2Contour integration
IB Complex Analysis
2.6 Singularities
The next thing to study is singularities of holomorphic functions. These are
places where the function is not defined. There are many ways a function can
be ill-defined. For example, if we write
f(z) =
1 − z
1 − z
,
then on the face of it, this function is not defined at
z
= 1. However, elsewhere,
f
is just the constant function 1, and we might as well define
f
(1) = 1. Then we
get a holomorphic function. These are rather silly singularities, and are singular
solely because we were not bothered to define f there.
Some singularities are more interesting, in that they are genuinely singular.
For example, the function
f(z) =
1
1 − z
is actually singular at
z
= 1, since
f
is unbounded near the point. It turns out
these are the only possibilities.
Proposition
(Removal of singularities)
.
Let
U
be a domain and
z
0
∈ U
. If
f
:
U \{z
0
} → C
is holomorphic, and
f
is bounded near
z
0
, then there exists an
a such that f(z) → a as z → z
0
.
Furthermore, if we define
g(z) =
(
f(z) z ∈ U \ {z
0
}
a z = z
0
,
then g is holomorphic on U .
Proof. Define a new function h : U → C by
h(z) =
(
(z −z
0
)
2
f(z) z 6= z
0
0 z = z
0
.
Then since
f
is holomorphic away from
z
0
, we know
h
is also holomorphic away
from z
0
.
Also, we know
f
is bounded near
z
0
. So suppose
|f
(
z
)
| < M
in some
neighbourhood of z
0
. Then we have
h(z) − h(z
0
)
z −z
0
≤ |z −z
0
|M.
So in fact
h
is also differentiable at
z
0
, and
h
(
z
0
) =
h
0
(
z
0
) = 0. So near
z
0
,
h
has a Taylor series
h(z) =
X
n≥0
a
n
(z −z
0
)
n
.
Since we are told that a
0
= a
1
= 0, we can define a g(z) by
g(z) =
X
n≥0
a
n+2
(z −z
0
)
n
,
defined on some ball
B
(
z
0
, ρ
), where the Taylor series for
h
is defined. By
construction, on the punctured ball
B
(
z
0
, ρ
)
\{z
0
}
, we get
g
(
z
) =
f
(
z
). Moreover,
g(z) → a
2
as z → z
0
. So f(z) → a
2
as z → z
0
.
Since g is a power series, it is holomorphic. So the result follows.
This tells us the only way for a function to fail to be holomorphic at an
isolated point is that it blows up near the point. This won’t happen because
f
fails to be continuous in some weird ways.
However, we are not yet done with our classification. There are many ways
in which things can blow up. We can further classify these into two cases — the
case where
|f
(
z
)
| → ∞
as
z → z
0
, and the case where
|f
(
z
)
|
does not converge as
z → z
0
. It happens that the first case is almost just as boring as the removable
ones.
Proposition.
Let
U
be a domain,
z
0
∈ U
and
f
:
U \{z
0
} → C
be holomorphic.
Suppose
|f
(
z
)
| → ∞
as
z → z
0
. Then there is a unique
k ∈ Z
≥1
and a unique
holomorphic function g : U → C such that g(z
0
) 6= 0, and
f(z) =
g(z)
(z − z
0
)
k
.
Proof.
We shall construct
g
near
z
0
in some small neighbourhood, and then
apply analytic continuation to the whole of
U
. The idea is that since
f
(
z
) blows
up nicely as z → z
0
, we know
1
f(z)
behaves sensibly near z
0
.
We pick some
δ >
0 such that
|f
(
z
)
| ≥
1 for all
z ∈ B
(
z
0
;
δ
)
\ {z
0
}
. In
particular, f(z) is non-zero on B(z
0
; δ) \ {z
0
}. So we can define
h(z) =
(
1
f(z)
z ∈ B(z
0
; δ) \ {z
0
}
0 z = z
0
.
Since
|
1
f(z)
| ≤
1 on
B
(
z
0
;
δ
)
\{z
0
}
, by the removal of singularities,
h
is holomorphic
on
B
(
z
0
, δ
). Since
h
vanishes at the
z
0
, it has a unique definite order at
z
0
, i.e.
there is a unique integer
k ≥
1 such that
h
has a zero of order
k
at
z
0
. In other
words,
h(z) = (z − z
0
)
k
`(z),
for some holomorphic ` : B(z
0
; δ) → C and `(z
0
) 6= 0.
