LECTURE-19 : ZEROES OF HOLOMORPHIC FUNCTIONS
VED V. DATAR∗
A complex number a is called a zero of a holomorphic function f : Ω → C
if f (a) = 0. A basic fact is that zeroes of holomorphic functions are isolated.
This follows from the following theorem.
Theorem 0.1. Let f : Ω → C be a holomorphic function that is not identi-
cally zero, and let a ∈ Ω be a zero of f . Then there exists a disc D around
a, a non vanishing holomorphic function g : D → C (that is, g(z) 6= 0 for
all z ∈ D), and a unique positive integer n such that
f (z) = (z − a)n g(z).
The positive integer n is called the order of the zero at a. The main
tool that we need in order to prove this is the strong principle of analytic
continuation.
Lemma 0.1. Let Ω ⊂ C be a connected domain, and f be holomorphic on
Ω. If there is a p ∈ Ω such that
f k (p) = 0
for all k = 0, 1, 2 · · · , then f (z) = 0 for all z ∈ Ω.
Proof. Let E ⊂ Ω defined by
E = {z ∈ Ω | f (n) (z) = 0, ∀ n = 0, 1, 2, · · · }.
Then by hypothesis p ∈ E, and hence E is non-empty. It is enough to show
that E = Ω. The proof is completed by proving three claims.
Claim 1. E is an open subset.
To see this, let a ∈ E. Since Ω is open, there exists a disc Dε (a) such
that Dε (a) ⊂ Ω. By analyticity, for any z ∈ Dε (a),
∞
X
f (z) = an (z − a)n .
n=0
But then an = f (n) (a)/n!, and since a ∈ E this implies that an = 0 for
all n = 0, 1, 2, · · · . Hence f (z) = 0 for all z ∈ Dε (a), and in particular
f (n) (z) = 0 for all z ∈ Dε (a) and all n, and hence Dε (a) ⊂ E. That is, every
point in E has an open neighborhood which also belongs to E, which shows
that E is open.
Date: 24 August 2016.
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Claim 2. E is closed in Ω. That is, if {zk } a sequence of points in E such
that
lim zk = a ∈ Ω,
k→∞
then a ∈ E.
We need to show that E contains all it’s limit points. Let a ∈ Ω be a
limit point of a sequence {zk } in E. Then for any n and k, f (n) (zk ) = 0.
Since f (n) is continuous, this implies that
f (n) (a) = lim f (n) (zk ) = 0,
k→∞
which in turn implies that a ∈ E. Hence E is closed in Ω.
Claim 3. E = Ω.
If not, then there is a q ∈ Ω \ E. Since Ω is connected, there exists a
continuous path γ : [0, 1] → Ω such that γ(0) = p and γ(1) = q. Let
T = sup {γ(t) ∈ E}.
t∈[0,1]
Then T is the first time that the curve leaves E. Since E is closed, γ(T ) ∈ E.
To see this, let tn → T − . Then since γ is continuous, γ(tn ) → γ(T ), and
hence γ(T ) ∈ Ω is a limit point for the sequence {γ(tn )}. Each point of the
sequence γ(tn ) belongs to E and since E is closed in Ω, this implies that
γ(T ) ∈ E. But now, E is also open. So there is a disc Dε (γ(T )) ⊂ E.
Since γ is continuous, γ −1 (Dε (γ(T ))), which contains T , is also open. In
particular, there is a t > T such that t ∈ γ −1 (Dε (γ(T ))) or equivalently
γ(t) ∈ Dε (γ(T )). So we have found a t > T such that γ(t) ∈ E which
contradicts the maximality of T and hence there can be no point q ∈ Ω \
E.
Remark 0.1. For readers familiar with some point-set topology, the above
argument should be familiar. Namely, if Ω is a connected set, then the only
subsets that are both open and closed in Ω, are the empty set and Ω itself.
Proof of Theorem 0.1. By the lemma, if f is not identically zero, there
exists an n such that f (k) (a) = 0 for k = 0, 1, · · · , n − 1 but f (n) (a) 6= 0. Let
Dε (a) be a disc such that Dε (a) ⊂ Ω. Then f has a power series expansion
in the disc centered at z = a with the first n terms vanishing. So we write
f (z) = an (z − a)n + an+1 (z − a)n+1 + · · ·
with an 6= 0. The the theorem is proved with
g(z) = an + an+1 (z − a) + · · · .
As an immediate corollary to the theorem we have the following.
Corollary 0.1. Let f : Ω → C holomorphic.
(1) The set of zeroes of f is isolated. That is, for every zero a, there
exists a small disc Dε (a) centered at a such that f (z) 6= 0 for all
z ∈ Dε (a) \ {a}.
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(2) (weak principle of analytic continuation) If Ω is connected, and f, g :
Ω → C are holomorphic such that for some open subset U ⊂ Ω,
f (z) = g(z),
for all z ∈ U , then
f (z) = g(z)
for all z ∈ Ω.
Proof. By the theorem, if a ∈ Ω is a root, then there exists a disc D around
a, and integer m, and a holomorphic function g : D → C such that g(a) 6= 0
and
f (z) = (z − a)m g(z).
Since g(a) 6= 0, by continuity, there is a small disc Dε (a) ⊂ D such that
g(z) 6= 0 for any z ∈ Dε (a). But then on this disc (z − a) is also not zero
anywhere except at a, and hence for any z ∈ Dε (a) \ {a}, f (z) 6= 0 exactly
what we wished to prove. Part (2) is a trivial consequence of the Lemma.
Example 0.1. Even though the zeroes are isolated, they could converge to
the boundary. For instance, consider the holomorphic function
π
f (z) = sin
z
∗
on C . Clearly z = 1/n is a sequence of zeroes. They are isolated, but
converge to z = 0 which is not in the domain.
∗ Department of Mathematics, UC Berkeley
E-mail address: vvdatar@berkeley.edu
