2.3 Lecture 6: 2.1-5 Complex sequences and (power)...

126 CHAPTER 2. CHAPTER II

so

R(z) = 1

/

R

(

1

z

)

=|z|2m

|z|2n

∣

∣

∣

∣

∣

1

/

R

(

1

z

)

∣

∣

∣

∣

∣

2an

∏kj=1 (z − αj |z|2)nj

bn

∏ℓj=1 (z − βj|z|2)mj

.

It follows then that R(z) = R(z) when |z| = 1. Thus, these two rationalfunctions must be identical. This means there is a correspondence betweenroots αj and poles βj = 1/αj given by reflection about the unit circle. Theconjecture is established.

2.3 Lecture 6: § 2.1-5 Complex sequences

and (power) series

2.3.1 Complex sequences

A complex sequence is denoted {αj}∞j=1 or α1, α2, α3, . . .. The sequence{αj}∞j=1 ⊂ C converges to L ∈ C if for any ǫ > 0, there is some N > 0 suchthat

j > N implies |αj − L| < ǫ.

In this case, we write αj → L as j ր ∞ or

limjր∞

αj = L.

We say αn → ∞ as j ր ∞ if for any M > 0, there is some N such that

j > N implies |αj| > M.

Definition 6 {αj}∞j=1 is Cauchy if for any ǫ > 0, there is some N so that

j, k > N implies |αj − αk| < ǫ.

General Fact: If αj → L in any metric space, then {αj}∞j=1 is Cauchy.Proof: For any ǫ > 0, there is some N such that

j, k > N implies d(αj, L), d(αk, L) < ǫ/2

implies d(αj, αk) ≤ d(αj , L) + d(αk, L) < ǫ. 2

2.3. LECTURE 6: § 2.1-5 COMPLEX SEQUENCES AND (POWER) SERIES 127

Remember that a metric space is called complete if every Cauchy se-quence in the space converges in the space. Also remember that C and R arecomplete. Now, let us prove these facts.Proof that C is complete (using the fact that R is complete):

Say {zj}∞j=1 is Cauchy. For any ǫ > 0, there is some N such that

j, k > N implies |zj − zk| < ǫ.

Let Re(zj) = aj and Im(zj) = bj . Then

|aj − ak| ≤ |zj − zk| and |bj − bk| ≤ |zj − zk|,

so {aj}∞j=1 ⊂ R and {bj}∞j=1 ⊂ R are Cauchy. Therefore, since R is complete,there are limits aj → A and bj → B. Also,

|zj − (A + iB)| = |aj − A + i(bj − B)| ≤ |aj − A| + |bj − B|.

Therefore, zj → L = A + iB ∈ C. 2

Proof that R is complete (using the least upper bound property):Say {aj}∞j=1 is Cauchy.Consider the set

Ak = {aj : j ≥ k} = {ak, ak+1, ak+2, . . .}.

There are two possibilities: Ak is bounded above or not. In the first case,there is a well-defined least upper bound for Ak; that is the least upper boundproperty. In this case, let us denote the least upper bound by l.u.b. Ak.Furthermore, it will be noted that if Ak is bounded above for one k, then thesets Ak are all bounded above by l.u.b. A1. There is another possibility. IfAk is not bounded above, then the following holds:

For any M > 0 and any N > 0, there is some j > N with|aj| > M .

Consequently, Ak is not bounded above for any k, and we see the two possi-bilities are mutually exclusive. In this case we write sup Ak = +∞. If eitherpossibility holds, we define

sup Ak =

{

+∞, if Ak is not bounded abovel.u.b. Ak ∈ R if Ak is bounded above.


Notice that in the unbounded case, sup Ak = +∞ for all k, and in thebounded case sup Ak+1 ≤ sup Ak ≤ · · · ≤ sup A1 for all k. We conclude thatthe sequence

{sup Ak}∞k=1

is either a non-increasing sequence or is identically +∞. In view of thisobservation, we set

lim supjր∞

aj = limjր∞

aj =

{

+∞, if sup Ak = +∞limkր∞ sup Ak otherwise.

Note that lim aj is well-defined in R ∪ {±∞}.A similar discussion applies to lower bounds for arbitrary sets with

inf Ak =

{

−∞, if Ak is not bounded belowg.l.b. Ak ∈ R if Ak is bounded below.

and

lim infjր∞

aj = limjր∞

aj =

{

−∞, if inf Ak = −∞limkր∞ inf Ak otherwise.

Note that lim aj also takes values in R ∪ {±∞}.These are general defintions that apply to any and all real sequences. We

have the following result:

Theorem 10 Given any sequence {aj}∞j=1 ⊂ R,

1. The numberslimjր∞

aj and limjր∞

aj

are well defined in the extended real line [−∞,∞],

2. These numbers satisfy

limjր∞

aj ≤ limjր∞

aj,

3. The limitlimjր∞

aj

exists (as an extended real number) if and only if

limjր∞

aj = limjր∞

aj .


In this case, the limit is the common value:

limjր∞

aj = L = limjր∞

aj = limjր∞

aj .

Proof: The first assertion is established in the discussion above. The secondassertion follows from the fact that

inf Ak ≤ supAk for all k.

For the third assertion, say the limit L exists and is finite. Then for anyǫ > 0, there is some N such that j > N implies |aj −L| < ǫ. This mean thatfor k > N ,

L − ǫ ≤ infAk ≤ sup Ak ≤ L + ǫ.

