Chap
.
16
Laurent Series
Residue Integration
This is the last portion of our discussions of series. Laurent series have no significant counterpart in calculus.
Their importance in complex analysis results primarily from an integration method based on these series.
Sec
.
16
.
1
Laurent Series
A given function
f
u
z
U
may have several Laurent series with the same center
z
0
. Of these series, most important
is the one that converges in a neighborhood of
z
0
(except at
z
0
itself). Its negative powers form what is called
the
principal part
of the singularity of
f
u
z
U
(or of that Laurent series) at
z
0
; see Theorem 1.
Problem Set 16
.
1
.
Page 707
1
.
Laurent series
. You have, by using the sum formula of the geometric series, the Laurent series,
f
u
z
U
u
1
z
4
u
z
5
u
1
z
4
u
1
u
z
U
u
1
z
4
1
u
1
u
z
U
u
1
z
4
U
n
u
0
±
z
n
u
U
n
u
0
±
z
n
u
4
u
1
z
4
U
1
z
3
U
1
z
2
U
1
z
U
1
U
z
U
±
.
The principal part of the singularity of
f
u
z
U
at
z
u
0 is
z
u
4
U
z
u
3
U
z
u
2
U
z
u
1
.
7
.
Singularity at
z
²
0
. In Prob. 1 the singularity was at
z
u
0. Not much changes when it is at a
z
²
0,
here,
z
u
1. Write
e
z
u
ee
z
u
1
and develop
e
z
u
1
by using the familiar Maclaurin series (12) in Sec. 15.4
with
z
replaced by
z
u
1,
e
z
u
1
u
U
n
u
0
±
u
z
u
1
U
n
n
!
.
Multiply this by
e
/
u
z
u
1
U
to obtain the answer on p. A40.
11
.
Laurent series
. Use the sum formula for the geometric series to obtain
1
u
z
U
i
U
2
u
u
z
U
i
U
u
u
1
u
z
U
i
Uu
1
u
u
z
U
i
UU
u
u
1
z
U
i
U
n
u
0
±
u
z
U
i
U
n
u
u
1
u
z
U
i
U
u
1
u
u
z
U
i
U
u
u
z
U
i
U
2
u
±
.
15
.
Taylor and Laurent Series
. By the sum formula for the geometric series with
z
3
instead of
z
you
obtain
1
1
u
z
3
u
U
n
u
0
±
z
3
n
u

z

±
1
U
.
Similarly, the Laurent series converging for 
z

²
1 is obtained by the following trick, which you should
remember
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Laurent Series. Residue Integration
149
1
1
u
z
3
u
1
u
z
3
1
u
1
z
3
u
1
u
z
3
U
n
u
0
±
1
z
3
n
u
u
1
z
3
u
1
z
6
u
1
z
9
u
u
.
23
.
Laurent series
. Develop
sin
z
u
sin
UU
z
U
u
2
±
u
u
2
±
u
sin
U
z
U
u
2
±
cos
u
2
u
cos
U
z
U
u
2
±
sin
u
2
.
Using cos
u
/2
u
0 and sin
u
/2
u
1, you obtain by the familiar Maclaurin series for the cosine
u
U
n
u
0
±
U
u
1
±
n
z
U
u
2
2
n
U
2
n
±
!
.
Multiply this by
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 Spring '12
 M.lee

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