6Complex power series
IA Analysis I
6 Complex power series
Before we move on to integration, we first have a look at complex power series.
This will allow us to define the familiar exponential and trigonometric functions.
Definition (Complex power series). A complex power series is a series of the
form
∞
X
n=0
a
n
z
n
.
when z ∈ C and a
n
∈ C for all n. When it converges, it is a function of z.
When considering complex power series, a very important concept is the
radius of convergence. To make sense of this concept, we first need the following
lemma:
Lemma. Suppose that
P
a
n
z
n
converges and
|w| < |z|
, then
P
a
n
w
n
converges
(absolutely).
Proof. We know that
|a
n
w
n
| = |a
n
z
n
| ·
w
z
n
.
Since
P
a
n
z
n
converges, the terms a
n
z
n
are bounded. So pick C such that
|a
n
z
n
| ≤ C
for every n. Then
0 ≤
∞
X
n=0
|a
n
w
n
| ≤
∞
X
n=0
C
w
z
n
,
which converges (geometric series). So by the comparison test,
P
a
n
w
n
converges
absolutely.
It follows that if
P
a
n
z
n
does not converge and
|w| > |z|
, then
P
a
n
w
n
does
not converge.
Now let
R
=
sup{|z|
:
P
a
n
z
n
converges
}
(
R
may be infinite). If
|z| < R
,
then we can find
z
0
with
|z
0
| ∈
(
|z|, R
] such that
P
∞
n
a
n
z
n
0
converges. So by
lemma above,
P
a
n
z
n
converges. If
|z| > R
, then
P
a
n
z
n
diverges by definition
of R.
Definition (Radius of convergence). The radius of convergence of a power series
P
a
n
z
n
is
R = sup
n
|z| :
X
a
n
z
n
converges
o
.
{z : |z| < R} is called the circle of convergence.
a
.
If
|z| < R
, then
P
a
n
z
n
converges. If
|z| > R
, then
P
a
n
z
n
diverges. When
|z| = R, the series can converge at some points and not the others.
Example.
∞
X
n=0
z
n
has radius of convergence of 1. When
|z|
= 1, it diverges
(since the terms do not tend to 0).
Example.
∞
X
n=0
z
n
n
has radius of convergence 1, since the ratio of (
n
+ 1)th term
to nth is
z
n+1
/(n + 1)
z
n
/n
= z ·
n
n + 1
→ z.
So if
|z| <
1, then the series converges by the ratio test. If
|z| >
1, then eventually
the terms are increasing in modulus.
If
z
= 1, then it diverges (harmonic series). If
|z|
= 1 and
z 6
= 1, it converges
by Abel’s test.
Example. The series
∞
X
n=1
z
n
n
2
converges for |z| ≤ 1 and diverges for |z| > 1.
As evidenced by the above examples, the ratio test can be used to find the
radius of convergence. We also have an alternative test based on the nth root.
Lemma. The radius of convergence of a power series
P
a
n
z
n
is
R =
1
lim sup
n
p
|a
n
|
.
Often
n
p
|a
n
| converges, so we only have to find the limit.
Proof.
Suppose
|z| <
1
/ lim sup
n
p
|a
n
|
. Then
|z|lim sup
n
p
|a
n
| <
1. Therefore
there exists N and ε > 0 such that
sup
n≥N
|z|
n
p
|a
n
| ≤ 1 − ε
by the definition of lim sup. Therefore
|a
n
z
n
| ≤ (1 − ε)
n
for every
n ≥ N
, which implies (by comparison with geometric series) that
P
a
n
z
n
converges absolutely.
On the other hand, if
|z|lim sup
n
p
|a
n
| >
1, it follows that
|z|
n
p
|a
n
| ≥
1 for
infinitely many
n
. Therefore
|a
n
z
n
| ≥
1 for infinitely many
n
. So
P
a
n
z
n
does
not converge.
Example. The radius of convergence of
z
n
2
n
is 2 because
n
p
|a
n
|
=
1
2
for every
n
.
So lim sup
n
p
|a
n
| =
1
2
. So 1/ lim sup
n
p
|a
n
| = 2.
But often it is easier to find the radius convergence from elementary methods
such as the ratio test, e.g. for
P
n
2
z
n
.
a
Note to pedants: yes it is a disc, not a circle