Math 411: Honours Complex Variables - University of Alberta

holds for some a0,a1,a2,... ∈ C and all z ∈ Br(z0). As z0 is a cluster point of Z(f),
there exists a sequence (zk) ∞ k=1 in Z(f)\{z0} such that z0 = lim zk. Inductively, we
k→∞
find that an = 0 for all n ∈ N0: given that an = 0 for n = 0,1,2,m−1 we see that
f(zk)
0 = lim
k→∞ (zk −z0) m
∞�
= lim an(zk −z0) n−m = am,
k→∞
n=m
on noting, in view of Theorems 3.1 and 6.3, that the uniform convergence of the
power series on, say, Br/2[z0] justifies the interchange of limits. Hence f(z) = 0 for
all z ∈ Br(z0). That is, Br(z0) ⊂ Z(f) so that Br(z0) ⊂ V.
We now claim that D \V is also open. Let z0 ∈ D \V, and ǫ > 0 be such that
Bǫ(z0) ⊂ D and (Bǫ(z0)\{z0})∩Z(f) = ∅. For any z ∈ Bǫ(z0)\{z0} and δ > 0 such
that Bδ(z) ⊂ Bǫ(z0)\{z0}, we thus have (Bδ(z) \{z}) ∩Z(f) = ∅. It follows that
z /∈ V. Consequently, Bǫ(z0) ⊂ D\V.
In summary, V and D\V are both open and clearly satisfy D = V ∪(D\V) and
V ∩(D \V) = ∅. The connectedness of D yields D = V and thus Z(f) = D.
Examples.
1. There is no non-zero entire function f: C → C such that f � �
1 = 0 for n ∈ N.
n
For any entire function f with this property, 0 is a cluster point of Z(f). By
the Identity Theorem, this means f ≡ 0.
2. Since R has cluster points in C, the holomorphic extensions of exp, cos, and sin
from R to C are unique. For analogous reasons, Log is the only holomorphic
extension of log to C−.
3. There is no entire function f such that
� �
1
f =
n
1
n
and
�
f − 1
�
=
n
1
n2 for n ∈ N. In the view of the identity theorem, the first condition necessitates
that f(z) = z whereas the second one implies that f(z) = (−z) 2 for all z ∈ C.
Lemma 7.1. Let D ⊂ C be open, let f: D → C be holomorphic, and let z0 ∈ D and
r > 0 be such that Br[z0] ⊂ D. Suppose that
Then f has a zero in Br(z0).
|f(z0)|< inf |f(z)|.
z∈∂Br(z0)
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