126 CHAPTER 2. CHAPTER II

so

R̃(z) = 1

/

R

(

1

z̄

)

= |z|2m |z|2n

∣

∣

∣

∣

∣

1

/

R

(

1

z̄

)

∣

∣

∣

∣

∣

2 an ∏k

j=1 (z − αj |z|2) nj

bn ∏ℓ

j=1 (z − βj|z|2) mj

.

It follows then that R̃(z) = R(z) when |z| = 1. Thus, these two rational functions must be identical. This means there is a correspondence between roots αj and poles βj = 1/ᾱj given by reflection about the unit circle. The conjecture 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 such that

j > N implies |αj − L| < ǫ. In this case, we write αj → L as j ր ∞ or

lim jր∞

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 are complete. 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 bound property. 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 the sets Ak are all bounded above by l.u.b. A1. There is another possibility. If Ak 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 either possibility holds, we define

sup Ak =

{

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

128 CHAPTER 2. CHAPTER II

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

{sup Ak}∞k=1 is either a non-increasing sequence or is identically +∞. In view of this observation, we set

lim sup jր∞

aj = lim jր∞

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 below g.l.b. Ak ∈ R if Ak is bounded below.

and

lim inf jր∞

aj = lim jր∞

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 numbers

lim jր∞

aj and lim jր∞

aj

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

2. These numbers satisfy

lim jր∞

aj ≤ lim jր∞

aj,

3. The limit lim jր∞

aj

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

lim jր∞

aj = lim jր∞

aj .

2.3. LECTURE 6: § 2.1-5 COMPLEX SEQUENCES AND (POWER) SERIES 129

In this case, the limit is the common value:

lim jր∞

aj = L = lim jր∞

aj = lim jր∞

aj .

Proof: The first assertion is established in the discussion above. The second assertion 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 any ǫ > 0, there is some N such that j > N implies |aj −L| < ǫ. This mean that for k > N ,

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

L − ǫ ≤ lim jր∞

aj ≤ lim jր∞

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 show lim aj = limkր∞ inf Ak = +∞, but this is obvious sicne for any M , there is some N for which j > N implies aj > M , hence when k > N , we have inf 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 which k > 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 cases lim aj = lim aj = ±∞ as an exercise. 2

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

130 CHAPTER 2. CHAPTER II

Returning to the Cauchy sequence {aj}∞j=1, and taking ǫ = 1, there is some 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

−∞ < lim jր∞

aj ≤ lim jր∞

aj < ∞

in particular. Assume, by way of contradiction that

lim jր∞

aj < lim jր∞

aj .

Then we can take

ǫ = 1

2

(

lim jր∞

aj − lim jր∞

aj

)

and there is some N such that

k, ℓ > N implies |ak − al| < ǫ = 1

2

(

lim jր∞

aj − lim jր∞

aj

)

.

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

ak < lim jր∞

aj + ǫ

2 and aℓ > lim

jր∞ aj −

ǫ

2 .

Thus,

|ak − aℓ| = aℓ − ak > 2ǫ − ǫ

2 − ǫ

2 = ǫ.

This is a contradiction, so

lim jր∞

aj = lim jր∞

aj .

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

k > N implies L − ǫ < aj < L + ǫ,

i.e., |aj − L| < ǫ. 2

2.3. LECTURE 6: § 2.1-5 COMPLEX SEQUENCES AND (POWER) SERIES 131

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 down does not imply anything about the individual terms α1, α2, α3, . . . nor is any assertion 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 limit L ∈ C, then we write

∞ ∑

j=1

αj = L.

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

∞ ∑

j=1

αj = lim kր∞

k ∑

j=1

αj = L,

we say the series converges.

132 CHAPTER 2. CHAPTER II

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