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The Real And Complex Number Systems

Integers

1.1

Prove that there is no largest prime.

Proof : Suppose p is the largest prime. Then p! + 1 is NOT a prime. So, there exists a prime q such that

q |p! + 1 ⇒ q |1 which is impossible. So, there is no largest prime.

Remark: There are many and many proofs about it. The proof that we give comes from Archimedes 287-212 B. C. In addition, Euler Leonhard (1707-1783) find another method to show it. The method is important since it develops to study the theory of numbers by analytic method. The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 91-93. (Chinese Version)

1.2

If n is a positive integer, prove the algebraic identity an− bn = (a − b)

n−1

X

k=0

akbn−1−k

Proof : It suffices to show that

xn− 1 = (x − 1)

n−1

X

k=0

xk.

(2)

Consider the right hand side, we have

(x − 1)

n−1

X

k=0

xk=

n−1

X

k=0

xk+1

n−1

X

k=0

xk

=

n

X

k=1

xk

n−1

X

k=0

xk

= xn− 1.

1.3

If 2n− 1 is a prime, prove that n is prime. A prime of the form 2p− 1, where p is prime, is called a Mersenne prime.

Proof : If n is not a prime, then say n = ab, where a > 1 and b > 1. So, we have

2ab− 1 = (2a− 1)

b−1

X

k=0

(2a)k

which is not a prime by Exercise 1.2. So, n must be a prime.

Remark: The study of Mersenne prime is important; it is related with so called Perfect number. In addition, there are some OPEN prob- lem about it. For example, is there infinitely many Mersenne nem- bers? The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 13-15. (Chinese Version)

1.4

If 2n+ 1 is a prime, prove that n is a power of 2. A prime of the form 22m+ 1 is called a Fermat prime. Hint. Use exercise 1.2.

Proof : If n is a not a power of 2, say n = ab, where b is an odd integer.

So,

2a+ 1

2ab+ 1

and 2a+ 1 < 2ab+ 1. It implies that 2n+ 1 is not a prime. So, n must be a power of 2.

Remark: (1) In the proof, we use the identity

x2n−1+ 1 = (x + 1)

2n−2

X

k=0

(−1)kxk.

(3)

Proof : Consider (x + 1)

2n−2

X

k=0

(−1)kxk=

2n−2

X

k=0

(−1)kxk+1+

2n−2

X

k=0

(−1)kxk

=

2n−1

X

k=1

(−1)k+1xk+

2n−2

X

k=0

(−1)kxk

= x2n+1+ 1.

(2) The study of Fermat number is important; for the details the reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 15. (Chinese Version)

1.5

The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, ... are defined by the recur- sion formula xn+1 = xn+ xn−1, with x1 = x2 = 1. Prove that (xn, xn+1) = 1 and that xn= (an− bn) / (a − b) , where a and b are the roots of the quadratic equation x2− x − 1 = 0.

Proof : Let d = g.c.d. (xn, xn+1) , then

d |xn and d |xn+1 = xn+ xn−1 . So,

d |xn−1 . Continue the process, we finally have

d |1 . So, d = 1 since d is positive.

Observe that

xn+1= xn+ xn−1, and thus we consider

xn+1= xn+ xn−1, i.e., consider

x2 = x + 1 with two roots, a and b.

If we let

Fn = (an− bn) / (a − b) ,

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then it is clear that

F1 = 1, F2 = 1, and Fn+1 = Fn+ Fn−1 for n > 1.

So, Fn= xn for all n.

Remark: The study of the Fibonacci numbers is important; the reader can see the book, Fibonacci and Lucas Numbers with Applications by Koshy and Thomas.

1.6

Prove that every nonempty set of positive integers contains a small- est member. This is called the well–ordering Principle.

Proof : Given (φ 6=) S (⊆ N ) , we prove that if S contains an integer k, then S contains the smallest member. We prove it by Mathematical Induction of second form as follows.

As k = 1, it trivially holds. Assume that as k = 1, 2, ..., m holds, consider as k = m + 1 as follows. In order to show it, we consider two cases.

(1) If there is a member s ∈ S such that s < m + 1, then by Induction hypothesis, we have proved it.

(2) If every s ∈ S, s ≥ m + 1, then m + 1 is the smallest member.

Hence, by Mathematical Induction, we complete it.

Remark: We give a fundamental result to help the reader get more. We will prove the followings are equivalent:

(A. Well–ordering Principle) every nonempty set of positive integers contains a smallest member.

