52.6. COUNTING ZEROS 1665

andn(η ,ak) = n(γ,ak) , n( f ◦ γ,α) = n( f ◦η ,α) (52.6.7)

for k = 1, · · · ,m. Thenn( f ◦ γ,α) = n( f ◦η ,α)

=1

2πi

∫f◦η

dww−α

=1

2πi

∫ b

a

f ′ (η (t))f (η (t))−α

η′ (t)dt

=1

2πi

∫η

f ′ (z)f (z)−α

dz

=m

∑k=1

n(η ,ak)

By Theorem 52.6.1. By 52.6.7, this equals ∑mk=1 n(γ,ak) which proves the theorem.

The next theorem is incredible and is very interesting for its own sake. The followingpicture is descriptive of the situation of this theorem.

f

aa1

a2

a3

a4

B(a,ε)

zα

B(α,δ )

Theorem 52.6.3 Let f : B(a,R)→ C be analytic and let

f (z)−α = (z−a)m g(z) , ∞ > m≥ 1

where g(z) ̸= 0 in B(a,R) . ( f (z)−α has a zero of order m at z = a.) Then there existε,δ > 0 with the property that for each z satisfying 0 < |z−α|< δ , there exist points,

{a1, · · · ,am} ⊆ B(a,ε) ,

such thatf−1 (z)∩B(a,ε) = {a1, · · · ,am}

and each ak is a zero of order 1 for the function f (·)− z.

Proof: By Theorem 51.5.3 f is not constant on B(a,R) because it has a zero of orderm. Therefore, using this theorem again, there exists ε > 0 such that B(a,2ε)⊆ B(a,R) andthere are no solutions to the equation f (z)−α = 0 for z ∈ B(a,2ε) except a. Also assumeε is small enough that for 0 < |z−a| ≤ 2ε, f ′ (z) ̸= 0. This can be done since otherwise, a