For example, if aj ∈ Ak, then −ǫ < aj − L, so L − ǫ < aj which impliesL − ǫ ≤ infAk. Taking this one step further

L − ǫ ≤ limjր∞

aj ≤ limjր∞

aj ≤ L + ǫ.

Since ǫ > 0 is arbitrary we have lim aj = lim aj = L. If lim aj = +∞,then clearly −∞ < inf Ak < sup Ak = +∞ for all k. We only need to showlim aj = limkր∞ inf Ak = +∞, but this is obvious sicne for any M , thereis some N for which j > N implies aj > M , hence when k > N , we haveinf Ak ≥ M . The case lim aj = −∞ is similar.

For the converse, assume first

lim aj = lim aj ∈ R

and call the common value L. For any ǫ > 0, there is some N for whichk > N implies

L − ǫ < inf Ak ≤ supAk < L + ǫ.

This means, in particular, that for j > N we have

L − ǫ < aj < L + ǫ.

So |L− aj| < ǫ and lim aj exists and is L ∈ R. We leave the remaining caseslim aj = lim aj = ±∞ as an exercise. 2

Nexercise 23 Show that if lim aj = lim aj = ±∞, then lim aj = ±∞.


Returning to the Cauchy sequence {aj}∞j=1, and taking ǫ = 1, there issome N such that

j > N implies − |aN | − 1 < |aj| < |aN | + 1.

(|aj − aN | < ǫ implies | |aj| − |aN | | < ǫ.)It follows that {aj}∞j=1 is bounded (above and below) and

−∞ < limjր∞

aj ≤ limjր∞

aj < ∞

in particular.Assume, by way of contradiction that

limjր∞

aj < limjր∞

aj .

Then we can take

ǫ =1

2

(

limjր∞

aj − limjր∞

aj

)

and there is some N such that

k, ℓ > N implies |ak − al| < ǫ =1

2

(

limjր∞

aj − limjր∞

aj

)

.

On the other hand, there are some k and ℓ (with k, ℓ > N) such that

ak < limjր∞

aj +ǫ

2and aℓ > lim

jր∞aj −

ǫ

2.

Thus,

|ak − aℓ| = aℓ − ak > 2ǫ − ǫ

2− ǫ

2= ǫ.

This is a contradiction, so

limjր∞

aj = limjր∞

aj .

We claim that this common value is the limit of the sequence. In fact, forany ǫ > 0, there is some N such that

k > N implies L − ǫ < aj < L + ǫ,

i.e., |aj − L| < ǫ. 2


Theorem 11 (Cauchy’s comparison for sequences) If {zj}∞j=1 ⊂ C and{wj}∞j=1 ⊂ C are sequences such that

1. {zj}∞j=1 converges, and

2. There is some N such that

|wj − wk| ≤ |zj − zk| for all j, k > N,

then {wj}∞j=1 converges.

2.3.2 Complex series

A formal series is just a sequence {αj}∞j=1 which we write with the notation

∞∑

j=1

αj = α1 + α2 + α3 + · · · .

One can think of the “+” signs, more or less, as commas; writing this downdoes not imply anything about the individual terms α1, α2, α3, . . . nor is anyassertion that

∑

αj is associated with any particular complex number.The partial sums

sk =k∑

j=1

αj

of a formal series are well-defined complex numbers, and they define a se-quence of partial sums. If this sequence (of partial sums) has a limitL ∈ C, then we write

∞∑

j=1

αj = L.

That is, we associate the formal series with the complex number L. If

∞∑

j=1

αj = limkր∞

k∑

j=1

αj = L,

we say the series converges.


Theorem 12 A formal series∑

αj converges if and only if for any ǫ > 0,there is some N such that

∣

∣

∣

∣

∣

N+k∑

j=N

αj

∣

∣

∣

∣

∣

< ǫ for all k ≥ 0. (2.17)

Proof:

|sk − sℓ| =

∣

∣

∣

∣

∣

k∑

j=ℓ+1

αj

∣

∣

∣

∣

∣

.

Corollary 1 If∑

αj converges, then |αj| → 0 as j ր ∞.

Proof: Take k = 0 in (2.17). 2

Theorem 13 If∑

|αj | converges (in R), then∑

αj converges. In this sit-uation

∑

αj is said to be absolutely convergent.

Proof:

|sk − sℓ| =

∣

∣

∣

∣

∣

k∑

j=ℓ+1

αj

∣

∣

∣

∣

∣

≤k∑

j=ℓ+1

|αj| =k∑

j=1

|αj| −ℓ∑

j=1

|αj|. 2

2.3.3 Sequences and series of functions

If fj(z) for j = 1, 2, 3, . . . defines a sequnce of functions on the same domainU , we say the sequence has a pointwise limit f : U → C if for each fixedz ∈ U ,

limjր∞

fj(z) = f(z).

The convergence is said to be uniform if for each ǫ > 0, there is some Nsuch that

j > N implies |fj(z) − f(x)| < ǫ for all x ∈ U .

It is clear that uniform convergence implies pointwise convergence.

Nexercise 24 Give three fundamentally different examples of sequences offunctions which converge pointwise but not uniformly.

Theorem 14 If {fj}∞j=1 is a sequence of continuous functions and fj → funiformly on U , then f is continuous.


Proof: Let ǫ > 0. Then there is some N such that

j > N implies |fj(z) − f(x)| < ǫ/3 for all x ∈ U .

