Zeroes of Holomorphic Functions

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 identically 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)ng(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 n f(z) = an(z − a) . n=0 (n) But then an = f (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 (n) f (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. 1 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 (n) limit point of a sequence {zk} in E. Then for any n and k, f (zk) = 0. Since f (n) is continuous, this implies that (n) (n) f (a) = lim f (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. −1 Since γ is continuous, γ (Dε(γ(T ))), which contains T , is also open. In −1 particular, there is a t > T such that t ∈ γ (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 n n+1 f(z) = an(z − a) + an+1(z − a) + ··· 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}. 2 (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)mg(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: [email protected]