(B. Mathematical Induction of first form) Suppose that S (⊆ N ) , if S satisfies that

(1). 1 in S

(2). As k ∈ S, then k + 1 ∈ S.

Then S = N.

(C. Mathematical Induction of second form) Suppose that S (⊆ N ) , if S satisfies that

(1). 1 in S

(2). As 1, ..., k ∈ S, then k + 1 ∈ S.

(5)

Then S = N.

Proof : (A ⇒ B): If S 6= N, then N − S 6= φ. So, by (A), there exists the smallest integer w such that w ∈ N − S. Note that w > 1 by (1), so we consider w − 1 as follows.

Since w − 1 /∈ N − S, we know that w − 1 ∈ S. By (2), we know that w ∈ S which contadicts to w ∈ N − S. Hence, S = N.

(B ⇒ C): It is obvious.

(C ⇒ A): We have proved it by this exercise.

Rational and irrational numbers

1.7 Find the rational number whose decimal expansion is 0.3344444444....

Proof: Let x = 0.3344444444..., then x = 3

10+ 3 102 + 4

103 + ... + 4

10n + .., where n ≥ 3

= 33 102 + 4

103

 1 + 1

10 + ... + 1 10n + ..



= 33 102 + 4

103

 1

1 −101



= 33 102 + 4

900

= 301 900.

1.8 Prove that the decimal expansion of x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form 2n5m, where m and n are nonnegative integers.

Proof: (⇐)Suppose that x = 2nk5m, if n ≥ m, we have k5n−m

2n5n = 5n−mk 10n .

So, the decimal expansion of x will end in zeros. Similarly for m ≥ n.

(⇒)Suppose that the decimal expansion of x will end in zeros (or in nines).

(6)

For case x = a0.a1a2· · · an. Then x =

Pn

k=010n−kak

10n =

Pn

k=010n−kak 2n5n . For case x = a0.a1a2· · · an999999 · · · . Then

x = Pn

k=010n−kak

2n5n + 9

10n+1 + ... + 9

10n+m + ...

= Pn

k=010n−kak

2n5n + 9 10n+1

X

j=0

10−j

= Pn

k=010n−kak 2n5n + 1

10n

= 1 +Pn

k=010n−kak

2n5n .

So, in both case, we prove that x is a rational number whose denominator is of the form 2n5m, where m and n are nonnegative integers.

1.9 Prove that√ 2 +√

3 is irrational.

Proof: If √ 2 +√

3 is rational, then consider

√

3 +√ 2 √

3 −√ 2

= 1 which implies that √

3 −√

2 is rational. Hence, √

3 would be rational. It is impossible. So, √

2 +√

3 is irrational.

Remark: (1)√

p is an irrational if p is a prime.

Proof : If √

p ∈ Q, write √

p = ab, where g.c.d. (a, b) = 1. Then b2p = a2 ⇒ p

a2 ⇒ p |a (*)

Write a = pq. So,

b2p = p2q2 ⇒ b2 = pq2 ⇒ p

b2 ⇒ p |b . (*’)

By (*) and (*’), we get

p |g.c.d. (a, b) = 1 which implies that p = 1, a contradiction. So, √

p is an irrational if p is a prime.

(7)

Note: There are many and many methods to prove it. For example, the reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 19-21. (Chinese Version)

(2) Suppose a, b ∈ N. Prove that√ a +√

b is rational if and only if, a = k2 and b = h2 for some h, k ∈ N.

Proof : (⇐) It is clear.

(⇒) Consider

√ a +√

b

 √ a −√

b



= a2− b2, then √

a ∈ Q and √

b ∈ Q. Then it is clear that a = h2 and b = h2 for some h, k ∈ N.

1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b) / (cx + d) is usually irrational. When do exceptions occur?

Proof: We claim that (ax + b) / (cx + d) is rational if and only if ad = bc.

(⇒)If (ax + b) / (cx + d) is rational, say (ax + b) / (cx + d) = q/p. We consider two cases as follows.

(i) If q = 0, then ax + b = 0. If a 6= 0, then x would be rational. So, a = 0 and b = 0. Hence, we have

ad = 0 = bc.

(ii) If q 6= 0, then (pa − qc) x + (pb − qd) = 0. If pa − qc 6= 0, then x would be rational. So, pa − qc = 0 and pb − qd = 0. It implies that

qcb = qad ⇒ ad = bc.

