268 CHAPTER 11. FUNCTIONS OF ONE COMPLEX VARIABLE

11.6 Counting Zeros, Open Mapping TheoremThe open mapping theorem is perhaps a little digression from the main emphasis of thisbook, but it is such a marvelous result, that it seems a shame not to include it. It comesfrom a remarkable formula which counts the number of zeros of an analytic function insidea ball.

Definition 11.6.1 Let U be a nonempty open set in C. Then it is called connectedif for any z,w ∈U, there exists a continuous one to one piecewise linear γ : [0,1]→U suchthat γ (0) = z and γ (1) = w. A connected open set will be called a region in this section.

By Theorem 6.5.8 the above implies the usual definition of a connected set. To go theother way, suppose the usual definition and let z ∈U be given. Let S denote those pointsreachable by a continuous one to one piecewise linear curve from z. Show S is open. Nowshow that those points of U which are not reachable by such a curve is also open. Thus Uis the disjoint union of open sets. One of them must be empty if U is connected accordingto the usual definition. This is really very easy if you use convexity of balls. It is not thepurpose of this book to belabor higher dimensional considerations and the above definitionis sufficiently descriptive.

Here is a very useful equivalence.

Proposition 11.6.2 Let U be a region as just defined and suppose f is an analyticfunction defined on U. Then the following are equivalent

1. There exists z ∈U such that f (n) (z) = 0 for all n = 0,1,2, ....

2. f (z) = 0 for all z ∈U.

Proof: 1.⇒ 2. Let z be as in 1. Then, since U is open, B(z,r) ⊆U for small enoughpositive r. It follows from Corollary 11.2.11 that for w in this ball,

f (w) =∞

∑k=0

f (k) (z)k!

(w− z)k = 0

and so f is zero on B(z,r). This shows that near z, f (n) (w) = 0 for all n an integer largerthan or equal to 0. Letting S denote z ∈U such that f (n) (z) = 0 for all n, this shows thatS is an open subset of U . If S is all of U then this was what was to be shown. Oth-erwise, let z ∈ S and w ∈ U \ S. Let γ (t) go from z to w,γ (0) = z,γ (1) = w. Then letT ≡ sup{t ∈ [0,1] : γ (s) ∈ S for s≤ t} and suppose T < 1. Then let tn be an increasing se-quence converging to T, 0 = f (γ (tn))→ f (γ (T )) and so f (γ (T )) = 0. However, eachderivative of f is continuous also and so the same reasoning shows that f (n) (γ (T )) = 0 foreach n ≥ 1. Hence γ (T ) ∈ S. But this violates the definition of T because γ (t) ∈ S for allt ∈ [T,T + ε] for suitably small ε due to what was just shown that S is open. Hence T = 1and so w ∈ S. Thus U \S = /0.

2.⇒ 1. This is obvious. If f (z) = 0 for all z, then all derivatives are also 0 on the openset U .

Next is one more equivalence.

Theorem 11.6.3 Suppose f is analytic on a region U (open and connected). Thefollowing are equivalent.