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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 13
Theme 10. Zeros, singularities and residues.
We now know that the value of the integral of a function with one singularity around a
simple closed contour containing that singularity is controlled by the residue of the function
when it is expanded about its singular point. In this section we see how to generalise this
to functions with more than one singularity. We would also like to develop a technique for
extracting the residue
without
computing the series.
To do this, we need to introduce the notions of zeros and poles.
Example:
Consider the function
f
(
z
) = sin
z
at
z
= 0.
Definition:
Suppose
f
has a Taylor series about
z
0
such that
f
(
z
)=
a
m
(
z
-
z
0
)
m
+
a
m
+1
(
z
-
z
0
)
m
+1
+
a
m
+2
(
z
-
z
0
)
m
+2
+
···
=(
z
-
z
0
)
m
(
a
m
+
a
m
+1
(
z
-
z
0
)+
a
m
+2
(
z
-
z
0
)
2
+
···
)
,
where
m
≥
1 and
a
m
6
= 0. Then
z
0
is a
zero of order
m
of
f
. A zero of order 1 is called a
simple zero
.
Example:
Find the orders of the zeros of
f
(
z
) = sin
z
.
13
Page 14
Lemma:
(Test for a zero of order
m
at
z
0
) If
f
is analytic at
z
0
,
f
(
z
0
) = 0 and lim
z
→
z
0
f
(
z
)
(
z
-
z
0
)
m
exists and is
6
= 0, then
f
has a zero of order
m
at
z
0
.
Example:
Show that
f
(
z
) = sin
z
has a simple zero at
z
=
π
using the lemma.
Example:
Show that
f
(
z
) = cos
z
+ 1 has a zero of order 2 at
z
=
π
using the lemma.
Singularities:
Definition:
A point
z
0
is a
singular point
of a function
f
if
f
is not analytic at
z
0
but
every neighbourhood of
z
0
contains a point where
f
is analytic.
There are two types of singularities:
(1)
branch points and branch cuts
(e.g.
-
2 is a singular point of
f
(
z
) = Log
z
);
(2)
isolated singularities
Here
f
is analytic at
all
points in some neighbourhood of
z
0
.
(e.g. 1 and 0 are isolated singularities of
f
(
z
)=
1
z
(
z
-
1)
).
There are three types of these, depending on the Laurent series
···
+
a
-
m
(
z
-
z
0
)
m
+
···
+
a
-
2
(
z
-
z
0
)
2
+
a
-
1
z
-
z
0
+
a
0
+
a
1
(
z
-
z
0
)+
a
2
(
z
-
z
0
)
2
+
···
(
*
)
for
f
about
z
0
valid on 0
<
|
z
-
z
0
|
<R
.
14
Page 15
Definition:
Suppose that
z
0
is a singular point of
f
.
The
principal part
of (
*
) is the
series
···
+
a
-
m
(
z
-
z
0
)
m
+
···
+
a
-
2
(
z
-
z
0
)
2
+
a
-
1
z
-
z
0
of terms with negative powers.
Removable Singularities:
If the principal part is zero (
···
=
a
-
m
=
···
=
a
-
1
= 0), then
f
has a
removable sin-
gularity at
z
0
.
Example:
Let
f
(
z
)=
sin
z
z
. The Laurent series is
sin
z
z
=
1
z
z
-
z
3
3!
+
z
5
5!
-···
=1
-
z
2
3!
+
z
4
5!
-···
if
z
6
=0
.
The principal part is 0, so
f
has a removable singularity at 0.
The point of the term “removable” is that we can define a function
g
which is analytic
at
z
0
and agrees with
f
everywhere else by
g
(
z
)=
(
sin
z
z
if
z
6
=0
1
if
z
= 0.
Poles:
(a) If the principal part is
a
-
1
z
-
z
0
, with
a
-
1
not zero, then
f
has a
simple pole
at
z
=
z
0
and
the residue is simply
a
-
1
.
We can test for simple poles by noting that
(
z
-
z
0
)
f
(
z
)=
a
-
1
+
a
0
(
z
-
z
0
)+
a
1
(
z
-
z
0
)
2
+
....
and so taking the limit as
z
goes to
z
0
we have
lim
z
→
z
0
(
z
-
z
0
)
f
(
z
)=
a
-
1
.
Hence
z
0
is a simple pole of
f
is lim
z
→
z
0
(
z
-
z
0
)
f
(
z
) exists and is non-zero. Moreover, the value
of this limit, if it exists, gives the residue.
Example:
Let
f
(
z
)=
e
2
z
z
-
3
15
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