Class 1/28 1 Zeros of an Analytic Function

Math 752 Spring 2011

Class 1/28

1 Zeros of an analytic function

Towards the fundamental theorem of algebra and its statement for analytic functions.

Deﬁnition 1. Let f : G → C be analytic and f(a) = 0. a is said to have multiplicity m ≥ 1 if there exists an analytic function g : G → C with g(a) 6= 0 so that f(z) = (z − a)mg(z).

Deﬁnition 2. If f is analytic in C it is called entire. An entire function has a power series expansion with inﬁnite radius of convergence.

Theorem 1 (Liouville’s Theorem). If f is a bounded entire function then f is constant.

0 Proof. Assume |f(z)| ≤ M for all z ∈ C. Use Cauchy’s estimate for f to obtain that |f 0(z)| ≤ M/R for every R > 0 and hence equal to 0.

Theorem 2 (Fundamental theorem of algebra). For every non-constant polynomial there exists a ∈ C with p(a) = 0. Proof. Two facts: If p has degree ≥ 1 then

lim p(z) = ∞ z→∞ where the limit is taken along any path to ∞ in C∞. (Sometimes also written as |z| → ∞.) If p has no zero, its reciprocal is therefore entire and bounded. Invoke Liouville’s theorem.

Corollary 1. If p is a polynomial with zeros aj (multiplicity kj) then p(z) = k k km c(z − a1) 1 (z − a2) 2 ...(z − am) . Proof. Induction, and the fact that p(z)/(z − a) is a polynomial of degree n − 1 if p(a) = 0.

1 The zero function is the only analytic function that has a zero of inﬁnite order. (Compare with the real situation, e.g, x 7→ e−1/x2 at the origin.)

Theorem 3. G connected open set, f : G → C analytic. Equivalent: 1. f ≡ 0,

2. there is a ∈ G with f (n)(a) = 0 for all n ≥ 0,

3. the zeros of f in G have a limit point.

Proof. The ﬁrst statement implies the other two. We show that the third implies the second: Assume that the zeros of f have a limit point a and that f (n)(a) 6= 0 for some n. Let R > 0 be such that f is analytic in a ball of radius R about a. Then deﬁne g by

f(z) = (z − a)n−1g(z), (1) and use power series expansions to show that g is analytic in this ball with g(a) 6= 0. Since a is a limit point of zeros, (1) shows that g(a) = 0, a contradiction. Note that the second statement implies the ﬁrst in a disk right away. (Power series expansions!). For general open sets, consider A = {z ∈ G : f (n)(z) = 0 for all n}. Every limit point of A in G lies in A (since the derivatives are continuous). But A is also open in G: For z ∈ A there exists a disk with center z in G, and a power series expansion shows that f is the zero function in the disk. Hence the disk is a subset of A. It follows that A = G.

Corollary 2. f ≡ g in an open, connected G iﬀ the set of points with f(z) = g(z) has a limit point.

Corollary 3. f analytic and not identically zero in an open, connected G, and a ∈ G with f(a) = 0. Then there exists integer n ≥ 1 and analytic n g : G → C with g(a) 6= 0 and f(z) = (z − a) g(z). Proof. Let n be so that f (k)(a) = 0 for k < n and f (n)(a) = 0. Deﬁne g(z) = (z − a)−nf(z) and note that g0(z) exists for all z ∈ G\{a}. Use a power series expansion of f to show that g is analytic at a as well.

Corollary 4. For every zero a of an analytic f there exists a punctured disk with center a on which f is non-zero.

2 Theorem 4 (Maximum Modulus Theorem). If G is a region (open, connected subset of C) and f : G → C analytic such that there exists a ∈ G with |f(a)| ≥ |f(z)| for all z ∈ G, then f is constant.

The proof will be given next time.