Complex Analysis
Chapter III. Elementary Properties and Examples of Analytic
Functions
III.1. Power Series—Proofs of Theorems
November 12, 2017
() Complex Analysis November 12, 2017 1/9
Table of contents
1 Proposition III.1.1
2 Theorem III.1.3
3 Proposition III.1.4
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Proposition III.1.1
Proposition III.1.1
P
Proposition III.1.1. If an converges absolutely, then the series
converges.
Proof. Let ε > 0. Let zn be thePpartial sum of ∞
P
k=1 ak :
∞
zn = a1 + a2 + · · · + an . Since n=1 |an | converges by hypothesis, then
there is N ∈ N such that
∞ ∞
N−1
X X X
|ak | − |an | = |an | < ε.
k=1 n=1 n=N
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Proposition III.1.1
Proposition III.1.1
P
Proposition III.1.1. If an converges absolutely, then the series
converges.
Proof. Let ε > 0. Let zn be thePpartial sum of ∞
P
k=1 ak :
∞
zn = a1 + a2 + · · · + an . Since n=1 |an | converges by hypothesis, then
there is N ∈ N such that
∞ ∞
N−1
X X X
|ak | − |an | = |an | < ε.
k=1 n=1 n=N
So if m > k ≥ N then by the Triangle Inequality,
∞
m
X Xm X
|zm − zk | = an ≤ |an | ≤ |an | < ε,
n=k+1 n=k+1 n=N
and so {zn } is a Cauchy sequence of complex numbers.
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Proposition III.1.1
Proposition III.1.1
P
Proposition III.1.1. If an converges absolutely, then the series
converges.
Proof. Let ε > 0. Let zn be thePpartial sum of ∞
P
k=1 ak :
∞
zn = a1 + a2 + · · · + an . Since n=1 |an | converges by hypothesis, then
there is N ∈ N such that
∞ ∞
N−1
X X X
|ak | − |an | = |an | < ε.
k=1 n=1 n=N
So if m > k ≥ N then by the Triangle Inequality,
∞
m
X Xm X
|zm − zk | = an ≤ |an | ≤ |an | < ε,
n=k+1 n=k+1 n=N
and so {zn } is a Cauchy sequence of complex numbers. Since C is
P∞ II.3.6, then zn → z for some z ∈ C. That is,
complete by Proposition
there is z ∈ C with n=1 an = z.
() Complex Analysis November 12, 2017 3/9
Proposition III.1.1
Proposition III.1.1
P
Proposition III.1.1. If an converges absolutely, then the series
converges.
Proof. Let ε > 0. Let zn be thePpartial sum of ∞
P
k=1 ak :
∞
zn = a1 + a2 + · · · + an . Since n=1 |an | converges by hypothesis, then
there is N ∈ N such that
∞ ∞
N−1
X X X
|ak | − |an | = |an | < ε.
k=1 n=1 n=N
So if m > k ≥ N then by the Triangle Inequality,
∞
m
X Xm X
|zm − zk | = an ≤ |an | ≤ |an | < ε,
n=k+1 n=k+1 n=N
and so {zn } is a Cauchy sequence of complex numbers. Since C is
P∞ II.3.6, then zn → z for some z ∈ C. That is,
complete by Proposition
there is z ∈ C with n=1 an = z.
() Complex Analysis November 12, 2017 3/9
Theorem III.1.3
Theorem III.1.3
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r . Moreover, R is the only number having
properties (a) and (b). R is called the radius of convergence
of the power series.
Proof. Without loss of generality, a = 0.
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Theorem III.1.3
Theorem III.1.3
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r . Moreover, R is the only number having
properties (a) and (b). R is called the radius of convergence
of the power series.
Proof. Without loss of generality, a = 0.
(a) If |z| < R, there is r with |z| < r < R. Then 1/R < 1/r and so there
exists N ∈ N such that for all n ≥ N, |an |1/n < 1/r (by definition of
lim|an |1/n ).
() Complex Analysis November 12, 2017 4/9
Theorem III.1.3
Theorem III.1.3
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r . Moreover, R is the only number having
properties (a) and (b). R is called the radius of convergence
of the power series.
Proof. Without loss of generality, a = 0.
(a) If |z| < R, there is r with |z| < r < R. Then 1/R < 1/r and so there
exists N ∈ N such that for all n ≥ N, |an |1/n < 1/r (by definition of
lim|an |1/n ). So for n ≥ N, |an | < 1/r n and |an z n | < (|z|/r )n . Next,
∞ N−1 ∞ N−1 ∞
|z| n
X X X X X
n n n n
|an z | = |an z | + |an z | < |an z | + .
r
n=0 n=0 n=N n=0 n=N
() Complex Analysis November 12, 2017 4/9
Theorem III.1.3
Theorem III.1.3
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r . Moreover, R is the only number having
properties (a) and (b). R is called the radius of convergence
of the power series.
Proof. Without loss of generality, a = 0.
