Complex Analysis
Prof. Pranav Haridas
Kerala School of Mathematics
Lecture No – 3.1
Power Series
In the last week, we defined the notion of complex differentiability. We saw that
complex differentiable functions also satisfy the laws of calculus, namely linearity, the
product rule, quotient rule and also the chain rule. Thereafter, we saw a few examples
of complex differentiable functions and in particular we noted that polynomials in the
variable z are complex differentiable in the entire complex plane; they are entire func-
tions.
This week we begin by discussing the notion of power series. Power series is an in-
finite degree variant of a polynomial inside its disk of convergence. A power series be-
haves very similar to polynomials both analytically and algebraically. Let us start this
lecture by defining a power series around a point z 0 in the complex plane.
(Refer Slide Time: 01:11)
1
2
Let z 0 ∈ C. A formal power series around z 0 with complex coefficients is a formal
expansion,
∞
a n (z − z 0 )n
X
n=0
where a n ∈ C and z is an indeterminate.
We could ask whether a formal power series converges at a given point z ∈ C.
∞
a n (z − z 0 )n converges absolutely.
P
For example, at z 0 , the formal power series
n=0
∞
z n , then for any z with |z| ≥ 1,
P
Another example is,consider the geometric series,
n=0
∞
n
z n does not converge.
P
the summands, |z | > 1 and does not converge to 0. Hence
n=0
(Refer Slide Time: 05:18)
Radius of Convergence
∞
a n (z − z 0 )n be a formal power series around z 0 . We define the radius of con-
P
Let
n=0
vergence R of the formal power series to be the number in [0, ∞] given by
R := lim inf |a n |−1/n
n→0
.
3
The radius of convergence, as the name suggests, is a quantity which tells us some-
thing about the convergence of the given formal power series which is captured in the
next proposition.
(Refer Slide Time: 07:41)
∞
a n (z − z 0 )n be a power series around z 0 . Let R be the radius of
P
P ROPOSITION 1. Let
n=0
∞
a n (z − z 0 )n converges
P
convergence of the power series. Then for z ∈ D(z 0 , R), the series
n=0
absolutely. For r < R, the series converges uniformly on D(z 0 , r ). Furthermore, if |z − z 0 | >
∞
a n (z − z 0 )n diverges.
P
R, then
n=0
(The disc D(z 0 , R) is called the disc of convergence of the formal power series).
P ROOF. Let z ∈ C such that |z − z 0 | > R. Then ∃ infinitely many n ∈ N such that
|a n |−1/n < |z − z 0 | =⇒ |a n |−1/n |z − z 0 | > 1 =⇒ |a n (z − z 0 )n | > 1
for infinitely many n ∈ N → (∗). Since the summands does not converge to 0(by (∗)), we
∞
a n (z − z 0 )n does not converge.
P
can conclude that
n=0
Let z ∈ D(z 0 , R) =⇒ |z − z 0 | < r < R for some r > 0. Since R = lim inf |a n |−1/n , ∃ N ∈ N
n→∞
−1/n n
such that ∀n > N , |a n | > r =⇒ |a n |r < 1∀n > N .
4
(Refer Slide Time: 17:19)
For n > N ,
|z − z 0 |n
|a n (z − z 0 )n | = |a n r n |
rn
¶n
|z − z 0 |
µ
≤
r
|z − z 0 | n
µ ¶
n
X X
a n (z − z 0 ) ≤ .
n>N n>N r
∞ ∞
|a n (z − z 0 )n | converges. That is a n (z − z 0 )n converges absolutely.
P P
Hence
n=0 n=0
Let r < R and R 1 be such that r < R 1 < R. ∃N ∈ N such that ∀n > N , |a n |−1/n > R 1 > r . For
n > 1 and z ∈ D(z 0 , r ),
n ¶n
|z − z 0 | r
µ
n
|a n (z − z 0 ) | = |a n R 1n | ≤
R 1n R1
∞ r n X r n
µ ¶ µ ¶
is a convergent series, given ε > 0, ∃ N0 > N such that < ε.
