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 2 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 2 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 2 Ω such that f k(p) = 0 for all k = 0; 1; 2 ··· , then f(z) = 0 for all z 2 Ω. Proof. Let E ⊂ Ω defined by E = fz 2 Ω j f (n)(z) = 0; 8 n = 0; 1; 2; · · · g: Then by hypothesis p 2 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 2 E. Since Ω is open, there exists a disc D"(a) such that D"(a) ⊂ Ω. By analyticity, for any z 2 D"(a), 1 X n f(z) = an(z − a) : n=0 (n) But then an = f (a)=n!, and since a 2 E this implies that an = 0 for all n = 0; 1; 2; ··· . Hence f(z) = 0 for all z 2 D"(a), and in particular (n) f (z) = 0 for all z 2 D"(a) and all n, and hence D"(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 fzkg a sequence of points in E such that lim zk = a 2 Ω; k!1 then a 2 E. We need to show that E contains all it's limit points. Let a 2 Ω be a (n) limit point of a sequence fzkg 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!1 which in turn implies that a 2 E. Hence E is closed in Ω. Claim 3. E = Ω. If not, then there is a q 2 Ω n 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 fγ(t) 2 Eg: t2[0;1] Then T is the first time that the curve leaves E. Since E is closed, γ(T ) 2 E. − To see this, let tn ! T . Then since γ is continuous, γ(tn) ! γ(T ), and hence γ(T ) 2 Ω is a limit point for the sequence fγ(tn)g. Each point of the sequence γ(tn) belongs to E and since E is closed in Ω, this implies that γ(T ) 2 E. But now, E is also open. So there is a disc D"(γ(T )) ⊂ E. −1 Since γ is continuous, γ (D"(γ(T ))), which contains T , is also open. In −1 particular, there is a t > T such that t 2 γ (D"(γ(T ))) or equivalently γ(t) 2 D"(γ(T )). So we have found a t > T such that γ(t) 2 E which contradicts the maximality of T and hence there can be no point q 2 Ω n 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 D"(a) be a disc such that D"(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 D"(a) centered at a such that f(z) 6= 0 for all z 2 D"(a) n fag. 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 2 U, then f(z) = g(z) for all z 2 Ω. Proof. By the theorem, if a 2 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 D"(a) ⊂ D such that g(z) 6= 0 for any z 2 D"(a). But then on this disc (z − a) is also not zero anywhere except at a, and hence for any z 2 D"(a) n fag, 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] 3.