(a) If |z| < R, there is r with |z| < r < R. Then 1/R < 1/r and so there
exists N ∈ N such that for all n ≥ N, |an |1/n < 1/r (by definition of
lim|an |1/n ). So for n ≥ N, |an | < 1/r n and |an z n | < (|z|/r )n . Next,
∞ N−1 ∞ N−1 ∞
|z| n
X X X X X
n n n n
|an z | = |an z | + |an z | < |an z | + .
r
n=0 n=0 n=N n=0 n=N
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Theorem III.1.3
Theorem III.1.3 (continued 1)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(a) if |z − a| < R, the series converges absolutely,
(b) if |z − a| > R, the series diverges, and
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r . Moreover, R is the only number having
properties (a) and (b). R is called the radius of convergence
of the power series.
Proof (continued). Next,
∞ N−1 ∞ N−1 ∞
|z| n
X X X X X
n n n n
|an z | = |an z | + |an z | < |an z | + .
r
n=0 n=0 n=N n=0 n=N
P∞ n
Since |z|/r < 1, then n=N an z converges absolutely, and the power
series converges absolutely for |z| < R.
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Theorem III.1.3
Theorem III.1.3 (continued 2)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r .
Proof (continued). (c) Suppose r < R and choose ρ such that
r < ρ < R. As in the proof of (a), let N ∈ N be such that |an | < 1/ρn for
all n ≥ N. Then if |z| ≤ r , we have |an z n | < (r /ρ)n for all n ≥ N, and
(r /ρ) < 1.
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Theorem III.1.3
Theorem III.1.3 (continued 2)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r .
Proof (continued). (c) Suppose r < R and choose ρ such that
r < ρ < R. As in the proof of (a), let N ∈ N be such that |an | < 1/ρn for
all n ≥ N. Then if |z| ≤ r , we have |an z n | < (r /ρ)n for all n ≥ N, and
(r /ρ) < 1. Now, the Weierstrass M-Test says (Theorem II.6.2): Let
un : X → C be a function from a metric P space X to C such thatP
|un (x)| ≤ Mn for all x ∈ X and suppose ∞ n=1 Mn < ∞. Then
∞
n=1 un is
uniformly convergent.
() Complex Analysis November 12, 2017 6/9
Theorem III.1.3
Theorem III.1.3 (continued 2)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r .
Proof (continued). (c) Suppose r < R and choose ρ such that
r < ρ < R. As in the proof of (a), let N ∈ N be such that |an | < 1/ρn for
all n ≥ N. Then if |z| ≤ r , we have |an z n | < (r /ρ)n for all n ≥ N, and
(r /ρ) < 1. Now, the Weierstrass M-Test says (Theorem II.6.2): Let
un : X → C be a function from a metric P space X to C such thatP
|un (x)| ≤ Mn for all x ∈ X and suppose ∞ n=1 Mn < ∞. Then
∞
n=1 un is
uniformly convergent. So with M = (r /ρ) n , we see that the series
P∞ P∞ n
u = a z n converges uniformly on {z | |z| ≤ r } (and so does
P∞ n=N n n=N n
n
n=0 an z ), by the Weierstrass M-Test.
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Theorem III.1.3
Theorem III.1.3 (continued 2)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(c) if 0 < r < R then the series converges uniformly on
|z − a| ≤ r .
Proof (continued). (c) Suppose r < R and choose ρ such that
r < ρ < R. As in the proof of (a), let N ∈ N be such that |an | < 1/ρn for
all n ≥ N. Then if |z| ≤ r , we have |an z n | < (r /ρ)n for all n ≥ N, and
(r /ρ) < 1. Now, the Weierstrass M-Test says (Theorem II.6.2): Let
un : X → C be a function from a metric P space X to C such thatP
|un (x)| ≤ Mn for all x ∈ X and suppose ∞ n=1 Mn < ∞. Then
∞
n=1 un is
uniformly convergent. So with M = (r /ρ) n , we see that the series
P∞ P∞ n
u = a z n converges uniformly on {z | |z| ≤ r } (and so does
P∞ n=N n n=N n
n
n=0 an z ), by the Weierstrass M-Test.
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Theorem III.1.3
Theorem III.1.3 (continued 3)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(b) if |z − a| > R, the series diverges, and
Proof (continued). (b) Let |z| > R and choose r with |z| > r > R.
Then 1/r < 1/R. So, by the definition of lim, there are infinitely many
n ∈ N such that 1/r < |an |1/n . For these n, |an z n | > (|z|/r )n and since
|z|/r > 1, these terms are unbounded and hence the series diverges for
such z.
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Theorem III.1.3
Theorem III.1.3 (continued 3)
∞
X
Theorem III.1.3. If an (z − a)n , define the number R as
n=0
1
R = lim |an |1/n (so 0 ≤ R ≤ ∞). Then
(b) if |z − a| > R, the series diverges, and
Proof (continued). (b) Let |z| > R and choose r with |z| > r > R.
Then 1/r < 1/R. So, by the definition of lim, there are infinitely many
n ∈ N such that 1/r < |an |1/n . For these n, |an z n | > (|z|/r )n and since
|z|/r > 1, these terms are unbounded and hence the series diverges for
such z.
