# Complex Analysis: #8 Zeros of Analytic Functions

**Deﬁnition 7**

Let f : G → ℂ be an analytic function defined in a region G. A point z

_{0}∈ G with f(z

_{0}) = 0 is called a

*zero*of the function.

So let z

_{0}∈ G be a zero of the analytic function f : G → ℂ. As usual, without loss of generality, we may assume that z

_{0}= 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 c

_{0}= 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) = c

_{1}≠ 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 z

_{0}∈ G be such that f(z

_{0}) = 0 while f '(z

_{0}) ≠ 0. Then there exists an ∈ > 0 such that B(z

_{0}, ∈ ) ⊂ G and the only zero of f in B(z

_{0}, ∈ ) is the single number z

_{0}.

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 ℝ

^{2}. The mapping f is then totally differentiable, and the derivative at z

_{0}is not singular; thus it is a local bijection around z

_{0}.

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 z

_{0}∈ G be a zero of f of order k. Then there exists an ∈ > 0 such that in the open disc B(z

_{0}, ∈ ) of radius ∈ around z

_{0}we have f(z) = (h(z))

^{k}where h : B(z

_{0}, ∈ ) → ℂ is analytic with a simple zero at z

_{0}.

*Proof*

For sufficiently small ∈ > 0, we can write

say, for z ∈ B(z

_{0}, ∈ ). Here g : B(z

_{0}, ∈ ) → ℂ is analytic, and g(z

_{0}) = c

_{k}≠ 0. Thus there are k distinct k-th roots of the number g(z

_{0}) = 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 z

_{1}be one of these k-th roots of g(z

_{0}). Now consider the particular polynomial function ϕ

_{k}'(z) = z

^{k}. We know that ϕ

_{k}is an entire function, and that ϕ

_{k}'(z

_{1}) = kz

_{1}

^{k−1}≠ 0, since z

_{1}≠ 0. So there is a neighborhood U

_{1}of z

_{1}, and a neighborhood V

_{1}of g(z

_{0}), such that ϕ

_{k}: U

_{1}→ V

_{1}is a bijection [ϕ

_{k}is totally differentiable and non-singular at z

_{1}], with ϕ

_{k}(z

_{1}) = g(z

_{0}). Let ϕ

_{k}

^{−1}: V

_{1}→ U

_{1}be the inverse mapping. Now choose ∈ > 0 so small that g(B(z

_{0}, ∈ )) ⊂ V

_{1}. Then take h(z) = (z − z

_{0}) · ϕ

_{k}

^{−1}(g(z)). This defines a function h : B(z

_{0}, ∈ ) → 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 z

_{0}∈ G be a zero of order k. Then there exists an ∈

_{0}> 0 and an open neighborhood U

_{∈0}⊂ G of z

_{0}with f(U

_{∈0}) = B(0, ∈

_{0}). Within U

_{∈0},z

_{0}is the only zero of f, and if w ≠ 0 in B(0, ∈

_{0}) then there are precisely k different points v

_{1}, . . ., 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 u

_{1}, . . ., 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(z

_{0}, ∈

_{0}). This is impossible, owing to the fact that there are only k different k-th roots of w.

## No comments:

## Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.

Comments will only be published after moderation