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Complex Analysis 3.pdf
Complex_Analysis_3.pdf
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Complex Analysis 3.pdf-Theme 8. Taylor and Laurent...
Complex_Analysis_3.pdf-Theme 8. Taylor and Laurent Series.
Complex Analysis 3.pdf-Theme 8. Tay...
Complex_Analysis_3.pdf-Theme 8. Taylor and Laurent Series.
Page 9
Example:
Find the Laurent series for
f
(
z
)=
1
(
z
-
5)
2
(
z
-
4)
about 5 that converges when
z
= 7.
Lemma:
(Properties of power series – we have used many of these already)
1. (Uniqueness) If a series
∞
X
n
=
-∞
c
n
(
z
-
z
0
)
n
converges to
f
(
z
) at all points
z
in some annular domain
D
about
z
0
, then it is the
Laurent series expansion for
f
about
z
0
.
2. If two power series converge on a domain
D
then we can add, subtract, multiply and
divide them as if we had polynomials, and the resulting series will converge as expected.
3. If a power series converges on a domain
D
then we can differentiate or integrate term
by term, and the resulting series will converge as expected.
9
Page 10
Theme 9.
Integrals and Laurent Series.
Now that we have constructed these rather strange series for complex functions, we ask the
question: What use are they? We have already seen that we can evaluate certain complex
integrals around closed contours, using CIF. But there are still many integrals we cannot do.
For example:
I
|
z
|
=
π
2
1
sin
z
dz.
The following remarkable theorem, tells us how to evaluate integrals using Laurent series. It
is based on the following observation:
Let
γ
be a simple closed contour around
z
0
, oriented anti-clockwise. Then
Z
γ
(
z
-
z
0
)
n
dz
=
0
if
n
≥
0
2
πi
if
n
=
-
1
0
if
n
≤-
2
by the Cauchy–Goursat Theorem (
n
≥
0) and Cauchy’s Integral formulae (
n
≤-
1).
Theorem:
Suppose that
f
has one singularity at the point
z
0
, and that
f
has a Laurent
series about the point
z
=
z
0
of the form
f
(
z
)=
...
+
a
-
2
(
z
-
z
0
)
2
+
a
-
1
z
-
z
0
+
a
0
+
a
1
(
z
-
z
0
)+
...
If
γ
is any simple closed contour containing
z
0
, then
I
γ
f
(
z
)
dz
=2
πia
-
1
.
Thus, the value of the integral is controlled by the number
a
-
1
in the Laurent series for
f
,
when it is expanded about the singular point
z
0
.
Proof:
10
Page 11
The number
a
-
1
in the above theorem is called the
residue
if
f
at
z
0
. We use the notation:
a
-
1
= Res
z
=
z
0
f
(
z
)
.
Example:
Use Laurent series to find
I
|
z
|
=3
e
z
z
-
1
dz.
Example:
Find the first three non-zero terms of the Laurent series for
f
(
z
)=
1
sin
z
and
hence find
I
|
z
|
=
π
2
1
sin
z
dz.
11
Page 12
Example:
Find
I
|
z
|
=2
sin
z
z
2
-
1
dz.
We now generalise the theorem relating integrals and residues to deal with functions which
have more than one singularity.
Theorem:
(Residue Theorem) Let
γ
be a simple closed contour oriented anti-clockwise.
If
f
is analytic on and inside
γ
except for a finite number of isolated singularities
z
1
,...,z
n
inside
γ
, then
Z
γ
f
(
z
)
dz
=2
πi
n
X
k
=1
Res
z
=
z
k
f.
Proof:
By the Cauchy-Goursat Theorem,
Z
γ
f
(
z
)
dz
=
Z
γ
1
f
(
z
)
dz
+
Z
γ
2
f
(
z
)
dz
+
···
+
Z
γ
n
f
(
z
)
dz,
where the
γ
k
are non-intersecting circles around
z
k
, inside
γ
, oriented anti-clockwise.
If
f
(
z
)=
∑
∞
n
=
-∞
c
k
n
(
z
-
z
k
)
n
is the Laurent series for
f
about
z
k
, then
Z
γ
k
f
(
z
)
dz
=
∞
X
n
=
-∞
Z
γ
k
c
k
n
(
z
-
z
k
)
n
=2
πic
k
-
1
=2
πi
Res
z
=
z
k
f.
12
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