Fix j > N and fix z0 ∈ U . Since fj is continuous (at z0), there is some δ > 0such that

|z − z0| < δ (and z ∈ U) implies |fj(z) − fj(z0)| < ǫ/3.

Therefore, for the same points z ∈ U with |z − z0| < δ, we have

|f(z) − f(z0)| ≤ |f(z) − fj(z)| + |fj(z) − fj(z0)| + |fj(z0) − f(z0)| < ǫ.

This means f is also continuous at z0. Since z0 was arbitrary, we have shownf is continuous on U . 2

Definition 7 (The uniform Cauchy condition) The sequence of func-tions {fj}∞j=1 on the common domain U is said to be uniformly Cauchy iffor any ǫ > 0, there is some N such that

j, k > N implies |fj(z) − fk(z)| < ǫ for all z ∈ U .

Nexercise 25 Show that a sequence of functions {fj}∞j=1 on the commondomain U converges uniformly to a function f : U → C if and only if thesequence is uniformly Cauchy.

Theorem 15 (Cauchy’s comparison for sequences of functions) If {fj}∞j=1

and {gj}∞j=1 are sequences of functions on the common domain U satisfying

1. {gj}∞j=1 converges uniformly on U to a function g : U → C, and

2. There is some N such that

|fj(z) − fk(z)| ≤ |gj(z) − gk(z)| for all j, k > N and all z ∈ U ,

then {fj}∞j=1 converges uniformly to a function f : U → C.

The result is often used with the functions gj taken to be constants; then therequired uniform convergence is just the convergence of the sequence {gj}.

As with sequences and series of constants, this kind of comparison maybe applied to series of functions. A sequence {gj}∞j=1 of functions is said tobe a majorant of the sequence {fj}∞j=1 if there is some N such that

|fj(z)| ≤ gj(z) for j > N and all z ∈ U .

Notice the functions gj are required to be non-negative.


Theorem 16 If a majorant series

∞∑

j=1

gj(z)

converges (uniformly) on U , then the series∑

fj is absolutely convergent onU (and uniformly convergent).

2.3.4 § 2.3 exercises

Exercise 1

A convergent sequence is bounded.Solution: If zj → L, then there is some N so that

j > N implies |zj − L| < 1.

LetM = max

1≤j≤N|zj − L|.

Then for all j

|zj| ≤ |zj − L| + |L| ≤ (M + 1) + |L| < ∞.

Exercise 2

If lim zj = L, then

limk→∞

1

k

k∑

j=1

aj = L.

Let ǫ > 0. There is some N so that j > N implies |aj − L| < ǫ/4. Also, forany finite n,

limk→∞

1

k

n∑

j=1

aj = 0.

In particular, there is some K > N such that k > K implies

∣

∣

∣

∣

∣

1

k

N∑

j=1

aj

∣

∣

∣

∣

∣

<ǫ

2.


Therefore, for all k > K,∣

∣

∣

∣

∣

1

k

k∑

j=1

aj − L

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

∣

1

k

N∑

j=1

aj

∣

∣

∣

∣

∣

+k − N

k

∣

∣

∣

∣

∣

1

k − N

k∑

j=N+1

(aj − L)

∣

∣

∣

∣

∣

<ǫ

2+

k − N

k

ǫ

4.

Finally,

limkր∞

k − N

k= 1

so we may take k large enough so that (k − N)/k < 1 + ǫ. We may alsoassume ǫ < 1, so ǫ2 < ǫ and there is some M so that k > M implies

∣

∣

∣

∣

∣

1

k

k∑

j=1

aj − L

∣

∣

∣

∣

∣

<ǫ

2+ (1 + ǫ)

ǫ

4=

ǫ

2+

ǫ

4+

ǫ2

4<

ǫ

2+

ǫ

4+

ǫ

4= ǫ.

Exercise 3

Show the sum of an absolutely convergent series does not change if the termsare rearranged.

Let∑

αj denote our series, so we know

∞∑

j−1

|αj | < ∞.

Let σ : N → N be any rearrangement of the natural numbers, that is, asone-to-one and onto function. Then we wish to show

∞∑

j=1

ασ(j) =∞∑

j=1

αj .

Recall that∑

αj converges because the partial sums satisfy∣

∣

∣

∣

∣

k∑

j=1

αj −ℓ∑

j=1

αj

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

∣

k∑

j=1

|αj| −ℓ∑

j=1

|αj|∣

∣

∣

∣

∣

and the sequence of partial sums for∑ |αj| is Cauchy (because that sequence

converges). Let L =∑

αj and M =∑

|αj|. For any ǫ > 0, there is some N0

for which k > N0 implies∣

∣

∣

∣

∣

k∑

j=1

αj − L

∣

∣

∣

∣

∣

<ǫ

2and

∣

∣

∣

∣

∣

k∑

j=1

|αj | − M

∣

∣

∣

∣

∣

<ǫ

2.


Since

M =

∞∑

j=1

|αj|

the last inequality may be written as∣

∣

∣

∣

∣

∞∑

j=k+1

|αj |∣

∣

∣

∣

∣

<ǫ

2.

Fix k = N0 + 1 > N0. Since σ is onto, there is some N for which

A = {1, 2, . . . , k} ⊂ {σ(1), σ(2), . . . , σ(N)} = B.

If ℓ > N , then

∣

∣

∣

∣

∣

ℓ∑

j=1

ασ(ℓ) − L

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

k∑

j=1

αj +∑

j∈B\A

αj − L

∣

∣

∣

∣

∣

∣

≤∣

∣

∣

∣

∣

k∑

j=1

αj − L

∣

∣

∣

∣

∣

+

∞∑

j=k+1

|αj|

< ǫ/2 + ǫ/2 = ǫ.