X
. Since
n=0 R 1 n≥N0 R 1
(Refer Slide Time: 23:24)
5
Then,
∞ N0
a n (z − z 0 )n − a n (z − z 0 )n | = | a n (z − z 0 )n |
X X X
|
n=0 n=0 n>N0
¶n
r
µ
X
<
n≥N0 R1
< ε.
(Refer Slide Time: 24:26)
6
∞
a n (z − z 0 )n be a formal power series with radius of conver-
P
P ROPOSITION 2. Let
n=0
gence R. Assume that a n is non-zero for n sufficiently large. Then
|a n | |a n |
lim inf ≤ R ≤ lim sup .
n→∞ |a n+1 | n→∞ |a n+1 |
|a n |
P ROOF. Let R 1 = lim inf . Let r < R 1 . Then ∃N ∈ N such that ∀n > N ,
n→∞ |a n+1 |
|a n |
> r =⇒ |a n+1 |r < |a n |.
|a n+1 |
(Refer Slide Time: 27:45)
7
For z ∈ D(z 0 , r ) and n > N ,
|z − z 0 | n+1
µ ¶
n+1
|a n+1 (z − z 0 ) | = |a n+1 |
r
¶n+1
n |z − z 0 |
µ
< |a n r |
r
..
.
¶n+1
N ¯ |z − z 0 |
µ
¯ ¯
< aN r
¯ .
r
¯ X |z − z 0 | n
µ ¶
n N¯
P ¯
Hence |a n (z − z 0 ) | ≤ a N r
¯ .
n≥N n≥N r
∞
a n (z − z 0 )n converges =⇒ D(z 0 , r ) ⊆ {z : |z − z 0 | ≤ R} =⇒ r ≤ R. Since this is
P
Hence
n=0
true for all r < R 1 , we have R 1 ≤ R → (∗).
|a n |
Let R 2 = lim sup . Let r > R 2 and let z ∈ C be such that |z − z 0 | > r . Since r >
n→∞ |a n+1 |
R 2 , ∃N ∈ N such that ∀n > N ,
|a n |
< r =⇒ |a n+1 r | > |a n |.
|a n+1 |
8
(Refer Slide Time: 35:38)
Then for n > N ,
|a n+1 (z − z 0 )n+1 | > |a n+1 r n+1 | > |a n r n | · · · > |a N r N | = M (say)
∞
a n (z − z 0 )n does not converge.
P
Since the summands do not converge to 0,
n=0
That is, {z : |z − z 0 | > r } ⊆ {z : |z − z 0 | ≥ R} =⇒ D(z 0 , R) ⊆ {z : |z − z 0 | ≤ r } =⇒ R ≤ r .
Since this is true for all r > R 2 =⇒ R ≤ R 2 → (∗∗).
|a n | |a n |
By (∗) and (∗∗), lim inf ≤ R ≤ lim sup .
n→∞ |a n+1 | n→∞ |a n+1 |
E XAMPLE 3.
∞ zn
• ez =
X
n=0 n!
1 |a n | (n + 1)!
Here a n = . Now, = = (n + 1). Hence the radius of conver-
n! |a n+1 | n!
gence is infinite.
(Refer Slide Time: 38:12)
9
∞
n!z n .
X
•
n=0
|a n | (n! 1
Then = = → 0 as n → ∞. Hence the series converges
|a n+1 | (n + 1)! (n + 1)
only at the origin.
∞
z n has radius of convergence 1. For |z| = 1, the series diverges. For z ∈
P
•
n=0
D(0, 1), we have
∞ ∞
n
z n+1
X X
z z =
n=0 n=0
∞
zn
X
=
n=1
∞
z n − 1.
X
=
n=0
∞ 1
zn =
P
On D(0, 1), we have .
n=0 1−z
(Refer Slide Time: 41:10)
10
1 1
On C \ {1}, the function is holomorphic. Now note that is a func-
1−z 1−z
∞
zn.
