2Contour integration
IB Complex Analysis
2.5 Zeroes
Recall that for a polynomial
p
(
z
), we can talk about the order of its zero at
z
=
a
by looking at the largest power of (
z − a
) dividing
p
. A priori, it is not
clear how we can do this for general functions. However, given that everything
is a Taylor series, we know how to do this for holomorphic functions.
Definition
(Order of zero)
.
Let
f
:
B
(
a, r
)
→ C
be holomorphic. Then we
know we can write
f(z) =
∞
X
n=0
c
n
(z − a)
n
as a convergent power series. Then either all
c
n
= 0, in which case
f
= 0 on
B
(
a, r
), or there is a least
N
such that
c
N
6
= 0 (
N
is just the smallest
n
such
that f
(n)
(a) 6= 0).
If N > 0, then we say f has a zero of order N.
If f has a zero of order N at a, then we can write
f(z) = (z − a)
N
g(z)
on B(a, r), where g(a) = c
N
6= 0.
Often, it is not the actual order that is too important. Instead, it is the
ability to factor f in this way. One of the applications is the following:
Lemma
(Principle of isolated zeroes)
.
Let
f
:
B
(
a, r
)
→ C
be holomorphic and
not identically zero. Then there exists some 0
< ρ < r
such that
f
(
z
)
6
= 0 in the
punctured neighbourhood B(a, ρ) \ {a}.
Proof. If f(a) 6= 0, then the result is obvious by continuity of f .
The other option is not too different. If
f
has a zero of order
N
at
a
, then
we can write
f
(
z
) = (
z − a
)
N
g
(
z
) with
g
(
a
)
6
= 0. By continuity of
g
,
g
does
not vanish on some small neighbourhood of
a
, say
B
(
a, ρ
). Then
f
(
z
) does not
vanish on B(a, ρ) \ {a}.
A consequence is that given two holomorphic functions on the same domain,
if they agree on sufficiently many points, then they must in fact be equal.
Corollary
(Identity theorem)
.
Let
U ⊆ C
be a domain, and
f, g
:
U → C
be
holomorphic. Let
S
=
{z ∈ U
:
f
(
z
) =
g
(
z
)
}
. Suppose
S
contains a non-isolated
point, i.e. there exists some
w ∈ S
such that for all
ε >
0,
S ∩ B
(
w, ε
)
6
=
{w}
.
Then f = g on U.
Proof.
Consider the function
h
(
z
) =
f
(
z
)
− g
(
z
). Then the hypothesis says
h
(
z
)
has a non-isolated zero at
w
, i.e. there is no non-punctured neighbourhood of
w
on which
h
is non-zero. By the previous lemma, this means there is some
ρ >
0
such that h = 0 on B(w, ρ) ⊆ U .
Now we do some topological trickery. We let
U
0
= {a ∈ U : h = 0 on some neighbourhood B(a, ρ) of a in U},
U
1
= {a ∈ U : there exists n ≥ 0 such that h
(n)
6= 0}.
Clearly,
U
0
∩U
1
=
∅
, and the existence of Taylor expansions shows
U
0
∪U
1
=
U
.
Moreover,
U
0
is open by definition, and
U
1
is open since
h
(n)
(
z
) is continuous
near any given
a ∈ U
1
. Since
U
is (path) connected, such a decomposition can
happen if one of
U
0
and
U
1
is empty. But
w ∈ U
0
. So in fact
U
0
=
U
, i.e.
h
vanishes on the whole of U. So f = g.
In particular, if two holomorphic functions agree on some small open subset
of the domain, then they must in fact be identical. This is a very strong result,
and is very false for real functions. Hence, to specify, say, an entire function, all
we need to do is to specify it on an arbitrarily small domain we like.
Definition
(Analytic continuiation)
.
Let
U
0
⊆ U ⊆ C
be domains, and
f
:
U
0
→
C
be holomorphic. An analytic continuation of
f
is a holomorphic function
h : U → C such that h|
U
0
= f, i.e. h(z) = f (z) for all z ∈ U
0
.
By the identity theorem, we know the analytic continuation is unique if it
exists.
Thus, given any holomorphic function
f
:
U → C
, it is natural to ask how
far we can extend the domain, i.e. what is the largest
U
0
⊇ U
such that there is
an analytic continuation of f to U
0
.
There is no general method that does this for us. However, one useful trick
is to try to write our function
f
in a different way so that it is clear how we can
extend it to elsewhere.
Example. Consider the function
f(z) =
X
n≥0
z
n
= 1 + z + z
2
+ ···
defined on B(0, 1).
By itself, this series diverges for
z
outside
B
(0
,
1). However, we know well
that this function is just
f(z) =
1
1 − z
.
This alternative representation makes sense on the whole of
C
except at
z
= 1.
So we see that
f
has an analytic continuation to
C \ {
1
}
. There is clearly no
extension to the whole of C, since it blows up near z = 1.
Example. Alternatively, consider
f(z) =
X
n≥0
z
2
n
.
Then this again converges on
B
(0
,
1). You will show in example sheet 2 that
there is no analytic continuation of f to any larger domain.
Example. The Riemann zeta function
ζ(z) =
∞
X
n=1
n
−z
defines a holomorphic function on
{z
:
Re
(
z
)
>
1
} ⊆ C
. Indeed, we have
|n
−z
|
=
|n
Re(z)
|
, and we know
P
n
−t
converges for
t ∈ R
>1
, and in fact does so
uniformly on any compact domain. So the corollary of Morera’s theorem tells us
that ζ(z) is holomorphic on Re(z) > 1.
We know this cannot converge as
z →
1, since we approach the harmonic
series which diverges. However, it turns out
ζ
(
z
) has an analytic continuation to
C \ {1}. We will not prove this.
At least formally, using the fundamental theorem of arithmetic, we can
expand n as a product of its prime factors, and write
ζ(z) =
Y
primes p
(1 + p
−z
+ p
−2z
+ ···) =
Y
primes p
1
1 − p
−z
.
If there were finitely many primes, then this would be a well-defined function
on all of
C
, since this is a finite product. Hence, the fact that this blows up at
z = 1 implies that there are infinitely many primes.