Exercise 4

Discuss completely the convergence and uniform convergence of the sequence

{jzj}∞j=1.

For |z| < 1, the sequence converges to zero because

jzj = (j1/jz)j and j1/j → 1 (as shown two different ways in class).

This means that for some fixed δ > 0 with 0 < δ < 1 and N large we have|j1/jz| < 1 − δ < 1 for j > N . Taking j > N

|(j1/jz)j | ≤ (1 − δ)j → 0 as j → ∞.

This estimate shows the convergence is uniform on any disk Br(0) with r < 1.Similarly, for any fixed N > 0 and ǫ > 0, there is some δ > 0 so that for anyθ the complex number z = (1 − δ)(cos θ + i sin θ) satisfies

|NzN | = [N1/N (1 − δ)]N > ǫ.


This shows the convergence is not uniform on the entire disk B1(0).Finally, for |z| ≥ 1,

|jzj| ≥ j → ∞so the sequence converges to the point at infinity.

2.3.5 § 2.4 Complex power series

Everything we have said about series so far applies to the series of the specialform

∞∑

j=0

aj(z − z0)j

with z0, a0, a1, a2, . . . ∈ C. As we know, taking z0 = 0 and aj = 1 for all j,we get the geometric series with ratio z. In fact, we know

(1 − z)(1 + z + · + zk) = 1 − zk+1,

so if z 6= 1, we have that the partial sums satisfy

k∑

j−1

zj =1 − zk+1

1 − z.

If |z| < 1, then∞∑

j=1

zj =1

1 − z. (2.18)

If |z| ≥ 1, then the individual terms in the series satisfy |zj | = |z|j ≥ 1,so the series cannot converge. This particular series, the geometric series,will be used for comparison in general. It is worth special attention for thisand other reasons. The right side in (2.18) is a rational function

f(z) =1

1 − z

with a pole at z = 1. The function f makes sense, i.e., is finite valued,everywhere in C except for z = 1. Note what has happened here carefully:The formal geometric series defines the function f for |z| < 1 (and nowhereelse), but the function has an analytic continuation beyond the meaningof the series.


Consider another example:

∞∑

j=1

1

jzj . (2.19)

Since |zj/j| ≤ |z|j, this series is also absolutely convergent in the unit disk|z| < 1. The boundary behavior is somewhat more interesting. If z =1, the series becomes the harmonic series and diverges. If z = −1, theseries becomes the alternating harmonic series and converges. It willgenerally be difficult to determine whether or not a series converges on theboundary of the region (disk) where the series is absolutely convergent, butit is worth noting that the outcome can be either convergence or divergence.The geometric series is atypical in this sense. We will show the series in (2.19)also defines an analytic function in the open unit disk. Just for reference thefunction defined by (2.19) is

f(z) = 1 + ln

(

1

1 − z

)

.

Theorem 17 (radius of convergence) Given any formal power series

∑

aj(z − z0)j ,

there is a well-defined non-negative extended real number

H = limjր∞

j

√

|aj| (2.20)

such that if 0 < H < ∞, then the series converges absolutely for |z − z0| <R = 1/H, and if |z − z0| > R, then

lim |aj(z − z0)| = +∞ so the series cannot converge.

If H = 0, then the series converges absolutely in C, and we say the functiondefined by the series is entire. (Here, R = +∞.)

If H = ∞, then the series converges for z = z0 and only for z = z0. Inthis case, R = 0.

The value H is named after Hadamard, and the non-negative extendedreal number R is called the radius of convergence of the series.


Proof: Assume 0 < |z − z0| < R. Then

lim j

√

|aj| <1

|z − z0|,

and there is some ǫ > 0 such that for j large2

j

√

|aj | <1

|z − z0|− ǫ.

That is,

|aj(z − z0)j| < (1 − ǫ|z − z0|)j < 1.

Since r = 1 − ǫ|z − z0| is a fixed positive number with r < 1, the convergentgeometric series

∑

rj is a majorant for∑

|aj(z − z0)j|. The series converges

absolutely.

If R < ∞ and |z − z0| > R = 1/ lim j√

|aj|, then

lim j

√

|aj| >1

|z − z0|.

It follows that for some ǫ > 0 there are infinitely many j with

j

√

|aj | >1

|z − z0|+ ǫ =

1 + ǫ|z − z0||z − z0|

.

For these j,

|aj(z − z0)j| ≥ (1 + ǫ|z − z0|)j.

Since (1+ ǫ|z− z0|)j ր +∞ as j ր ∞, we know that for any M and any N ,there is some j > N with

|aj(z − z0)j | ≥ (1 + ǫ|z − z0|)j > M.

In particular,

lim |aj(z − z0)j | = +∞. 2

We can sharpen this result:

2He Guo suggests the use of (1 − ǫ)/|z − z0| here. Then r = 1 − ǫ < 1.


Theorem 18 If r < R with

F =

∞ if H = lim j√

|aj| = 01/H if 0 < H < ∞0 if H = ∞

,

then∑

aj(z − z0)j converges absolutely and uniformly for |z − z0| ≤ r.

Proof: Consider z with 0 ≤ |z − z0| < r < R. We then have

lim j

√

|aj| <1

r.

