Complex Analysis 3.pdf-Theme 8. Taylor a...
Complex_Analysis_3.pdf-Theme 8. Taylor and Laurent Series.
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Complex Analysis 3.pdf-Theme 8. Tay...
Complex_Analysis_3.pdf-Theme 8. Taylor and Laurent Series.
##### Page 4
Example:
Let
g
(
z
)=
z
2
e
5
z
. Find the Taylor series for
g
z
0
= 0.
Example:
Find the Taylor series for sin
z
z
0
=
π
Example:
Let
h
(
z
)=
10
(
z
-
i
)(
z
+ 3)
. Find the Taylor series for
h
z
0
= 2, that is,
in powers of (
z
-
2).
What is its radius of convergence?
We have the partial fractions
decomposition
h
(
z
)=
10
(
z
-
i
)(
z
+ 3)
=
3
-
i
z
-
i
+
i
-
3
z
+3
,
4

##### Page 5
Example:
What is the radius of convergence for the Taylor series for sec
z
the ﬁrst three non-zero terms of this series.
5

##### Page 6
We can now deduce a version of L’Hˆ
opital’s Rule.
Theorem:
(L’Hˆ
opital’s Rule) If
f
and
g
are analytic at
z
0
and
f
(
z
0
)=0=
g
(
z
0
) = 0 but
g
0
(
z
0
)
6
= 0, then
lim
z
z
0
f
(
z
)
g
(
z
)
= lim
z
z
0
f
0
(
z
)
g
0
(
z
)
.
Since
f
and
g
are analytic at
z
0
, Taylor’s Theorem gives
f
(
z
)=
X
n
=0
a
n
(
z
-
z
0
)
n
and
g
(
z
)=
X
n
=0
b
n
(
z
-
z
0
)
n
for
z
near
z
0
. Since
a
0
=
f
(
z
0
) = 0 and
b
0
=
g
(
z
0
) = 0 we have
f
(
z
)=
a
1
(
z
-
z
0
)+
a
2
(
z
-
z
0
)
2
+
a
3
(
z
-
z
0
)
3
+
···
g
(
z
)=
b
1
(
z
-
z
0
)+
b
2
(
z
-
z
0
)
2
+
b
3
(
z
-
z
0
)
3
+
···
for
z
near
z
0
.
Note that
g
0
(
z
0
)=
b
1
6
= 0. Now
lim
z
z
0
f
(
z
)
g
(
z
)
= lim
z
z
0
a
1
(
z
-
z
0
)+
a
2
(
z
-
z
0
)
2
+
a
3
(
z
-
z
0
)
3
+
···
b
1
(
z
-
z
0
)+
b
2
(
z
-
z
0
)
2
+
b
3
(
z
-
z
0
)
3
+
···
= lim
z
z
0
a
1
+
a
2
(
z
-
z
0
)+
a
3
(
z
-
z
0
)
2
+
···
b
1
+
b
2
(
z
-
z
0
)+
b
3
(
z
-
z
0
)
2
+
···
=
a
1
b
1
= lim
z
z
0
f
0
(
z
)
g
0
(
z
)
.
6

##### Page 7
Laurent Series
If a function
f
fails to be analytic at a point
z
0
then we cannot ﬁnd a Taylor series converging
to
f
in any neigbourhood of
z
0
, but it is often possible to ﬁnd a series representation for
f
involving both positive and negative powers of
z
-
z
0
.
Example:
Consider the function
f
(
z
)=
e
z
z
2
.
Theorem:
(Laurent’s Theorem) Suppose that a function
f
is analytic throughout the an-
nulus
D
=
{
z
:
R
1
<
|
z
-
z
0
|
<R
2
}
. Let
γ
be any simple closed contour around
z
0
, oriented
anti-clockwise, lying entirely in
D
. For each
z
D
we have
f
(
z
)=
···
+
b
2
(
z
-
z
0
)
2
+
b
1
z
-
z
0
+
a
0
+
a
1
(
z
-
z
0
)+
a
2
(
z
-
z
0
)
2
+
···
(
)
where
a
n
=
1
2
πi
Z
γ
f
(
z
)
(
z
-
z
0
)
n
+1
dz
for
n
=0
,
1
,
2
···
b
n
=
1
2
πi
Z
γ
f
(
z
)
(
z
-
z
0
)
-
n
+1
dz
for
n
=0
,
1
,
2
···
We call (
) the
Laurent series for
f
z
0
.
In practice, we hardly ever use the for-
mulas for the coeﬃcients
a
n
and
b
n
– we manipulate known series instead (just as we do to
ﬁnd Taylor series).
If
f
is analytic on the disc
|
z
-
z
0
|
<R
2
then
b
n
=
1
2
πi
Z
γ
f
(
z
)(
z
-
z
0
)
n
-
1
dz
=0
for
n
=0
,
1
,
2
···
,
7

##### Page 8
by the Cauchy-Goursat Theorem and (
) reduces to a Taylor series about
z
0
. So a Taylor
series is a Laurent series (but not vice versa).
Example:
Find the Laurent series for
f
(
z
)=
e
1
z
Example:
Let
f
be the function
f
(
z
)=
1
(
z
-
1)(
z
-
4)
=
-
1
/
3
z
-
1
+
1
/
3
z
-
4
.
a) Find the largest open annuli or discs centred at
i
in which
f
is analytic;
b) ﬁnd the Laurent series for
f
i
which is valid when
z
= 2;
8

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