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 51.6.1Let Ω be an open set in ℂ and let γ :
[a,b]
→ Ω be closed,continuous, bounded variation, and n
(γ,z)
= 0 for all z
∈∕
Ω. Suppose also that f isanalytic on Ω having zeros a_{1},
⋅⋅⋅
,a_{m}where the zeros are repeated according tomultiplicity, and suppose that none of these zeros are on γ^{∗}. Then
-1- ∫ f′(z) m∑
2πi γ f (z)dz = n(γ,ak).
k=1
Proof: Let f
(z)
= ∏_{j=1}^{m}
(z − aj)
g
(z)
where g
(z)
≠0 on Ω. Note that some of the
a_{j} could be repeated. Hence
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.
PICT
Theorem 51.6.2Let Ω be a region, let γ :
[a,b]
→ Ω be closed continuous, andbounded variation such that n
(γ,z)
= 0 for all z
∕∈
Ω. Also suppose f : Ω → ℂ is analyticand 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 theirmultiplicities as zeros of the function f − α Then
m∑
n (f ∘γ,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))| =
||∫ 1 ′ ||
|| f (γ(s)+ λ(γ (t)− γ(s)))(γ (t)− γ (s))dλ||
0
≤ 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. By
uniform 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
--(-----ε)-
γ([t,t+ δ]) ⊆ B γ(t),2 ⊆ Ω.
Let C ≥ max
{|f′(z)| : z ∈ ∪p B-(γ-(t-), ε)}
j=1 j 2
where t_{j}≡
-j
p
(b− a)
+ a. Then from what
was just shown,
p∑−1
V (f ∘γ,[a,b]) ≤ V (f ∘ γ,[tj,tj+1])
j=0
p∑−1
≤ C V (γ,[tj,tj+1]) < ∞
j=0
showing that f ∘ γ is bounded variation as claimed. Now from Theorem 50.7.15 there
exists η ∈ C^{1}
([a,b])
such that
η(a) = γ(a) = γ (b) = η(b),η ([a,b]) ⊆ Ω,
and
n (η,ak) = n (γ,ak) ,n (f ∘ γ,α ) = n(f ∘η,α) (51.6.7)
because it has a zero
of order m. Therefore, using this theorem again, there exists ε > 0 such that
B
(a,2ε)
⊆ B
(a,R )
and there are no solutions to the equation f
(z)
− α = 0 for
z ∈B
(a,2ε)
except a. Also assume ε is small enough that for 0 <
|z − a|
≤ 2ε, f^{′}
(z)
≠0.
This can be done since otherwise, a would be a limit point of a sequence of
points, z_{n}, having f^{′}
(zn )
= 0 which would imply, by Theorem 50.5.3 that f^{′} = 0
on B
(a,R )
, 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.