There is some ǫ > 0 such that for j large

j

√

|aj | <1

r− ǫ =

1 − ǫr

r, and |aj | <

(

1 − ǫr

r

)j

.

Hence, |aj(z − z0)j| ≤ (1 − ǫr)j, and

∑

aj(z − z0) is (uniformly) majorizedby∑

(1 − ǫr)j. 2

Theorem 19 (analytic functions defined by power series) The func-tion

f(z) =

∞∑

j=0

aj(z − z0)j defined for |z − z0| < R

is analytic with

f ′(z) =

∞∑

j=0

jaj(z − z0)j−1. (2.21)

Proof: First of all

lim j

√

j|aj | = lim j

√

|aj |,

so the radius of convergence of the series in (2.21) is also R. We have usedthe facts

1. The real sequence j√

j → 1 as j ր ∞, and

2. If {aj}∞j=1 is a real sequence and aj → a, then

lim ajbj = a lim bj .


Ahlfors gives a nice proof of the first fact:

j√

j > 1 so write j√

j = 1 + δ.

Then

j = (1+δ)j =

j∑

ℓ=0

(

jℓ

)

δℓ = 1+jδ+j(j − 1)

2δ2+

j∑

ℓ=3

(

jℓ

)

δℓ ≥ 1+j(j − 1)

2δ2.

Therefore,

δ2 ≤ 2

j→ 0.

Nexercise 26 If {aj}∞j=1 is a real sequence and aj → a, then

lim ajbj = a lim bj .

In order to more clearly express what’s happening in the next step (and saveon typing), let us specialize to z0 = 0. Having made that simplification, letus restrict to a ball Br(0) ⊂ BR(0) where the convergence is uniform (forboth sequences).

f(ζ) − f(z)

ζ − z−

∞∑

j=1

jajzj−1 =

1

ζ − z

(

k∑

j=0

aj(ζj − zj) +

∞∑

j=k+1

aj(ζj − zj)

)

−∞∑

j=1

jajzj−1

=1

ζ − z

k∑

j=0

aj(ζj − zj) −

k∑

j=1

jajzj−1

1

ζ − z

∞∑

j=k+1

aj(ζj − zj) −

∞∑

j=k+1

jajzj−1.

Estimating the second term with the difference quotient with the usual fac-toring of ζj − zj , we see∣

∣

∣

∣

∣

1

ζ − z

∞∑

j=k+1

aj(ζj − zj)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∞∑

j=k+1

aj(ζj−1 + ζj−2z + · · ·+ ζzj−2 + zj−1)

∣

∣

∣

∣

∣

≤∞∑

j=k+1

j|aj|rj−1.


Since∑

j|aj |rj−1 converges, we know

limkր∞

∞∑

j=k+1

j|aj |rj−1 = 0.

To be precise, let ǫ > 0 be fixed. There is some N such that k > N implies

∣

∣

∣

∣

∣

1

ζ − z

∞∑

j=k+1

aj(ζj − zj)

∣

∣

∣

∣

∣

<ǫ

3.

If k > N , we also have (or can assume)

∣

∣

∣

∣

∣

∞∑

j=k+1

jajzj−1

∣

∣

∣

∣

∣

≤∞∑

j=k+1

j|aj |rj−1 <ǫ

3.

For the finite sum, we have already shown (using the usual factoring as inExam 1) that there is some δ > 0 such that |ζ − z| < δ implies

∣

∣

∣

∣

∣

1

ζ − z

k∑

j=0

aj(ζj − zj) −

k∑

j=1

jajzj−1

∣

∣

∣

∣

∣

<ǫ

3.

Fixing k large enough, we then choose δ, and we have shown

limζ→z

∣

∣

∣

∣

∣

f(ζ) − f(z)

ζ − z−

∞∑

j=1

jajzj−1

∣

∣

∣

∣

∣

= 0

for f(z) =∑

ajzj . That is,

f ′(z) =∞∑

j=1

jajzj−1

2

Nexercise 27 Rewrite the proof above with modifications to allow the centerof expansion to be an arbitrary z0 ∈ C instead of z0 = 0.

The following is really a corollary of the results above, but it is so impor-tant we state it as a theorem.


Theorem 20 (regularity for power series) If f(z) =∑

aj(z − z0)j is

defined by a power series with radius of convergence R > 0, then f is analytic(differentiable) and each derivative f (k)(z) for k = 1, 2, 3, . . . exists and isanalytic in BR(z0) = {z : |z − z0| < R}.

We have not shown regularity (that every analytic function has derivativesof all orders), but we have shown many analytic functions are regular—allthose represented by a complex power series. We will show every analyticfunction is represented (locally) by a power series.

Theorem 21 (Complex Taylor’s theorem) If f(z) =∑∞

j=0 aj(z − z0)j,

then

aj =f (j)

j!.

Proof: Evaluating the series at z = z0, we see f(z0) = a0. That’s onecoefficient. We know

f ′(z) =∞∑

j=1

jaj(z − z0)j , so f ′(z0) = a1.

That’s two coefficients. If

f (k)(z) =∞∑

j=k

j(j − 1) · · · (j − k + 1)aj(z − z0)j−k

(and it does), then

f (k+1)(z) =

∞∑

j=k+1

j(j − 1) · · · (j − k)aj(z − z0)j−k−1.

