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? let g ( z ) = ? a
z
…… …… ( ii )
? R. O. C of eq
n
(ii) = 1/ lim S up | a
|
/ = 1/R 1
? g ( z ) conv erg es fo r | z | /R 1
Let f 2(z) = g(1/z)
f 2 ( z ) ? a -kz
k
? f 2(z) cgs if g(1/z) cgs if
| |
i.e. if |z| > R 1
? f 2(z) cgs for |z| > R 1
f(z) = ? a
z
= f 1(z) + f 2(z)
? ? a
z
cgs for |z| > R 1 , |z|<R 2
Hence, the Laurent expansion ? a
z
cgs in the domain
D = {z : |z| > R 1 & |z| < R 2}
9.9 theorem : If f is analytic in the annulus A = {z |R 1 < |z| < R 2 +, a nd then f or an y z , f(z) has a
representation of the form
f(z) = ? a
z
is called Laurent expansion about origin.
Pro of , can see in the b o ok or in V ivek ’s note s
Corollary 9.10 If f is analytic in the annulus R 1 | z z 0| < R 2, and then f has a unique representation, f(z)
= ? a
( z z
)
where a k =
( )
( )
dz
an d C C ( z 0; R) with R 1 < R < R 2.
Proof in book & vivek notes
Example/pg 111 (i) find Laurent series expansion ? z .
f(z) =
( )
Sol
n
f(z) =
= z+2+
? z
which is Laurent expansion about origin ? z
(ii) f(z) =
( )
Sol
n
f(z) =
( )
is not analytic at 0 & 1 (where f is not defined is singularity)
? f ( z ) =
( z )
=
( z z
)
=
z z
Which is Laurent expansion about origin.
( iii ) f ( z ) z
( z )
w r it e L aur en t series exp an sion ab out z
S ol ution f ( z ) z
( z )
( z ) z
( z ) ( z )
( z ) , z -
( z )
, ( z ) -
( z )
* ( z ) 3 ( z )
( z )
+
z 3 ( z ) ( z )
f or | z |
( iv ) f in d the L aur en t series e xpan sion of exp (
* ab out z ( or igin ) .
Sol
n
f(z) = e
/
z
(
z
*
3 (
z
*
w hich is L aur en t exp .
Que. 7/113 find the Laurent exp. for
( )
z
z
… a
- z
- a
- z
- a
- z
- ?
a
z
a
z a
z
?
Comes in (ii) comes in (i)
Page 3
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For more 8130648819
? let g ( z ) = ? a
z
…… …… ( ii )
? R. O. C of eq
n
(ii) = 1/ lim S up | a
|
/ = 1/R 1
? g ( z ) conv erg es fo r | z | /R 1
Let f 2(z) = g(1/z)
f 2 ( z ) ? a -kz
k
? f 2(z) cgs if g(1/z) cgs if
| |
i.e. if |z| > R 1
? f 2(z) cgs for |z| > R 1
f(z) = ? a
z
= f 1(z) + f 2(z)
? ? a
z
cgs for |z| > R 1 , |z|<R 2
Hence, the Laurent expansion ? a
z
cgs in the domain
D = {z : |z| > R 1 & |z| < R 2}
9.9 theorem : If f is analytic in the annulus A = {z |R 1 < |z| < R 2 +, a nd then f or an y z , f(z) has a
representation of the form
f(z) = ? a
z
is called Laurent expansion about origin.
Pro of , can see in the b o ok or in V ivek ’s note s
Corollary 9.10 If f is analytic in the annulus R 1 | z z 0| < R 2, and then f has a unique representation, f(z)
= ? a
( z z
)
where a k =
( )
( )
dz
an d C C ( z 0; R) with R 1 < R < R 2.
Proof in book & vivek notes
Example/pg 111 (i) find Laurent series expansion ? z .
f(z) =
( )
Sol
n
f(z) =
= z+2+
? z
which is Laurent expansion about origin ? z
(ii) f(z) =
( )
Sol
n
f(z) =
( )
is not analytic at 0 & 1 (where f is not defined is singularity)
? f ( z ) =
( z )
=
( z z
)
=
z z
Which is Laurent expansion about origin.
( iii ) f ( z ) z
( z )
w r it e L aur en t series exp an sion ab out z
S ol ution f ( z ) z
( z )
( z ) z
( z ) ( z )
( z ) , z -
( z )
, ( z ) -
( z )
* ( z ) 3 ( z )
( z )
+
z 3 ( z ) ( z )
f or | z |
( iv ) f in d the L aur en t series e xpan sion of exp (
* ab out z ( or igin ) .
