Math 411: Honours Complex Variables - University of Alberta

48CHAPTER7. ELEMENTARYPROPERTIESOFHOLOMORPHICFUNCTIONS
Proof. Assume otherwise, i.e. f has no zero in Br(z0). The hypothesis implies that f
hasno zero on∂Br(z0), so that f has nozero inBr[z0]. Assume however, that foreach
R > 0 such that Br[z0] ⊂ BR(z0) ⊂ D, there is a zero of f in BR(z0). Then we have
a sequence (Rn) ∞ n=1
in (r,∞) with r = lim
n→∞ Rn such that Br[z0] ⊂ BRn(z0) ⊂ D and
Z(f)∩BRn(z0) �= ∅ for each n ∈ N. For each n ∈ N, pick zn ∈ Z(f)∩BRn(z0). Then
(zn) ∞ n=1 is bounded, and thus has a convergent subsequence (znk )∞k=1 with limit z′ .
Clearly, z ′ ∈ Z(f), and since lim Rnk
k→∞ = r, we have z′ ∈ Br[z0], which is impossible.
Consequently, f has no zero on some BR(z0) with Br[z0] ⊂ BR(z0) ⊂ D.
From the Cauchy Integral Formula, we obtain
1
|f(z0)| =
� � �
�
�
1 1 1 �
�
dζ�
2πi ∂Br(z0) f(ζ) ζ � −z0
≤ 1 1
2πr sup
2π ζ∈∂Br(z0) |f(ζ)|r =
1
infζ∈∂Br(z0)|f(ζ)|
and thus
which is a contradiction.
|f(z0)|≥ inf
ζ∈∂Br(z0) |f(ζ)|,
Theorem 7.2 (Open Mapping Theorem). Let D ⊂ C be open and connected, and let
f: D → C be holomorphic and not constant. Then f(D) ⊂ C is open and connected.
Proof. By the continuity of f, it is clear that f(D) is connected.
Let w0 ∈ f(D), and let z0 ∈ D be such that w0 = f(z0). Choose r > 0 such
that Br[z0] ⊂ D and such that {z ∈ Br[z0] : f(z) = w0} = {z0}. (This can be
accomplished with the help of the Identity Theorem.) Let ǫ = 1 inf |f(z)−w0|> 0.
2
∂Br(z0)
We claim that Bǫ(w0) ⊂ f(D). Let w ∈ Bǫ(w0). For z ∈ ∂Br(z0), we have
It follows that
By Lemma 7.1, this means that
|f(z)−w|≥ |f(z)−w0|−|w−w0|> 2ǫ−ǫ = ǫ.
|f(z0)−w|= |w−w0|< ǫ ≤ inf
z∈∂Br(z0) |f(z)−w|.
D → C, z ↦→ f(z)−w
has a zero in Br(z0). It follows that w ∈ f(D).
Theorem 7.3 (Maximum Modulus Principle). Let D ⊂ C be open and connected,
and let f: D → C be holomorphic such that the function
|f|: D → C, z ↦→ |f(z)|
attains a local maximum on D. Then f is constant.