X
tion which extends beyond D(0, 1) which is the disc of convergence of
n=0
X∞ zn
• 2
has a radius of convergence 1.
n=0 n
∞ zn
For z ∈ C such that |z| = 1,
X
2
converges.
n=0 n
∞
a n (z − z 0 )n be a power series with a positive radius
P
Abel’s Theorem: Let F (z) =
n=0
of convergence R and suppose z 1 = z 0 + Re i θ be a point such that F (z 1 ) converges. Then
lim F (z 0 + r e i θ ) = F (z 1 ).
r →R −
(Refer Slide Time: 45:18)
11
P ROOF. We may assume that F (z 1 ) = 0.
∞
b n (z − z 0 )n where b n = a n ∀n > 0, b 0 = a 0 − F (z 1 ). If
P
Define G(z) = F (z) − F (z 1 ) =
n=0
lim− G(z 0 + r e i θ ) = 0, then lim− (F (z 0 + r e i θ ) − F (z 1 )) = 0. Hence we shall assume that
r →R r →R
F (z 1 ) = 0.
We can also assume z 0 = 0.
(Refer Slide Time: 48:34)
12
∞
a n z n . If G(r e i θ ) = 0 and lim− G(r e i θ ) = 0, then reader should verify
P
Define G(z) =
n=0 r →R
that lim− G(r e i θ ) = lim− F (z 0 + r e i θ ) = 0. Hence we shall assume that z 0 = 0.
r →R r →R
Thus if we prove for the power series around 0, we can translate it can conclude that it is
true for all power series around any arbitrary point.
(Refer Slide Time: 50:40)
We may also assume that R = 1 and θ = 0.
Let G(z) = a n R n e −i nθ z n = F (Re −i nθ z).
X
1
The radius of convergence of G = lim inf |a n |−1/n R −n/n = lim inf |a n |−1/n = 1. If G(1) = 0
n→∞ R n→∞
and lim− G(r ) = 0, then lim− F (r e i θ ) = 0.
r →1 r →R
Hence to prove the theorem, it is enough to prove the following:
Let a n be a series converging to 0. Then lim− a n r n = 0.
P P
r →1
(Refer Slide Time: 54:42)
13
Let A n = a 0 + a 1 + · · · + a n . Given ε > 0, ∃ N such that ∀n > N , |A n | < 2ε .
m
a n r n = A 0 + (A 1 − A 0 )r + · · · + (A m − A m−1 )r m
X
n=0
m−1
A n (r n − r n+1 ) + A m r m
X
=
n=0
m−1
An r n + Am r m .
X
= (1 − r )
n=0
∞ m−1
a n r n = (1 − r ) A n r n → (∗).
P P
Hence
n=0 n=0
For r < 1 and given ε > 0 as above, R.H.S of (∗) becomes,
¯ ¯ ¯ ¯
¯ N ∞ ¯ ¯X N ¯ ∞
A n r n + (1 − r ) A n r n ¯ ≤ (1 − r ) ¯ A n r n ¯ + (1 − r ) |A n |r n
X X X
¯(1 − r )
¯ ¯ ¯ ¯
n=0 n=N n=0 n=N
¯ ¯ ¯ ¯
¯ ¯
¯X N ¯ ε rN
< (1 − r ) ¯ A n r n ¯ + (1 − r )
¯ ¯
¯n=0 ¯ 2 (1 − r )
¯ ¯
¯X N ¯ ε
< (1 − r ) ¯ An r n ¯ + .
¯ ¯
¯n=0 ¯ 2
(Refer Slide Time: 01:00:55)
14
N
A n r n is a continuous on [0, 1]. Hence ∃M such that
X
Then the function
¯ ¯ n=0
N
A n r n ¯¯ < M ∀ r ∈ [0, 1].
¯P ¯
¯
n=0
¯
ε ε ε
¯ ¯
∞
n¯
¯P ¯
Let r be such that (1 − r ) < . Then (1 − r ) ¯
¯ An r ¯ < M= .
2Mµ n=0
¶ 2M 2
∞
Hence lim− a n r n = lim− (1 − r ) A n r n = 0.
P P
r →1 r →1 n=0
