16.1 Open Mapping Theorem for Complex Valued Functions
The open mapping theorem is for an analytic function with values in ℂ. It is even more surprising result
than the theorem about the zeros of an analytic function. The following proof of this important theorem
uses an interesting local representation of the analytic function.
Theorem 16.1.1(Open mapping theorem) Let Ω be a region in ℂ and suppose f : Ω → ℂ isanalytic. Then f
(Ω )
is either a point or a region. In the case where f
(Ω)
is a region, it followsthat for each z_{0}∈ Ω, there exists an open set V containing z_{0} and m ∈ ℕ such that for allz ∈ V,
f (z) = f (z0)+ ϕ(z)m (16.1)
(16.1)
where ϕ : V → B
(0,δ)
is one to one, analytic and onto, ϕ
(z0)
= 0, ϕ^{′}
(z)
≠0 on V and ϕ^{−1}analytic onB
(0,δ)
. If f is one to one then m = 1 for each z_{0}and f^{−1} : f
(Ω)
→ Ω is analytic.
Proof: Suppose f
(Ω )
is not a point. Then if z_{0}∈ Ω it follows there exists r > 0 such that f
(z)
≠f
(z0)
for all z ∈ B
(z0,r)
∖
{z0}
. Otherwise, z_{0} would be a limit point of the set,
It only remains to verify the assertion about the case where f is one to one. If m > 1, then e^{}
2πmi
≠1 and
so for z_{1}∈ V,
2πi
e-m ϕ (z1) ⁄= ϕ(z1). (16.2)
(16.2)
But e^{}
2πi
m
ϕ
(z1)
∈ B
(0,δ)
and so there exists z_{2}≠z_{1}(since ϕ is one to one) such that ϕ
(z2)
= e^{2πi
m
}ϕ
(z1)
. But
then
( )m
ϕ (z2)m = e2πmiϕ(z1) = e2πiϕ(z1)m = ϕ (z1)m
implying f
(z2)
= f
(z1)
contradicting the assumption that f is one to one. Thus m = 1 and
f^{′}
(z)
= ϕ^{′}
(z)
≠0 on V. Since f maps open sets to open sets, it follows that f^{−1} is continuous and so
( )′ f−1(f (z1))− f−1(f (z))
f−1 (f (z)) = f(zl)im→f(z)-----f (z-)−-f (z)---
1 1
= lim ---z1 −-z--= -1′--. ■
z1→z f (z1)− f (z) f (z)
You can dispense with the appeal to the inverse function theorem by using Problem 21 on Page
1303.
One does not have to look very far to find that this sort of thing does not hold for functions mapping ℝ
to ℝ. Take for example, the function f
(x)
= x^{2}. Then f
(ℝ)
is neither a point nor a region. In fact f
(ℝ)
fails to be open.
Corollary 16.1.2Suppose in the situation of Theorem 16.1.1m > 1 for the local representationof f given in this theorem. Then there exists δ > 0 such that if w ∈ B
(f (z ),δ)
0
= f
(V )
for V anopen set containing z_{0}, then f^{−1}
(w)
consists of m distinct points in V. (f is m to one on V )
Proof: Let w ∈ B
(f (z0),δ)
. Then w = f
(^z)
where
^z
∈ V. Thus f
(^z)
= f
(z0)
+ ϕ
(^z)
^{m}. Consider the
m distinct numbers,
{ }
e2kmπiϕ (^z)
_{k=1}^{m}. Then each of these numbers is in B
(0,δ)
and so since ϕ maps V
one to one onto B
(0,δ)
, there are m distinct numbers in V ,
{zk}
_{k=1}^{m} such that ϕ
(zk)
= e^{2kπi
m
}ϕ
(^z)
. Then
( )m
f (zk) = f (z0)+ ϕ (zk)m = f (z0)+ e2kmπiϕ (^z)
m m
= f (z0)+ e2kπiϕ (z^) = f (z0)+ ϕ (^z) = f (^z) = w ■
Example 16.1.3Consider the open connected set D ≡ ℝ + i
(a− π,a + π)
. Then z → e^{z}is one to oneand analytic on D. It maps D onto ℂ ∖ l where l is the ray starting from 0 whose angle is a.Therefore, it has an analytic inverse defined on ℂ ∖ l. This is a branch of the logarithm. It is of theform
log(z) = ln|z|+ iarg (z)
a
where arg _{a}
(z)
is the angle in
(a− π,a+ π)
with the property that
ln|z|+iarga(z)
e = z
We usually let a = 0 and then the inverse is what is usually called the logarithm and is denoted by log . Asin Problem 10this is ln
(|z|)
+ iarg
(z)
where arg
(z)
is the angle between −π and π corresponding toz ∈ ℂ ∖ (−∞,0].
With the open mapping theorem, the maximum modulus theorem is fairly easy.
Theorem 16.1.4Let Ω be an open connected, bounded set in ℂ and let f : Ω → ℂ be analytic. Let∂Ω ≡Ω∖ Ω. Then
{ }
max |f (z)| : z ∈ ¯Ω = max {|f (z)| : z ∈ ∂Ω}
and if the maximum of
|f (z)|
is achieved at a point of Ω, then f is a constant.
Proof:Suppose f
(Ω)
is not a single point. That is, f is not constant. Then by the open mapping
theorem, f
(Ω )
is an open connected subset of ℂ and so z →
|f (z)|
has no maximum. Therefore, the
maximum of
|f (z)|
for z ∈Ω is on ∂Ω. If f
(Ω)
is a single point, then the equation still holds.
■