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Proposition III.1.4
Proposition III.1.4
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof. Without loss of generality, a = 0. Let α = lim |an /an+1 | and
suppose |z| < r < α. Then (by the definition of limit of a sequence) there
exists N ∈ N such that for all n ≥ N, we have |an /an+1 | > r .
() Complex Analysis November 12, 2017 8/9
Proposition III.1.4
Proposition III.1.4
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof. Without loss of generality, a = 0. Let α = lim |an /an+1 | and
suppose |z| < r < α. Then (by the definition of limit of a sequence) there
exists N ∈ N such that for all n ≥ N, we have |an /an+1 | > r . Let
B = |aN |r N and then |aN+1 |r N+1 = |aN+1 |rr N < |aN |r N = B (since
|aN | > |aN+1 |r ), |aN+2 |rr N+1 < |aN+1 |r N+1 < B, . . . , and |an r n | ≤ B for
all n ≥ N. This implies |an z n | = |an r n ||z|n /r n ≤ B|z|n /r n for all n ≥ N.
() Complex Analysis November 12, 2017 8/9
Proposition III.1.4
Proposition III.1.4
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof. Without loss of generality, a = 0. Let α = lim |an /an+1 | and
suppose |z| < r < α. Then (by the definition of limit of a sequence) there
exists N ∈ N such that for all n ≥ N, we have |an /an+1 | > r . Let
B = |aN |r N and then |aN+1 |r N+1 = |aN+1 |rr N < |aN |r N = B (since
|aN | > |aN+1 |r ), |aN+2 |rr N+1 < |aN+1 |r N+1 < B, . . . , and |an r n | ≤ B for
all n ≥PN. This implies |an z n | = |an r n ||z|n /r n ≤ B|z|n /r n for all n ≥ N.
Since ∞ n n
n=1 B|z| /r is a convergent geometric
P∞ series for |z| < r , then by
the Direct Comparison Test, the series n=1 |an z n | converges and the
original series converges absolutely. Since r < α is arbitrary, then α ≤ R.
() Complex Analysis November 12, 2017 8/9
Proposition III.1.4
Proposition III.1.4
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof. Without loss of generality, a = 0. Let α = lim |an /an+1 | and
suppose |z| < r < α. Then (by the definition of limit of a sequence) there
exists N ∈ N such that for all n ≥ N, we have |an /an+1 | > r . Let
B = |aN |r N and then |aN+1 |r N+1 = |aN+1 |rr N < |aN |r N = B (since
|aN | > |aN+1 |r ), |aN+2 |rr N+1 < |aN+1 |r N+1 < B, . . . , and |an r n | ≤ B for
all n ≥PN. This implies |an z n | = |an r n ||z|n /r n ≤ B|z|n /r n for all n ≥ N.
Since ∞ n n
n=1 B|z| /r is a convergent geometric
P∞ series for |z| < r , then by
the Direct Comparison Test, the series n=1 |an z n | converges and the
original series converges absolutely. Since r < α is arbitrary, then α ≤ R.
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Proposition III.1.4
Proposition III.1.4 (continued)
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof (continued). Next, suppose |z| > r > α. Then, as above, for some
N ∈ N, for all n ≥ N we have |an | < r |an+1 |. Again, with B = |aN r N |, for
n ≥ N we get |an r n | ≥ B = |aN r N | and |an z n | ≥ B|z|n /r n which diverges
to ∞ as n → ∞ since |z| > r . SoPan z n 6→ 0 and by the Test for
Divergence (for complex series), ∞ n
n=0 an z diverges. Since r > α is
arbitrary, then R ≤ α.
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Proposition III.1.4
Proposition III.1.4 (continued)
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof (continued). Next, suppose |z| > r > α. Then, as above, for some
N ∈ N, for all n ≥ N we have |an | < r |an+1 |. Again, with B = |aN r N |, for
n ≥ N we get |an r n | ≥ B = |aN r N | and |an z n | ≥ B|z|n /r n which diverges
to ∞ as n → ∞ since |z| > r . SoPan z n 6→ 0 and by the Test for
Divergence (for complex series), ∞ n
n=0 an z diverges. Since r > α is
arbitrary, then R ≤ α. Therefore R = α.
() Complex Analysis November 12, 2017 9/9
Proposition III.1.4
Proposition III.1.4 (continued)
∞
X
Proposition III.1.4. If an (z − a)n is a given power series with radius
n=0
of convergence R, then R = lim |an /an+1 |, if the limit exists.
Proof (continued). Next, suppose |z| > r > α. Then, as above, for some
N ∈ N, for all n ≥ N we have |an | < r |an+1 |. Again, with B = |aN r N |, for
n ≥ N we get |an r n | ≥ B = |aN r N | and |an z n | ≥ B|z|n /r n which diverges
to ∞ as n → ∞ since |z| > r . SoPan z n 6→ 0 and by the Test for
Divergence (for complex series), ∞ n
n=0 an z diverges. Since r > α is
arbitrary, then R ≤ α. Therefore R = α.
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