MTH 324 (Complex Analysis)
Lecture # 25( Laurents series )
Singularity
If a complex function
f
fails to be analytic at a point 0
z z
, then this point
is said to be singular point or singularity. The point 0
z z
is said to be an
isolated singularity of the function
f
if there exists a deleted neighborhood
R z z < <
0
0
through which
f
is analytic.
Remark
If 0
z z
is a singularity of a function
f
then
f
cannot be expanded as a
power series with 0
z
as center.
Laurents Theorem
Let
f
be analytic within the annular domain D defined by
R z z r < <
0 .
Then
f
has the series representation
k
k
k
z z a z f ) ( ) (
0
valid for
R z z r < <
0 . The coefficient k
a
are given by
,... 1 , 0 ,
) (
) (
2
1
1
0
t
+
k ds
z s
s f
i
a
C
k k
Proof.
Let 1
C
and 2
C
are concentric circles center at 0
z
and radii 1
r
and 2
R
,
where,
R R r r < < <
2 1 . By using Cauchys integral for after introducing a
crosscut between 1
C
and 2
C
, we have
1 2
) (
) (
2
1
) (
) (
2
1
) (
C C
ds
z s
s f
i
ds
z s
s f
i
z f
(25.1)
Working in similar way as we did in the proof of Taylors series, we obtain
0
0
) (
) (
) (
2
1
2
k
k
k
C
z z a ds
z s
s f
i
, (25.2)
where
... 2 , 1 , 0 ,
) (
) (
2
1
1
0
+
k ds
z s
s f
i
a
C
k k
We now solve the second integral in (25.1).
For this consider
( )
( )
) 3 . 25 ( ), (
) (
) (
...... 1
) (
) (
2
1
1
) (
) (
2
1
) ( ) (
) (
2
1
) (
) (
2
1
1
0
1
0
0
1
0
0
2
0
0
0
0
0
1
0
0
0
0 0
1
1
1 1
z R
z z
a
ds
s z z z
z s
z z
z s
z z
z s
z z
z s
z z
s f
i
ds
z z
z s
z z
s f
i
ds
z s z z
s f
i
ds
z s
s f
i
n
n
k
k
k
C
n
n
n
C
C C
+
'
,
_
+ +
,
_
,
_
'
where
n k ds
z s
s f
i
a
C
k k
... 2 , 1 ,
) (
) (
2
1
1
0
and
.
) )( (
) ( 2
1
) (
1
0
0
C
n
n n
ds
s z
z s s f
z z i
z R
Let
d z z
0 and M is the maximum of
) (z f
on 1
C
.
Using triangle inequality, we have,
( ) ( )
1 0 0
r d z s z z s z
.
Again by ML- inequality
n
C
n
n n
d
r
r d
Mr
ds
s z
z s s f
z z i
z R ) (
) )( (
) ( 2
1
) (
1
1
1 0
0
1
.
Clearly,
0 ) ( z R
n as
n
because
d r <
1
. So (25.3) becomes
1
0
) ( ) (
) (
2
1
1
k
k
k
C
z z
a
ds
z s
s f
i
. (25.4)
Using (25.2) and (25.4) in (125.1), we obtain the required Laurent series.
Remark
(1) Laurent series of
f
can be written as
+
1 0 0
0
) (
) ( ) (
k
k
k
k
k
k
z z
a
z z a z f
.
The first series in above expression in called analytic part of
f
and
the second series on right hand side is called principal part of
f
.
(2) If
0
k
a
for
,... 3 , 2 , 1 k
, then Laurent series reduces to Taylor series.
(3) If z is replaced by
z
1
and
f
is not changed then in Laurents series
expansion of
f
, k k
a a
for
,... 3 , 2 , 1 k
Example
Expand
) 1 (
1
) (
z z
z f
as a Laurent series expansion for the following annular
domains.
(a)
1 0 < <z
(b)
1 > z
(c)
1 1 0 < <z
(d)
1 1 > z
Solution
Clearly,
) 1 (
1
) (
z z
z f
have two isolated singularities at
0 z
and 1 z .
(a)
) 1 (
1
) (
z z
z f
can be written as
{ }
... 1
1
...... 1
1
) 1 (
1
) 1 (
1
) (
2
3 2
1
+ + + +
z z
z
z z z
z
z
z z z
z f
(b) As
)
1
1 (
1
) (
2
z
z
z f
, because
1
1
1 < >
z
z
.
. ..........
1 1
)
1
1 (
1
) (
3 2
1
2
+ +
z z
z z
z f
(c) In this case we have to find Laurent series expansion in the powers of
) 1 ( z
. Let 1 z w . Then
( )w w
z f
1
1
) (
+
for
1 0 < <w
. Now solve in
similar way as we did in part (a). Part (d) can be solved similarly.
Example
Expand z
e z f
3
) (
in a Laurent series valid for
< < z 0
.
Solution
As, .....
! 3 ! 2
1
3 2
+ + + +
z z
z e
z
So, for
< < z 0
, we have
.....
! 2
3 3
1
2
2 3
+ + +
z z
e
z
.
Example
Show that
)
1
( ) (
z
z Cos z f +
can be expanded as Laurent series given by
,
1
) (
1
0
,
_
+ +
k
k
k
k
z
z a a z f
where
d k Cos a
k
2
0
cos ) cos 2 (
2
1
, valid for
< < z 0
.
Solution
Given
)
1
( ) (
z
z Cos z f +
which remains same if we replace z by
z
1
. So in
Laurent series expansion of
f
, k k
a a
, for
,... 3 , 2 , 1 k
. Therefore,
.
1
) (
1
0
,
_
+ +
k
k
k
k
z
z a a z f
As
+
+
+
C
k
C
k k
ds
s
s
s Cos
i
k ds
s
s f
i
a
1
1
) (
)
1
(
2
1
,... 1 , 0 ,
) (
) (
2
1
Let
i
e s z C 1 :
Therefore,
d Sink Cos
i
d Cosk Cos
d ie
e
e e Cos
i
a
i
k i
i i
k
2
0
2
0
2
0
1
) cos 2 (
2
) cos 2 (
2
1
) (
) (
2
1
The second integral on right hand side can be shown zero using substitution
method, this completes the proof.