Therefore,

f (k)(z0) = k!ak and f (k+1)(z0) = (k + 1)!ak+1,

so that is all of them (by induction). 2


2.3.6 § 2.4 exercises

Exercise 1

Expand1

(1 − z)m

in powers of z.Let us consider the formal power series for the function

f(z) =1

(1 − z)m

which is clearly an analytic function for |z| < 1. We see f(0) = 1,

f ′(z) =m

(1 − z)m+1, f ′(0) = m.

f (j)(z) =m(m + 1) · · · (m + j − 1)

(1 − z)m+j, f (j)(0) =

(m + j − 1)!

(m − 1)!.

Thus,

f ∼∞∑

j=0

(m + j − 1)!

j!(m − 1)!xj =

1

(m − 1)!

∞∑

j=0

(m + j − 1)!

j!xj .

Notice

(

(m + j − 1)!

j!

)1/j

= (j + m − 1)1/j(j + m − 2)1/j · · · (j + 1)1/j → 1

since there are finitely many (m − 1) factors here. Thus, the radius of con-vergence is R = 1.

Exercise 2

Expand2z + 3

z + 1

in powers of z − 1. Find the radius of convergence.


Let us write

2z + 3

z + 1=

2(z − 1) + 5

z − 1 + 2

=1

2

2(z − 1) + 5

1 −(

−z−12

)

=1

2

[

2(z − 1)∞∑

j=0

(

−z − 1

2

)j

+ 5∞∑

j=0

(

−z − 1

2

)j]

=5

2+

∞∑

j=1

(−1)j

2j+1(z − 1)j.

The disk of convergence (because this is a geometric series) is determined by

∣

∣

∣

∣

z − 1

2

∣

∣

∣

∣

= 1.

That is to say, the radius of convergence is R = 2.

Exercise 5

If f(z) =∑

ajzj , what is

∑

j3ajzj?

f ′(z) =∑

jajzj−1 and f ′′(z) =

∑

j(j − 1)ajzj−2 =

∑

(j2 − j)ajzj−2.

Most importantly,

f ′′′(z) =∑

j(j − 1)(j − 2)ajzj−3 =

∑

(j3 − 3j2 + 2j)ajzj−3.

Thus,

z3f ′′′(z) =∑

(j3 − 3j2 + 2j)ajzj ,

and∑

j3ajzj = z3f ′′′(z) + 3z2f ′′(z) + f ′(z).

2.3.7 Abel limit theorem

An angle domain (or Stolz angle or sector) with vertex at z1 ∈ C andreference direction w ∈ S1 is defined by

Am = {z ∈ C : Re[(z − z1)w/|z − z1| > m}.


Figure 2.5: a convex angle domain

The number m is the cosine of the half angle of the domain. If m > 0, thenthe half angle is less than π/2 and the domain is convex. If −1 < m < 0,then the domain is said to be reentrant. We will only be dealing with convexsectors here.

The following result allows us to find the value of a convergent seriesat a boundary point by taking a limit from within the disk of convergence.Consequently, it gives a kind of one-sided continuity at the boundary whenthe series converges.

Theorem 22 (Abel limit theorem) If f(z) is defined in BR(z0) by

f(z) =∞∑

j=0

aj(z − z0)j

and z1 ∈ ∂BR(z0) satisfies

f(z1) =

∞∑

j=0

aj(z1 − z0)j ∈ C,

i.e., the series is convergent at z = z1, then for any ǫ > 0 and m > 0, thereis some δ > 0 such that

|f(z) − f(z1)| < ǫ for all z ∈ Bδ(z1) ∩ Am

where

Am =

{

z ∈ C : Re

[

(z − z1)(z0 − z1)

|z − z1||z0 − z1|

]

> m

}

.


Proof: We will give the proof in the case z0 = 0 ∈ C, R = 1, and z1 = 1. Inthis case, the angle domain becomes

Am =

{

z ∈ C : Re

[

(1 − z)

|1 − z|

]

=1 − Re(z)

|1 − z| > m

}

.

Figure 2.6: a convex angle domain at the boundary point z1 = 1

Lemma 12 For any µ with 0 < µ < 1, there is some δ1 > 0 such that

1 − |z||1 − z| > µm for z ∈ Bδ1(1) ∩ Am.

Proof: Let z = x+ iy ∈ Bδ1(1)∩Am and note that for δ1 small we must have0 < x < 1. Also, the condition

1 − Re(z)

|1 − z| > m

can be written as

1 − x > m√

(1 − x)2 + y2 which implies y2 <(1 − m2)

m2(1 − x)2.

From this, we see

1 − |z| = 1 −√

x2 + y2 > g(x) = 1 −√

x2 +1 − m2

m2(1 − x)2.


We can compute

g′(x) = −(

x − 1 − m2

m2(1 − x)

)

/√

x2 +(1 − m2)

m2(1 − x)2 .

Therefore g(z) satisfies

g(1) = 0, and g′(1) = −1.

On the other hand, h(x) = µ(1 − x) satisfies

h(1) = 0 and h′(1) = −µ > −1.

Thus, for some δ1 > 0, we have

h(x) < g(x) for 1 − δ1 < x < 1.

In particular, for z ∈ Bδ1(1) ∩ Am we have

1 − |z| > g(x) > h(x) = µ(1 − x) > µm|1 − z|. 2

Corollary 2 There is some δ1 > 0 and some M < ∞ such that

|1 − z|1 − |z| < M for z ∈ Bδ1(1) ∩ Am. (2.22)

Proof: M = 1/(µm). 2

Returning to the proof of the Abel limit theorem: Set

f(1) =∑

aj = α.

