1664 CHAPTER 52. THE OPEN MAPPING THEOREM

Ωfγ

a1

a2a3

f (γ([a,b]))

α

Theorem 52.6.2 Let Ω be a region, let γ : [a,b]→ Ω be closed continuous, and boundedvariation such that n(γ,z)= 0 for all z /∈Ω. Also suppose f : Ω→C is analytic and that α /∈f (γ∗) . Then f ◦ γ : [a,b]→ C is continuous, closed, and bounded variation. Also suppose{a1, · · · ,am}= f−1 (α) where these points are counted according to their multiplicities aszeros of the function f −α Then

n( f ◦ γ,α) =m

∑k=1

n(γ,ak) .

Proof: It is clear that f ◦ γ is continuous. It only remains to verify that it is of boundedvariation. Suppose first that γ∗ ⊆ B⊆ B⊆Ω where B is a ball. Then

| f (γ (t))− f (γ (s))|=∣∣∣∣∫ 1

0f ′ (γ (s)+λ (γ (t)− γ (s)))(γ (t)− γ (s))dλ

∣∣∣∣≤ C |γ (t)− γ (s)|

where C ≥max{| f ′ (z)| : z ∈ B

}. Hence, in this case,

V ( f ◦ γ, [a,b])≤CV (γ, [a,b]) .

Now let ε denote the distance between γ∗ and C \Ω. Since γ∗ is compact, ε > 0. Byuniform continuity there exists δ = b−a

p for p a positive integer such that if |s− t|< δ , then|γ (s)− γ (t)|< ε

2 . Then

γ ([t, t +δ ])⊆ B(

γ (t) ,ε

2

)⊆Ω.

Let C ≥ max{| f ′ (z)| : z ∈ ∪p

j=1B(γ (t j) ,

ε

2

)}where t j ≡ j

p (b−a) + a. Then from whatwas just shown,

V ( f ◦ γ, [a,b]) ≤p−1

∑j=0

V(

f ◦ γ,[t j, t j+1

])≤ C

p−1

∑j=0

V(γ,[t j, t j+1

])< ∞

showing that f ◦γ is bounded variation as claimed. Now from Theorem 51.7.15 there existsη ∈C1 ([a,b]) such that

η (a) = γ (a) = γ (b) = η (b) , η ([a,b])⊆Ω,