Now by continuity of
`
, there is some 0
< ε < δ
such that
`
(
z
)
6
= 0 for all
z ∈ B(z
0
, ε). Now define g : B(z
0
; ε) → C by
g(z) =
1
`(z)
.
Then g is holomorphic on this disc.
By construction, at least away from z
0
, we have
g(z) =
1
`(z)
=
1
h(z)
· (z − z
0
)
k
= (z − z
0
)
k
f(z).
g
was initially defined on
B
(
z
0
;
ε
)
→ C
, but now this expression certainly makes
sense on all of
U
. So
g
admits an analytic continuation from
B
(
z
0
;
ε
) to
U
. So
done.
We can start giving these singularities different names. We start by formally
defining what it means to be a singularity.
Definition
(Isolated singularity)
.
Given a domain
U
and
z
0
∈ U
, and
f
:
U \ {z
0
} → C holomorphic, we say z
0
is an isolated singularity of f.
Definition
(Removable singularity)
.
A singularity
z
0
of
f
is a removable singu-
larity if f is bounded near z
0
.
Definition
(Pole)
.
A singularity
z
0
is a pole of order
k
of
f
if
|f
(
z
)
| → ∞
as
z → z
0
and one can write
f(z) =
g(z)
(z − z
0
)
k
with g : U → C, g(z
0
) 6= 0.
Definition
(Isolated essential singularity)
.
An isolated singularity is an isolated
essential singularity if it is neither removable nor a pole.
It is easy to give examples of removable singularities and poles. So let’s look
at some essential singularities.
Example. z 7→ e
1/z
has an isolated essential singularity at z = 0.
Note that if
B
(
z
0
, ε
)
\{z
0
} → C
has a pole of order
h
at
z
0
, then
f
naturally
defines a map
ˆ
f : B(z
0
; ε) → CP
1
= C ∪ {∞}, the Riemann sphere, by
f(z) =
(
∞ z = z
0
f(z) z 6= z
0
.
This is then a “continuous” function. So a singularity is just a point that gets
mapped to the point ∞.
As was emphasized in IA Groups, the point at infinity is not a special point
in the Riemann sphere. Similarly, poles are also not really singularities from the
viewpoint of the Riemann sphere. It’s just that we are looking at it in a wrong
way. Indeed, if we change coordinates on the Riemann sphere so that we label
each point
w ∈ CP
1
by
w
0
=
1
w
instead, then
f
just maps
z
0
to 0 under the new
coordinate system. In particular, at the point
z
0
, we find that
f
is holomorphic
and has an innocent zero of order k.
Since poles are not bad, we might as well allow them.
Definition
(Meromorphic function)
.
If
U
is a domain and
S ⊆ U
is a finite or
discrete set, a function
f
:
U \ S → C
which is holomorphic and has (at worst)
poles on S is said to be meromorphic on U.
The requirement that
S
is discrete is so that each pole in
S
is actually an
isolated singularity.
Example.
A rational function
P (z)
Q(z)
, where
P, Q
are polynomials, is holomorphic
on
C \ {z
:
Q
(
z
) = 0
}
, and meromorphic on
C
. More is true — it is in fact
holomorphic as a function CP
1
→ CP
1
.
These ideas are developed more in depth in the IID Riemann Surfaces course.
As an aside, if we want to get an interesting holomorphic function with
domain
CP
1
, its image must contain the point
∞
, or else its image will be
a compact subset of
C
(since
CP
1
is compact), thus bounded, and therefore
constant by Liouville’s theorem.
At this point, we really should give essential singularities their fair share of
attention. Not only are they bad. They are bad spectacularly.
Theorem
(Casorati-Weierstrass theorem)
.
Let
U
be a domain,
z
0
∈ U
, and
suppose
f
:
U \ {z
0
} → C
has an essential singularity at
z
0
. Then for all
w ∈ C
,
there is a sequence z
n
→ z
0
such that f (z
n
) → w.
In other words, on any punctured neighbourhood
B
(
z
0
;
ε
)
\ {z
0
}
, the image
of f is dense in C.
This is not actually too hard to proof.
Proof. See example sheet 2.
If you think that was bad, actually essential singularities are worse than that.
The theorem only tells us the image is dense, but not that we will hit every
point. It is in fact not true that every point will get hit. For example
e
1
z
can
never be zero. However, this is the worst we can get
Theorem
(Picard’s theorem)
.
If
f
has an isolated essential singularity at
z
0
, then
there is some
b ∈ C
such that on each punctured neighbourhood
B
(
z
0
;
ε
)
\ {z
0
}
,
the image of f contains C \ {b}.
The proof is beyond this course.