(⇐)Suppose ad = bc. If a = 0, then b = 0 or c = 0. So, ax + b

cx + d =

 0 if a = 0 and b = 0

b

d if a = 0 and c = 0 . If a 6= 0, then d = bc/a. So,

ax + b

cx + d = ax + b

cx + bc/a = a (ax + b) c (ax + b) = a

c.

Hence, we proved that if ad = bc, then (ax + b) / (cx + d) is rational.

(8)

1.11

Given any real x > 0, prove that there is an irrational number between 0 and x.

Proof: If x ∈ Qc, we choose y = x/2 ∈ Qc. Then 0 < y < x. If x ∈ Q, we choose y = x/√

2 ∈ Q, then 0 < y < x.

Remark: (1) There are many and many proofs about it. We may prove it by the concept of Perfect set. The reader can see the book, Principles of Mathematical Analysis written by Walter Rudin, Theorem 2.43, pp 41. Also see the textbook, Exercise 3.25.

(2) Given a and b ∈ R with a < b, there exists r ∈ Qc, and q ∈ Q such that a < r < b and a < q < b.

Proof : We show it by considering four cases. (i) a ∈ Q, b ∈ Q. (ii) a ∈ Q, b ∈ Qc. (iii) a ∈ Qc, b ∈ Q. (iv) a ∈ Qc, b ∈ Qc.

(i) (a ∈ Q, b ∈ Q) Choose q = a+b2 and r = 1

2a + 1 − 1

2

 b.

(ii) (a ∈ Q, b ∈ Qc) Choose r = a+b2 and let c = 21n < b−a, then a+c := q.

(iii) (a ∈ Qc, b ∈ Q) Similarly for (iii).

(iv) (a ∈ Qc, b ∈ Qc) It suffices to show that there exists a rational number q ∈ (a, b) by (ii). Write

b = b0.b1b2· · · bn· ··

Choose n large enough so that

a < q = b0.b1b2· · · bn < b.

(It works since b − q = 0.000..000bn+1... ≤ 101n)

1.12 If a/b < c/d with b > 0, d > 0, prove that (a + c) / (b + d) lies bwtween the two fractions a/b and c/d

Proof: It only needs to conisder the substraction. So, we omit it.

Remark: The result of this exercise is often used, so we suggest the reader keep it in mind.

1.13

Let a and b be positive integers. Prove that√

2 always lies between the two fractions a/b and (a + 2b) / (a + b) . Which fraction is closer to √

2?

Proof : Suppose a/b ≤√

2, then a ≤√ 2b. So, a + 2b

a + b −√ 2 =

√2 − 1 √

2b − a

a + b ≥ 0.

(9)

In addition,

√

2 −a b

− a + 2b a + b −√

2



= 2√

2 − a

b +a + 2b a + b



= 2√

2 − a2+ 2ab + 2b2 ab + b2

= 1

ab + b2 h

2√ 2 − 2

ab + 2√

2 − 2

b2− a2i

≥ 1

ab + b2

"

 2√

2 − 2 a a

√2 + 2√

2 − 2

√a 2

2

− a2

#

= 0.

So, a+2ba+b is closer to √ 2.

Similarly, we also have if a/b >√

2, then a+2ba+b <√

2. Also, a+2ba+b is closer to √

2 in this case.

Remark: Note that a

b <√

2 < a + 2b a + b < 2b

a by Exercise 12 and 13.

And we know that a+2ba+b is closer to √

2. We can use it to approximate √ 2.

Similarly for the case

2b

a < a + 2b a + b <√

2 < a b. 1.14 Prove that√

n − 1 +√

n + 1 is irrational for every integer n ≥ 1.

Proof : Suppose that√

n − 1 +√

n + 1 is rational, and thus consider

√n + 1 +√

n − 1 √

n + 1 −√

n − 1

= 2 which implies that √

n + 1 −√

n − 1 is rational. Hence, √

n + 1 and√ n − 1 are rational. So, n − 1 = k2 and n + 1 = h2, where k and h are positive integer. It implies that

h = 3

2 and k = 1 2 which is absurb. So, √

n − 1 +√

n + 1 is irrational for every integer n ≥ 1.

(10)

1.15

Given a real x and an integer N > 1, prove that there exist integers h and k with 0 < k ≤ N such that |kx − h| < 1/N. Hint. Consider the N + 1 numbers tx − [tx] for t = 0, 1, 2, ..., N and show that some pair differs by at most 1/N.

