Math 411: Honours Complex Variables - University of Alberta

20 CHAPTER 3. POWER SERIES
Fix z ∈ BR(0) and let r ∈ (0,R) be such that z ∈ Br(0). Note that
�
f(w)−f(z) Sn(w)−Sn(z)
−g(z) = −S
w −z w−z
′ �
n(z) +(S ′ n(z)−g(z))+ Rn(w)−Rn(z)
w−z
for all w ∈ BR(0)\{z}. We shall see that each of the three summands on the righthand
side of this equation has modulus less than ǫ
, provided that n is sufficiently
3
large and w is sufficiently close to z.
We start with the last summand. First, note that
Rn(w)−Rn(z)
w −z
for all w ∈ BR(0)\{z} and also that
� �
� �
� �
� � =
�
� k� �
�
�
w k −z k
w −z
j=1
=
w k−j z j−1
∞�
k=n+1
�
�
�
�
� ≤
k�
j=1
w
ak
k −zk w −z
|w| k−j |z| j−1 ≤ kr k−1
for all w ∈ Br(0)\{z}. Since r < R, we have �∞ k=1k|ak|r k−1 < ∞. Consequently,
there exists n1 ∈ N such that �∞ k=n+1k|ak|r k−1 < ǫ
3 for all n ≥ n1 and therefore
� �
�
�
Rn(w)−Rn(z) �
�
� w−z � =
�
�
�
�
�
∞�
k=n+1
ak
w k −z k
w −z
�
�
�
�
� ≤
∞�
k=n+1
k|ak|r k−1 < ǫ
3
for all n ≥ n1 and all w ∈ Br(0)\{z}.
For the second summand, just note that lim S
n→∞ ′ n(z) = g(z); consequently, there
exists n2 ∈ N such that |S ′ ǫ
n (z)−g(z)|< for all n ≥ n2.
3
For the first summand, fix n ≥ max{n1,n2}. Since
Sn(w)−Sn(z)
lim
w→z w −z
there exists δ ∈ (0,r) such that
�
�
�
Sn(w)−Sn(z)
� w−z
−S ′ n (z)
= S ′ n (z),
for all w ∈ Bδ(z) ⊂ Br(0)\{z}. Consequently, we obtain for all w ∈ Bδ(z)\{z} that
� �
�
�
f(w)−f(z) �
� −g(z) �
w−z � ≤
�
�
�
Sn(w)−Sn(z)
� −S
w −z
′ n (z)
�
�
�
� +|S′ n (z)−g(z)|+
� �
�
�
Rn(w)−Rn(z) �
�
ǫ
� w−z � <
3 +ǫ
3 +ǫ = ǫ.
3
�
�
�
�
< ǫ
3
Since ǫ > 0 was arbitrary, we see that f ′ (z) exists and equals g(z).