Sol
n
f(z) = e
/
z
(
z
*
3 (
z
*
w hich is L aur en t exp .
Que. 7/113 find the Laurent exp. for
( )
z
z
… a
- z
- a
- z
- a
- z
- ?
a
z
a
z a
z
?
Comes in (ii) comes in (i)
Free coaching of B.Sc (h) maths & JAM
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( )
exp .
z
/
z ab out z
( )
z
ab out z ?
( )
( z )
?? ( ) ( ) z
z
z
( z
)
f(z) is not analytic at z = 0, z = ± i
? f ( z ) is a na ly ti c f or * z z +
? ( ) z
( z
)
z
( z
)
z
( z
z
z
)
z
z
z
z
) w hich is L aur en t exp an sion ab out ‘ ’
( b ) f ( z ) e
/ z f ( z ) is not an aly ti c at z ,
? f(z) i s a nalytic for {z : 0 < |z| <1}
f ( z ) e
( z )
6 z
(
z
*
7 , z z
z
-
[ ( z z
) (
z
z
z z
* (
z
z
z
z
* … … … ]
2 . / . / z . / z
. /
. /
3
( 3 * ( z z
) ( 3 * ( z
z
z
)
[ e ? z
( e ) ? z
] {
e 3 e 3 … … .
}
{ ? e z
? ( e ) z
}
( c ) f ( z ) z
( z ) ( z )
, f ( z ) is not an a ly ti c f or z ,
? f ( z ) is a na ly ti c f or * z | z | +
f ( z ) ( z ) ( z )
( z ) . z /
( z )
4 (
z * 5
( z )
( z )
w hich is r eq uir ed L aur en t exp an si on .
Self to find Laurent expansion, express the given fun
c
about the point.
Qu e /p g S ho w that if f is a na ly ti c in z an d “ od d ” ( i. e. f ( z ) f ( z ) ) then all the e v en t erm s in it s
Laurent expansion about z = 0 are 0
Sol
n
let f(z) be analytic ? z .
L et f ( z ) f ( z )
? C
( z )
? C
z
? C
( )
z
? C
z
( )
C
C
k
Page 4
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Free coaching of B.Sc (h) maths & JAM
For more 8130648819
? let g ( z ) = ? a
z
…… …… ( ii )
? R. O. C of eq
n
(ii) = 1/ lim S up | a
|
/ = 1/R 1
? g ( z ) conv erg es fo r | z | /R 1
Let f 2(z) = g(1/z)
f 2 ( z ) ? a -kz
k
? f 2(z) cgs if g(1/z) cgs if
| |
i.e. if |z| > R 1
? f 2(z) cgs for |z| > R 1
f(z) = ? a
z
= f 1(z) + f 2(z)
? ? a
z
cgs for |z| > R 1 , |z|<R 2
Hence, the Laurent expansion ? a
z
cgs in the domain
D = {z : |z| > R 1 & |z| < R 2}
9.9 theorem : If f is analytic in the annulus A = {z |R 1 < |z| < R 2 +, a nd then f or an y z , f(z) has a
representation of the form
f(z) = ? a
z
is called Laurent expansion about origin.
Pro of , can see in the b o ok or in V ivek ’s note s
Corollary 9.10 If f is analytic in the annulus R 1 | z z 0| < R 2, and then f has a unique representation, f(z)
= ? a
( z z
)
where a k =
( )
( )
dz
an d C C ( z 0; R) with R 1 < R < R 2.
Proof in book & vivek notes
Example/pg 111 (i) find Laurent series expansion ? z .
f(z) =
( )
Sol
n
f(z) =
= z+2+
? z
which is Laurent expansion about origin ? z
(ii) f(z) =
( )
Sol
n
f(z) =
( )
is not analytic at 0 & 1 (where f is not defined is singularity)
? f ( z ) =
( z )
=
( z z
)
=
z z
Which is Laurent expansion about origin.
( iii ) f ( z ) z
( z )
w r it e L aur en t series exp an sion ab out z
S ol ution f ( z ) z
( z )
( z ) z
( z ) ( z )
( z ) , z -
( z )
, ( z ) -
( z )
* ( z ) 3 ( z )
( z )
+
z 3 ( z ) ( z )
f or | z |
( iv ) f in d the L aur en t series e xpan sion of exp (
* ab out z ( or igin ) .
