Math 411: Honours Complex Variables - University of Alberta

It follows that {an(z−z0) n } ∞ n=1 does not converge to zero. Consequently, � ∞
n=0 an(z−
z0) n diverges.
Examples.
1. � ∞
n=0 zn : R = 1.
2. � ∞
n=0
zn : R = ∞. n!
3. � ∞
n=0 n!zn : R = 0.
4. �∞ z2n+1
n=0 (−1)n (2n+1)! and �∞ z2n
n=0 (−1)n : R = ∞. (2n)!
Theorem 3.2 (Term-by-Term Differentiation). Let � ∞
n=0 an(z − z0) n be a complex
power series with radius of convergence R. Then
f: BR(z0) → C, z ↦→
∞�
an(z −z0) n
n=0
is complex differentiable at each point z ∈ BR(z0) with
f ′ (z) =
∞�
nan(z −z0) n−1 .
n=1
Proof. Without loss of generality, suppose that z0 = 0.
We first show that � ∞
> limsup
n→∞
19
n=1nanz n−1 converges absolutely for each z ∈ BR(0).
�
n
|an|,
Let z ∈ BR(0), and choose r such that |z|< r < R. Since 1
r
there exists n0 ∈ N such that |an|< � �
1 n
for n ≥ n0 and thus
r
|nanz n−1 |< n
� �n−1 |z|
r r
for n ≥ n0. Since |z|
r < 1, we know from the Ratio Test that �∞ n
n=1 r
the Comparison Test then yields that � ∞
n=1 nanz n−1 converges absolutely.
In view of the foregoing, we may define
g: BR(0) → C, z ↦→
∞�
nanz n−1 .
n=1
� �n−1 |z|
< ∞; r
We shall devote the rest of the proof to showing that f is complex differentiable on
BR(0) with f ′ = g.
To this end, fix ǫ > 0, and define, for z ∈ BR(0) and n ∈ N,
Sn(z) :=
n�
akz k
k=0
and Rn(z) :=
∞�
k=n+1
akz k .