Chapter 9
Isolated Singularities and the Residue
Theorem
1
/r
2
has a nasty singularity at
r
=0
, but it did not bother Newton—the moon is far enough.
Edward Witten
9.1
Classifcation oF Singularities
What is the diference between the Functions
sin
z
z
,
1
z
4
, and exp
°
1
z
±
? All oF them are not de±ned at
0, but the singularities are oF a very diferent nature. ²or complex Functions there are three types
oF singularities, which are classi±ed as Follows.
Defnition 9.1.
IF
f
is holomorphic in the punctured disk
{
z
∈
C
:0
<

z
−
z
0

<R
}
For some
R>
0 but not at
z
=
z
0
then
z
0
is an
isolated singularity
oF
f
. The singularity
z
0
is called
(a)
removable
iF there is a Function
g
holomorphic in
{
z
∈
C
:

z
−
z
0

<R
}
such that
f
=
g
in
{
z
∈
C
:0
<

z
−
z
0

<R
}
,
(b) a
pole
iF lim
z
→
z
0

f
(
z
)

=
∞
,
(c)
essential
iF
z
0
is neither removable nor a pole.
Example 9.2.
The Function
sin
z
z
has a removable singularity at 0, as For
z
°
=0
sin
z
z
=
1
z
²
k
≥
0
(
−
1)
k
(2
k
+ 1)!
z
2
k
+1
=
²
k
≥
0
(
−
1)
k
(2
k
+ 1)!
z
2
k
.
and the power series on the righthand side represents an entire Function (you may meditate on the
Fact why it has to be entire).
Example 9.3.
The Function
1
z
4
has a pole at 0, as
lim
z
→
0
³
³
³
³
1
z
4
³
³
³
³
=
∞
.
93
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94
Example 9.4.
The function exp
°
1
z
±
does not have a removable singularity (consider, for example,
lim
x
→
0
+
exp
°
1
x
±
=
∞
). On the other hand, exp
°
1
z
±
approaches 0 as
z
approaches 0 from the
negative real axis. Hence lim
z
→
0
²
²
exp
°
1
z
±²
²
°
=
∞
, that is, exp
°
1
z
±
has an essential singularity at 0.
To get a feel for the diFerent types of singularities, we start with the following results.
Proposition 9.5.
Suppose
z
0
is a isolated singularity of
f
. Then
(a)
z
0
is removable if and only if
lim
z
→
z
0
(
z
−
z
0
)
f
(
z
) = 0;
(b)
if
z
0
is a pole then
lim
z
→
z
0
(
z
−
z
0
)
n
+1
f
(
z
)=0
for some positive integer
n
.
Remark.
The smallest possible
n
in (b) is the
order
of the pole. We will see in the proof that “near
the pole
z
0
” we can write
f
(
z
) as
h
(
z
)
(
z
−
z
0
)
n
for some function
h
which is holomorphic (and not zero)
at
z
0
. This is very similar to the game we played with zeros in Chapter 8:
f
has a zero of order (or
multiplicity)
m
at
z
0
if we can write
f
(
z
)=(
z
−
z
0
)
m
h
(
z
), where
h
is holomorphic and not zero
at
z
0
. We will make use of the notions of zeros and poles of certain orders quite extensively in this
chapter.
Proof.
(a) Suppose
z
0
is removable, and
g
is holomorphic on
D
R
(
z
0
), the open disk with radius
R
centered at
z
0
such that
f
=
g
for
z
°
=
z
0
. Then we can make use of the fact that
g
is continuous
at
z
0
:
lim
z
→
z
0
(
z
−
z
0
)
f
(
z
)= l
im
z
→
z
0
(
z
−
z
0
)
g
(
z
)=
g
(
z
0
)l
im
z
→
z
0
(
z
−
z
0
)=0
.
Conversely, suppose that lim
z
→
z
0
(
z
−
z
0
)
f
(
z
) = 0, and
f
is holomorphic on the punctured disk
ˆ
D
R
(
z
0
)=
D
R
(
z
0
)
\{
z
0
}
.Thende±ne
g
(
z
)=
³
(
z
−
z
0
)
2
f
(
z
)i
f
z
°
=
z
0
,
0i
f
z
=
z
0
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 lim, Laurent, residue theorem

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