Proof : Given N > 1, and thus consider tx − [tx] for t = 0, 1, 2, ..., N as follows. Since

0 ≤ tx − [tx] := at< 1,

so there exists two numbers ai and aj where i 6= j such that

|ai− aj| < 1

N ⇒ |(i − j) x − p| < 1

N, where p = [jx] − [ix] .

Hence, there exist integers h and k with 0 < k ≤ N such that |kx − h| < 1/N.

1.16

If x is irrational prove that there are infinitely many rational num- bers h/k with k > 0 such that |x − h/k| < 1/k2. Hint. Assume there are only a finite number h1/k1, ..., hr/kr and obtain a contradiction by apply- ing Exercise 1.15 with N > 1/δ, where δ is the smallest of the numbers

|x − hi/ki| .

Proof : Assume there are only a finite number h1/k1, ..., hr/kr and let δ = minri=1|x − hi/ki| > 0 since x is irrational. Choose N > 1/δ, then by Exercise 1.15, we have

1

N < δ ≤

x − h k

< 1 kN which implies that

1 N < 1

kN

which is impossible. So, there are infinitely many rational numbers h/k with k > 0 such that |x − h/k| < 1/k2.

Remark: (1) There is another proof by continued fractions. The reader can see the book, An Introduction To The Theory Of Numbers by Loo-Keng Hua, pp 270. (Chinese Version)

(2) The exercise is useful to help us show the following lemma. {ar + b : a ∈ Z, b ∈ Z} , where r ∈ Qc is dense in R. It is equivalent to {ar : a ∈ Z} , where r ∈ Qc is

dense in [0, 1] modulus 1.

(11)

Proof : Say {ar + b : a ∈ Z, b ∈ Z} = S, and since r ∈ Qc, then by Ex- ercise 1.16, there are infinitely many rational numbers h/k with k > 0 such that |kr − h| < 1k. Consider (x − δ, x + δ) := I, where δ > 0, and thus choos- ing k0 large enough so that 1/k0 < δ. Define L = |k0r − h0| , then we have sL ∈ I for some s ∈ Z. So, sL = (±) [(sk0) r − (sh0)] ∈ S. That is, we have proved that S is dense in R.

1.17

Let x be a positive rational number of the form x =

n

X

k=1

ak

k!,

where each ak is nonnegative integer with ak ≤ k − 1 for k ≥ 2 and an> 0.

Let [x] denote the largest integer in x. Prove that a1 = [x] , that ak = [k!x] − k [(k − 1)!x] for k = 2, ..., n, and that n is the smallest integer such that n!x is an integer. Conversely, show that every positive rational number x can be expressed in this form in one and only one way.

Proof : (⇒)First,

[x] =

"

a1+

n

X

k=2

ak k!

#

= a1+

" n X

k=2

ak k!

#

since a1 ∈ N

= a1 since

n

X

k=2

ak

k! ≤

n

X

k=2

k − 1 k! =

n

X

k=2

1

(k − 1)! − 1

k! = 1 − 1 n! < 1.

Second, fixed k and consider

k!x = k!

n

X

j=1

aj j! = k!

k−1

X

j=1

aj

j! + ak+ k!

n

X

j=k+1

aj j!

and

(k − 1)!x = (k − 1)!

n

X

j=1

aj

j! = (k − 1)!

k−1

X

j=1

aj

j! + (k − 1)!

n

X

j=k

aj j!.

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So,

[k!x] =

"

k!

k−1

X

j=1

aj

j! + ak+ k!

n

X

j=k+1

aj j!

#

= k!

k−1

X

j=1

aj

j! + ak since k!

n

X

j=k+1

aj j! < 1 and

k [(k − 1)!x] = k

"

(k − 1)!

k−1

X

j=1

aj

j! + (k − 1)!

n

X

j=k

aj j!.

#

= k (k − 1)!

k−1

X

j=1

aj

j! since (k − 1)!

n

X

j=k

aj j! < 1

= k!

k−1

X

j=1

aj j!

which implies that

ak = [k!x] − k [(k − 1)!x] for k = 2, ..., n.

Last, in order to show that n is the smallest integer such that n!x is an integer. It is clear that

n!x = n!

n

X

k=1

ak k! ∈ Z.

In addition,

(n − 1)!x = (n − 1)!

n

X

k=1

ak k!

