum of the Laurent series (around x=2) is equal to
\frac{1}{x+3}
But I can't find what the series is from this information alone.
Homework Equations
In the textbook, you have (for -1 < x < 1):
\frac{1}{1-x} = \sum_{n=0}^{\infty}x^n
and for |x|>1 I know...
Laurent series for f(z) = 1/(exp(z)-1)^2 ??
Homework Statement
Determine the Laurent series and residue for f(z) = \frac{1}{(e^{z} - 1)^{2}}.
Homework Equations
We know that the Taylor series expansion of e^{z} is = 1 + z + (z^2)/2! + ...
The Attempt at a Solution
I am soooo...
I am very confused about how to actually compute a Laurent series. Given an analytic function, we can write down its poles. Then, if I understand correctly, we have to write a Laurent series for each pole. What I'm confused about is the actual mechanics of writing one down. I know that for f(z)...
When doing complex contour integration one can use the C-R formula or the Laurent series and find the first coefficient of the principle part. What are the selection criteria for choosing these methods?
Regards,
Hob
Hello all,
I've got an exam tomorrow so any quick responses would be appreciated. I'm following the Boas section on Laurent series... Anyway, here's my problem:
In an example Boas starts with f(z) = 12/(z(2-z)(1+z), and then using partial fractions arrives at f(z) = (4/z)(1/(1+z) +...
[SOLVED] Laurent Series of z/(sin z)^2
Homework Statement
Find the first four terms of the Laurent series of f(z) = z/(\sin z)^2 about 0.
The attempt at a solution
I know that when z = 0, f(z) is undefined so it has a singularity there. This singularity is a pole because
\lim_{z \to 0}...
I'm so lost. I can follow the steps in the examples in the book, but all of the examples seem easier than this problem-- and in each case there is a point where you "spot" a similarity to the geometric series or something and then you can just patch it in... But, I just don't know how to make...