Sol
n
f(z) = e
/
z
(
z
*
3 (
z
*
w hich is L aur en t exp .
Que. 7/113 find the Laurent exp. for
( )
z
z
… a
- z
- a
- z
- a
- z
- ?
a
z
a
z a
z
?
Comes in (ii) comes in (i)
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
( )
exp .
z
/
z ab out z
( )
z
ab out z ?
( )
( z )
?? ( ) ( ) z
z
z
( z
)
f(z) is not analytic at z = 0, z = ± i
? f ( z ) is a na ly ti c f or * z z +
? ( ) z
( z
)
z
( z
)
z
( z
z
z
)
z
z
z
z
) w hich is L aur en t exp an sion ab out ‘ ’
( b ) f ( z ) e
/ z f ( z ) is not an aly ti c at z ,
? f(z) i s a nalytic for {z : 0 < |z| <1}
f ( z ) e
( z )
6 z
(
z
*
7 , z z
z
-
[ ( z z
) (
z
z
z z
* (
z
z
z
z
* … … … ]
2 . / . / z . / z
. /
. /
3
( 3 * ( z z
) ( 3 * ( z
z
z
)
[ e ? z
( e ) ? z
] {
e 3 e 3 … … .
}
{ ? e z
? ( e ) z
}
( c ) f ( z ) z
( z ) ( z )
, f ( z ) is not an a ly ti c f or z ,
? f ( z ) is a na ly ti c f or * z | z | +
f ( z ) ( z ) ( z )
( z ) . z /
( z )
4 (
z * 5
( z )
( z )
w hich is r eq uir ed L aur en t exp an si on .
Self to find Laurent expansion, express the given fun
c
about the point.
Qu e /p g S ho w that if f is a na ly ti c in z an d “ od d ” ( i. e. f ( z ) f ( z ) ) then all the e v en t erm s in it s
Laurent expansion about z = 0 are 0
Sol
n
let f(z) be analytic ? z .
L et f ( z ) f ( z )
? C
( z )
? C
z
? C
( )
z
? C
z
( )
C
C
k
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Let k = 2n (even)
( )
2n
C 2n C 2n
C 2n C 2n = 0 all even terms equal to zero.
9.13 PARTIAL FRACTION DECOMPOSITION OF RATIONAL FRACTION :
Any proper rational fun
c
R ( z ) P ( z )
Q ( z )
P ( z )
( z z
)
( z z
)
… … … ( z z
)
Wh ere P Q are p oly nomia l w it h d eg P d eg can be exp an d ed as a sum of p oly nomia l in
( z z
)
,
k , , … , n
F in d the L aur en t series exp an sion f or f ( z ) ( z ) ( z )
f or | z |
S oltio n f ( z ) ( z )
( z )
z
.
. z
/
0 z
1
z
( z
*
. z
/
( |
z
| |
z
| *
z
( z
z
* 4 z
z
5
z
z
z
z
z
…
Read More
|
27 docs|150 tests
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FAQs on Laurent Expansion - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is a Laurent expansion in mathematics? |  |
Ans. A Laurent expansion is a way to represent a complex function as a power series with both positive and negative powers of the variable. It is often used in complex analysis to study functions with singularities or poles.
| 2. How is a Laurent expansion different from a Taylor expansion? | |
Ans. While a Taylor expansion only includes non-negative powers of the variable, a Laurent expansion includes both positive and negative powers. This allows the Laurent expansion to capture the behavior of a function near singularities or poles, which cannot be represented accurately by a Taylor expansion.
| 3. What are the applications of Laurent expansions in mathematics? | |
Ans. Laurent expansions have various applications in complex analysis and related fields. They are used to analyze functions with singularities, calculate residues, evaluate complex integrals, solve differential equations, and study the behavior of functions near branch points.
| 4. How can I find the Laurent expansion of a given function? | |
Ans. To find the Laurent expansion of a function, one can use the method of partial fraction decomposition and the geometric series expansion. By expressing the function as a sum of simpler fractions and expanding each fraction as a power series, the Laurent expansion can be obtained.
| 5. Can a Laurent expansion be used to approximate any complex function? | |
Ans. No, a Laurent expansion can only be used to approximate analytic functions, which are functions that can be represented by a convergent power series. If a function has essential singularities or branch points, a Laurent expansion cannot accurately represent its behavior in those regions.
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