Let ǫ > 0. We wish to show |f(z) − f(1)| < ǫ for z ∈ Bδ(1) ∩ Am (for someδ > 0 yet to be identified).

Let us restrict, for the moment, to z ∈ Bδ1(1) ∩ Am. As usual, we havethe partial sums

sk = a0 + a1z + z2z2 + · · · + akz

k. (2.23)

Setbk = a0 + a1 + z2 + · · ·+ ak.

Then bk − bk−1 = ak. Notice the coefficients ak in (2.23) may be replaced bymeans of this relation. Let us modify this a little:

sk = sk − α = a0 − α + a1z + z2z2 + · · ·+ akz

k;


bk = bk − α = a0 − α + a1 + z2 + · · · + ak.

Then

sk = b0 + (b1 − b0)z + (b2 − b1)z2 + + · · ·+ (bk − bk−1)z

k

= b0(1 − z) + b1(1 − z)z + · · ·+ bk−1(1 − z)zk−1 + bkzk

= (1 − z)

k−1∑

j=0

bjzj + bkz

k.

On the other hand, bk → α as k → ∞, so bk → 0 as k → ∞. This means thelast term tends to zero, and the series

∑

bjzj has a finite limit:

f(z) − α = (1 − z)

∞∑

j=0

bjzj for z ∈ B1(0).

Again, since bj → 0, there is some N so that

j > N implies |bj | <ǫ

2M

where M comes from Corollary 2. Taking N as an integer, we can nowestimate as follows:

|f(z) − f(1)| = |f(z) − α|

=

∣

∣

∣

∣

∣

(1 − z)∞∑

j=0

bjzj

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

(1 − z)N∑

j=0

bjzj + (1 − z)

∞∑

j=N+1

bjzj

∣

∣

∣

∣

∣

≤ |1 − z|N∑

j=0

|bj | + |1 − z| ǫ

2M

∞∑

j=N+1

|z|j

≤ |1 − z|N∑

j=0

|bj | + |1 − z| ǫ

2M

∞∑

j=0

|z|j

= |1 − z|N∑

j=0

|bj| + |1 − z| ǫ

2M

1

1 − |z| .


Since∑N

j=0 |bj | is fixed, there is some δ < δ1 for which

z ∈ Bδ(1) implies |1 − z|N∑

j=0

|bj | <ǫ

2.

Also, from Corollary 2 we know

|1 − z|1 − |z| < M.

Therefore, we have that for z ∈ Bδ(1) ∩ Am there holds

|f(z) − f(1)| <ǫ

2+

ǫ

2= ǫ. 2

Nexercise 28 State and prove a version of Lemma 12 which applies in anarbitrary ball BR(z0) with an arbitrary boundary point z1 ∈ ∂BR(z0) and aconvex angle domain determined by w = (z0 − z1)/|z0 − z1|.

Nexercise 29 Fill in the details to show Abel’s theorem holds in the generalcase for z1 ∈ ∂BR(z0) as stated.

2.3.8 Ahlfors’ corner domains and generalized ellipses

Ahlfors suggests a domain of the form

Am =

{

z ∈ C :1 − |z||1 − z| ≥ m

}

constitutes a restriction to a corner domain. While Lemma 12 shows thisis essentially correct, the set Am is not a corner domain but a member of afamily of generalized ellipse domains which are interesting to consider.

Lemma 12 says that an arbitrarily small deformation of the domain Am

will contain a neighborhood of z = 1 in the domain Am. On the other hand,since

1 − Re(z)

|1 − z| ≥ 1 − |z||1 − z| ,

we know Am ⊂ Am. This means the domain Am indeed has a corner ofhalf-angle cos−1(m) at z = 1.


- 1.0 - 0.5 0.5 1.0

- 1.0

- 0.5

0.5

1.0

Figure 2.7: family of generalized ellipses |z| + m|1 − z| = 1 + ǫ for ǫ = 0.2and m = 0, 0.2, 0.4, . . . , 1

To see the rest of the boundary of Am it is convenient to consider themore general (and non-singular) domains

A(ǫ) = {z ∈ C : |z| + m|1 − z| < 1 + ǫ}

for ǫ > 0. We then have, for example, the closure

Am = ∩ǫ>0A(ǫ).

Notice that for m = 1, these are elliptical domains. In general, it is requiredthat the sum of the distance from z to the origin and some “fraction” mof the distance from z to z = 1 does not exceed a certain constant 1 + ǫ.Evidently, the boundary is characterized by the equation

|z| + m|1 − z| = 1 + ǫ. (2.24)

In particular, it is clear that Am ⊂ B1+ǫ(0) with the boundaries satisfying

∂Am ∩ ∂B1+ǫ(0) = {1 + ǫ}.

In particular, these domains are generalization of elliptical domains withinthe disk B1+ǫ(0). The origin is in each domain A(ǫ) and we can considerthe ray z(r) = r(cos θ + i sin θ). The condition |z(r)| + m|1 − z(r)| < 1 + ǫbecomes

r + m√

1 − 2r cos θ + r2 < 1 + ǫ. (2.25)


Taking θ = 0, this condition reads r +m|1− r| < 1+ ǫ. We see then that theentire interval 0 ≤ z ≤ 1 lies inside A(ǫ) and there is a boundary point withr = z(r) = 1 + ǫ/(1 + m) > 1. Taking θ = π we obtain the left endpoint ofthe real interval in A(ǫ), namely

z(r) = −r = −1 − m + ǫ

1 + m→ −1 − m

1 + m> −1 as ǫ → 0.

