• Complex Analysis: #8 Zeros of Analytic Functions

Definition 7
Let f : G → ℂ be an analytic function defined in a region G. A point z₀ ∈ G with f(z₀) = 0 is called a zero of the function.

So let z₀ ∈ G be a zero of the analytic function f : G → ℂ. As usual, without loss of generality, we may assume that z₀ = 0. As we have seen, we can choose some r > 0 such that B(0, r) = {z : |z| < r} ⊂ G and

for all such z ∈ B(0, r).

The fact that f(0) = 0 means that c₀ = 0. Let k > 0 be the smallest integer such that c_k ≠ 0. (If c_n = 0 for all n, then f is simply the constant function which is zero everywhere. This is not what we are interested in here so we will assume that some k exists with c_k ≠ 0.) The easiest case is then that k = 1. In this case, we have

and in particular f '(0) = c₁ ≠ 0. Could it be that for every ∈ > 0 there exists a complex number z_∈ with 0 < |z_∈| < ∈ and yet f(z_∈) = 0? But that would imply that

That is impossible, since f '(0) ≠ 0. Therefore we have:


Theorem 14
Let f : G → ℂ be analytic and let z₀ ∈ G be such that f(z₀) = 0 while f '(z₀) ≠ 0. Then there exists an ∈ > 0 such that B(z₀, ∈ ) ⊂ G and the only zero of f in B(z₀, ∈ ) is the single number z₀.

Of course, another way of thinking of these things — and remembering what was done in Analysis II — is to consider f to be a continuously differentiable mapping of G into ℂ, represented as ℝ². The mapping f is then totally differentiable, and the derivative at z₀ is not singular; thus it is a local bijection around z₀.

More generally, we might have k being greater than 1. In any case, the number k is called the order of the zero. A zero of order 1 is also called a simple zero.


Theorem 15
Let f : G → ℂ be analytic in the region G, and let z₀ ∈ G be a zero of f of order k. Then there exists an ∈ > 0 such that in the open disc B(z₀, ∈ ) of radius ∈ around z₀ we have f(z) = (h(z))^k where h : B(z₀, ∈ ) → ℂ is analytic with a simple zero at z₀.

Proof
For sufficiently small ∈ > 0, we can write

say, for z ∈ B(z₀, ∈ ). Here g : B(z₀, ∈ ) → ℂ is analytic, and g(z₀) = c_k ≠ 0. Thus there are k distinct k-th roots of the number g(z₀) = c_k. [For any w ≠ 0 in ℂ we have w = re^iθ say. Then each of the numbers ^k√r · e^{iθ/k + 2πil/k} , for l = 0, . . . , k − 1, is a different k-th root of w.] Let z₁ be one of these k-th roots of g(z₀). Now consider the particular polynomial function ϕ_k'(z) = z^k. We know that ϕ_k is an entire function, and that ϕ_k'(z₁) = kz₁^k−1 ≠ 0, since z₁ ≠ 0. So there is a neighborhood U₁ of z₁, and a neighborhood V₁ of g(z₀), such that ϕ_k : U₁ → V₁ is a bijection [ϕ_k is totally differentiable and non-singular at z₁], with ϕ_k(z₁) = g(z₀). Let ϕ_k⁻¹ : V₁ → U₁ be the inverse mapping. Now choose ∈ > 0 so small that g(B(z₀, ∈ )) ⊂ V₁. Then take h(z) = (z − z₀) · ϕ_k⁻¹(g(z)). This defines a function h : B(z₀, ∈ ) → ℂ which satisfies our conditions.[As in real analysis (the proof is the same here in complex analysis) we have the rule that if ϕ is an invertible differentiable function (with non-vanishing derivative), then ϕ⁻¹ is also differentiable, with derivative (ϕ⁻¹) '(z) = 1/ϕ '(ϕ⁻¹(z)).]


Theorem 16
Again let f : G → ℂ be analytic in the region G, and let z₀ ∈ G be a zero of order k. Then there exists an ∈₀ > 0 and an open neighborhood U_∈0 ⊂ G of z₀ with f(U_∈0) = B(0, ∈₀). Within U_∈0 ,z₀ is the only zero of f, and if w ≠ 0 in B(0, ∈₀) then there are precisely k different points v₁, . . ., v_k in U_∈0 with f(v_j) = w, for all j.

Proof
Since f is continuous and B(0, ∈₀) is open, it follows that U_∈0 = f ⁻¹(B(0, ∈₀)) is also open, regardless of how the number ∈₀ > 0 is chosen. So we begin by choosing an ∈₁ > 0 sufficiently small that we can use theorem 15 and write f(z) = (h(z))^k, for all z ∈ B(0, ∈₁). Since h has a simple zero at z ₀, and therefore the derivative at z ₀ is not zero (h '(z ₀) ≠ 0), there exists a neighborhood of z ₀ such that h is a bijection when restricted to the neighborhood. So let ∈ > 0 be chosen sufficiently small that B(0, ∈) is contained within the corresponding neighborhood of 0. Finally, with this ∈ , we take ∈₀ = ∈^k . Then if w ≠ 0 in B(0, ∈₀), we have k different k-th roots of w, lets call them u₁, . . ., u_k. They are all in B(0, ∈). Therefore each has a unique inverse under h, namely v_j = h⁻¹(u_j), for j = 1, . . . , k. Is it possible that some other point, v say, not equal to any of the v_j, also is such that f(v) = (h(v))^k = w? But then h(v) would also be a k-th root of w, not equal to any of the u_j, since after all, h⁻¹ is a bijection when restricted to B(z₀, ∈₀). This is impossible, owing to the fact that there are only k different k-th roots of w.