Topics In Analysis

51.6 Counting Zeros

The above proof of the open mapping theorem relies on the very important inverse function theorem from real analysis. There are other approaches to this important theorem which do not rely on the big theorems from real analysis and are more oriented toward the use of the Cauchy integral formula and specialized techniques from complex analysis. One of these approaches is given next which involves the notion of “counting zeros”. The next theorem is the one about counting zeros. It will also be used later in the proof of the Riemann mapping theorem.

Theorem 51.6.1 Let Ω be an open set in ℂ and let γ :

→ Ω be closed, continuous, bounded variation, and n

= 0 for all z

Ω. Suppose also that f is analytic on Ω having zeros a₁,

,a_m where the zeros are repeated according to multiplicity, and suppose that none of these zeros are on γ^∗. Then

-1- \int f'(z) m\sum 2πi γ f (z)dz = n(γ,ak). k=1

Proof: Let f

= ∏ _j=1^m

where g

≠0 on Ω. Note that some of the a_j could be repeated. Hence

f'(z) \summ --1--- g'(z) f (z) = z - aj + g(z) j=1

and so

-1-\int f'(z) \summ 1--\int g'(z) 2πi γf (z) dz = n (γ,aj)+ 2πi γ g(z)dz. j=1

But the function, z →

is analytic and so by Corollary 50.7.20, the last integral in the above expression equals 0. Therefore, this proves the theorem.

The following picture is descriptive of the situation described in the next theorem.

Theorem 51.6.2 Let Ω be a region, let γ :

→ Ω be closed continuous, and bounded variation such that n

= 0 for all z

Ω. Also suppose f : Ω → ℂ is analytic and that α

. Then f ∘γ :

→ ℂ is continuous, closed, and bounded variation. Also suppose

= f⁻¹

where these points are counted according to their multiplicities as zeros of the function f − α Then

m\sum n (f \circγ,α) = n(γ,ak). k=1

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

|f (γ (t))- f (γ (s))| =

||\int 1 ' || || f (γ(s)+ λ(γ (t)- γ(s)))(γ (t)- γ (s))dλ|| 0 \leq C |γ(t)- γ(s)|

where C ≥ max

. Hence, in this case,

V (f \circ γ,[a,b]) \leq CV (γ,[a,b]).

Now let ε denote the distance between γ^∗ and ℂ ∖ Ω. Since γ^∗ is compact, ε > 0. By uniform continuity there exists δ =

for p a positive integer such that if

< δ, then

. Then

--(-----ε)- γ([t,t+ δ]) \subseteq B γ(t),2 \subseteq Ω.

Let C ≥ max

where t_j ≡

+ a. Then from what was just shown,

p\sum-1 V (f \circγ,[a,b]) \leq V (f \circ γ,[tj,tj+1]) j=0 p\sum-1 \leq C V (γ,[tj,tj+1]) < \infty j=0

showing that f ∘ γ is bounded variation as claimed. Now from Theorem 50.7.15 there exists η ∈ C¹

such that

η(a) = γ(a) = γ (b) = η(b),η ([a,b]) \subseteq Ω,

and

n (η,ak) = n (γ,ak) ,n (f \circ γ,α ) = n(f \circη,α) (51.6.7)

(51.6.7)

for k = 1,

,m. Then

n(f \circγ,α) = n (f \circη,α )

\int = 1-- -dw-- 2πi f\circηw - α 1--\int b--f'(η-(t))- ' = 2πi a f (η (t))- αη (t)dt 1 \int f'(z) = --- --------dz 2πi ηf (z)- α \summ = n (η,ak) k=1

By Theorem 51.6.1. By 51.6.7, this equals ∑ _k=1^mn

which proves the theorem.

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

Theorem 51.6.3 Let f : B

→ ℂ be analytic and let

f (z) - α = (z - a)m g(z),\infty > m \geq 1

where g

≠0 in B

. (f

− α has a zero of order m at z = a.) Then there exist ε,δ > 0 with the property that for each z satisfying 0 <

< δ, there exist points,

{a1,\cdot\cdot\cdot,am } \subseteq B (a,ε) ,

such that

-1 f (z)\cap B(a,ε) = {a1,\cdot\cdot\cdot,am }

and each a_k is a zero of order 1 for the function f

− z.

Proof: By Theorem 50.5.3 f is not constant on B

because it has a zero of order m. Therefore, using this theorem again, there exists ε > 0 such that B

⊆ B

and there are no solutions to the equation f

− α = 0 for z ∈B

except a. Also assume ε is small enough that for 0 <

≤ 2ε, f^′

≠0. This can be done since otherwise, a would be a limit point of a sequence of points, z_n, having f^′

= 0 which would imply, by Theorem 50.5.3 that f^′ = 0 on B

, contradicting the assumption that f − α has a zero of order m and is therefore not constant. Thus the situation is described by the following picture.