Notice the limiting (intersection) value

z0 = −1 − m

1 + m

varies monotonically with m and limits to −1 as m ց 0 (corner angle in Am

and Am approaching π). The value z0 → 0 as m ր 1 (corner angle in Am

and Am approaching 0).Returning to (2.25) for general θ which we write as

+m√

1 − 2r cos θ + r2 < 1 + ǫ − r, (2.26)

we find r must satisfy

(1 − m2)r2 − 2(1 + ǫ − m2 cos θ)r + (1 + ǫ)2 − m2 > 0. (2.27)

Solving the equation (1 − m2)r2 − 2(1 + ǫ − m2 cos θ)r + (1 + ǫ)2 − m2 = 0we obtain two (positive) roots

r =1 + ǫ − m2 cos θ ±

√

(1 + ǫ − m2 cos θ)2 − [(1 + ǫ)2 − m2](1 − m2)

1 − m2.

As noted r = 0 corresponds to the origin within A(ǫ). The first root willcorrespond to the boundary of A(ǫ). For r greater than the second “+” root,the inequality in (2.27) will hold. On the other hand, the second root isgreater than

1 + ǫ − m2 cos θ

1 − m2=

1 − m2 cos θ

1 − m2+

ǫ

1 − m2> 1 + ǫ,

so the second “+” root produces a negative value on the right in (2.26) andis extraneous. The boundary of A(ǫ) is

∂A(ǫ) =

{

b −√

b2 − [(1 + ǫ)2 − m2](1 − m2)

1 − m2(cos θ + i sin θ) : 0 ≤ θ ≤ 2π.

}


where b = 1 + ǫ − m2 cos θ. When m = 0 we again obtain an ellipse whichis the circle |z| = 1 + ǫ. When m = 1 we have a nondegenerate ellipsewith focal points at z = 0 and z = 1 with major axis of length 1 + ǫ. Forthe intermediate values 0 < m < 1, we may numericaly plot the family ofgeneralized elliptical boundaries as indicated in Figure 2.7.

We may take any one of these generalized elliptical domains (with m fixed)and let ǫ tend to zero to obtain a family of nested domains and a limitingdomain with a corner at z = 1. To emphasize what we have said above, the

0.5 1.0

- 0.6

- 0.4

- 0.2

0.2

0.4

0.6

Figure 2.8: A corner domain |z| + m|1 − z| < 1 (dashed) obtained as thelimit of generalized elliptical domains |z| + m|1 − z| = 1 + ǫ; here m = 0.6and ǫ = 0.2, 0.5, 0.05 (and 0)

corner at z = 1 in this teardrop shaped domain Am has half angle cos−1(m),the same value as in the corner domain Am. Finally, we include a figureindicating the family of generalized elliptical domains as both parameters ǫand m vary. The collection of domains with ǫ ≡ 0 are all teardrop shapedcorner domains Am with the limit as m ր 1 the degenerate ellipse which isthe segment from the origin to z = 0, i.e., the Ahlfors corner domain withangle zero.

Open question: Is there a nice geometric interpretation (analogous to thatof the ellipse) of the curves bounding the generalized elliptical domains A(ǫ)?


- 2 - 1 1 2

- 2

- 1

1

2

- 2.0 - 1.5 - 1.0 - 0.5 0.5 1.0 1.5 2.0

- 2.0

- 1.5

- 1.0

- 0.5

0.5

1.0

1.5

2.0

- 1.5 - 1.0 - 0.5 0.5 1.0 1.5

- 1.5

- 1.0

- 0.5

0.5

1.0

1.5

- 1.5 - 1.0 - 0.5 0.5 1.0 1.5

- 1.5

- 1.0

- 0.5

0.5

1.0

1.5

- 1.0 - 0.5 0.5 1.0

- 1.0

- 0.5

0.5

1.0

Figure 2.9: generalized elliptical domains; figures are shown for ǫ = 1, ǫ = 0.8,eps = 0.6, ǫ = 0.4, and ǫ = 0 (the case ǫ = 0.2 is in Figure 2.7)

2.4 Lecture 7: § 3.1-4 Exponential and trigono-

metric functions

2.4.1 The real exponential and some comments on rigor

In discussing the derivative series

∞∑

j=1

jajzj−1

we used the fact that j1/j → 1 and j ր ∞, and we used Ahlfors’ directproof to show this fact. Of course, we have assumed here the real j-th root isa well-defined real function with familiar properties. (This function and itsproperties may be obtained rigorously using the ordering and completenessof the real numbers and techniques from calculus.) Another familiar proof ofthe fact that j1/j → ∞ as j ր ∞ runs as follows: Because ln j1/j = ln j/j,which is an indeterminate form ∞/∞, we can use L’Hopital’s rule to seelim ln j/j = lim 1/j = 0. Thus, by the continuity of the real exponentiallim j1/j = e0 = 1. This argument uses a number of facts from the calculusof real functions. Notably, we needed the existence of the real exponential(and logarithm), facts about derivatives, and L’Hoptial’s rule. While eachof these may be understood and justified to (hopefully) a satisfactory levelof rigor, Ahlfors wishes to present the real exponential (and trigonometricfunctions) as restrictions to the real line of their complex versions.

Partially to highlight some differences between the real approach andthe complex approach and partially because a rigorous proof of (the neededversion of) L’Hopital’s rule us usually not included in elementary calculus,

2.3 Lecture 6: 2.1-5 Complex sequences and (power)...

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