# What is Laurent series: Definition and 162 Discussions

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

f
(
z
)

{\displaystyle f(z)}
.

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1. ### A Laurent series for algebraic functions

Hi, I'm writting because I sort of had an idea that looks that it should work but, I did not find any paper talking about it. I was thinking about approximating something like algebraic functions. That is to say, a function of a complex variable z,(probably multivalued) that obeys something...
2. ### Multiplication of Taylor and Laurent series

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43. ### Harmonic function on annulus and finding Laurent series

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44. ### Partial Fractions in Laurent Series Expansion

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45. ### Need help finding a Laurent Series

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46. ### Laurent Series & Partial Fraction Decomposition.

Okay so the partial fraction decomposition theorem is that if f(z) is a rational function, f(z)=sum of the principal parts of a laurent expansion of f(z) about each root. I'm working through an example in my book, I am fine to follow it. (method 1 below) But instinctively , I would have...
47. ### Expand function in Laurent series

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48. ### Singularity points + Laurent series

Homework Statement Find and determine the type of singularity points for ##f(z)=\frac{\sin(3z)-3z}{z^5}##. Also calculate the regular and main part of Laurent series around those points.Homework Equations The Attempt at a Solution I am already having troubles with the first part. Singularity...
49. ### Question about understanding something in laurent series

So the question I got the represention for both partial fractions after I broke the functions into two partial fraction one I got as 1/3(z + 1) + 2/3(z - 2) and I got laurent series represention for both but I was wondering for |z| < 1 how can they both converge for |z| < 1 are we acctually...
50. ### Finding the Laurent Series of e^(1/(1-z)) for Residue Calculation

Homework Statement Hi! I need to find the laurent series of ##e^{1/(1-z)}## to get the residue at ##z=1##. Can somebody help me? The Attempt at a Solution https://scontent-a-ams.xx.fbcdn.net/hphotos-frc3/q71/s720x720/1461607_10201796752217165_1002449331_n.jpg I tried using the taylor series...