Tuesday, January 15, 2013

Complex Analysis: #8 Zeros of Analytic Functions

  • 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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 1

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 ck ≠ 0. (If cn = 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 ck ≠ 0.) The easiest case is then that k = 1. In this case, we have

Complex Analysis: #8 Zeros of Analytic Functions equation pic 2

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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 3

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

Complex Analysis: #8 Zeros of Analytic Functions equation pic 4

say, for z ∈ B(z0, ∈ ). Here g : B(z0, ∈ ) → ℂ is analytic, and g(z0) = ck ≠ 0. Thus there are k distinct k-th roots of the number g(z0) = ck. [For any w ≠ 0 in ℂ we have w = re say. Then each of the numbers kr · eiθ/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 ϕk'(z) = zk. We know that ϕk is an entire function, and that ϕk'(z1) = kz1k−1 ≠ 0, since z1 ≠ 0. So there is a neighborhood U1 of z1, and a neighborhood V1 of g(z0), such that ϕk : U1 → V1 is a bijection [ϕk is totally differentiable and non-singular at z1], with ϕk(z1) = g(z0). Let ϕk−1 : V1 → U1 be the inverse mapping. Now choose ∈ > 0 so small that g(B(z0, ∈ )) ⊂ V1. Then take h(z) = (z − z0) · ϕk−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 U0 ⊂ G of z0 with f(U0) = B(0, ∈0). Within U0 ,z0 is the only zero of f, and if w ≠ 0 in B(0, ∈0) then there are precisely k different points v1, . . ., vk in U0 with f(vj) = w, for all j.

Proof
Since f is continuous and B(0, ∈0) is open, it follows that U0 = 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, . . ., uk. They are all in B(0, ∈). Therefore each has a unique inverse under h, namely vj = h−1(uj), for j = 1, . . . , k. Is it possible that some other point, v say, not equal to any of the vj, 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 uj, 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.

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