= (n − 1)!

n−1

X

k=1

ak

k! +an

n

∈ Z since/ an n ∈ Z./ So, we have proved it.

(13)

(⇐)It is clear since every an is uniquely deermined.

Upper bounds

1.18 Show that the sup and the inf of a set are uniquely determined whenever they exists.

Proof : Given a nonempty set S (⊆ R) , and assume sup S = a and sup S = b, we show a = b as follows. Suppose that a > b, and thus choose ε = a−b2 , then there exists a x ∈ S such that

b < a + b

2 = a − ε < x < a which implies that

b < x

which contradicts to b = sup S. Similarly for a < b. Hence, a = b.

1.19 Find the sup and inf of each of the following sets of real numbers:

(a) All numbers of the form 2−p+ 3−q+ 5−r, where p, q, and r take on all positive integer values.

Proof : Define S = {2−p+ 3−q+ 5−r : p, q, r ∈ N }. Then it is clear that sup S = 12 + 13 +15, and inf S = 0.

(b) S = {x : 3x2− 10x + 3 < 0}

Proof: Since 3x2− 10x + 3 = (x − 3) (3x − 1) , we know that S = 13, 3 . Hence, sup S = 3 and inf S = 13.

(c) S = {x : (x − a) (x − b) (x − c) (x − d) < 0} , where a < b < c < d.

Proof: It is clear that S = (a, b) ∪ (c, d) . Hence, sup S = d and inf S = a.

1.20

Prove the comparison property for suprema (Theorem 1.16) Proof : Since s ≤ t for every s ∈ S and t ∈ T, fixed t0 ∈ T, then s ≤ t0 for all s ∈ S. Hence, by Axiom 10, we know that sup S exists. In addition, it is clear sup S ≤ sup T.

Remark: There is a useful result, we write it as a reference. Let S and T be two nonempty subsets of R. If S ⊆ T and sup T exists, then sup S exists and sup S ≤ sup T.

(14)

Proof : Since sup T exists and S ⊆ T, we know that for every s ∈ S, we have

s ≤ sup T.

Hence, by Axiom 10, we have proved the existence of sup S. In addition, sup S ≤ sup T is trivial.

1.21 Let A and B be two sets of positive numbers bounded above, and let a = sup A, b = sup B. Let C be the set of all products of the form xy, where x ∈ A and y ∈ B. Prove that ab = sup C.

Proof : Given ε > 0, we want to find an element c ∈ C such that ab − ε <

c. If we can show this, we have proved that sup C exists and equals ab.

Since sup A = a > 0 and sup B = b > 0, we can choose n large enough such that a − ε/n > 0, b − ε/n > 0, and n > a + b. So, for this ε0 = ε/n, there exists a0 ∈ A and b0 ∈ B such that

a − ε0 < a0 and b − ε0 < b0 which implies that

ab − ε0(a + b − ε0) < a0b0 since a − ε0 > 0 and b − ε0 > 0 which implies that

ab − ε

n(a + b) < a0b0 := c which implies that

ab − ε < c.

1.22 Given x > 0, and an integer k ≥ 2. Let a0 denote the largest integer

≤ x and, assumeing that a0, a1, ..., an−1 have been defined, let an denote the largest integer such that

a0+a1 k + a2

k2 + ... + an kn ≤ x.

Note: When k = 10 the integers a0, a1, ... are the digits in a decimal representation of x. For general k they provide a representation in the scale of k.

(a) Prove that 0 ≤ ai ≤ k − 1 for each i = 1, 2, ...

(15)

Proof : Choose a0 = [x], and thus consider [kx − ka0] := a1 then

0 ≤ k (x − a0) < k ⇒ 0 ≤ a1 ≤ k − 1 and

a0+ a1

k ≤ x ≤ a0+a1 k +1

k. Continue the process, we then have

0 ≤ ai ≤ k − 1 for each i = 1, 2, ...

and

a0 +a1 k + a2

k2 + ... + an

kn ≤ x < a0+a1 k + a2

k2 + ... + an kn + 1

kn. (*)

(b) Let rn= a0+ a1k−1+ a2k−2+ ... + ank−n and show that x is the sup of the set of rational numbers r1, r2, ...

Proof : It is clear by (a)-(*).

Inequality

1.23

Prove Lagrange’s identity for real numbers:

n

X

k=1

akbk

!2

=

n

X

k=1

a2k

! n X

k=1

b2k

!

− X

1≤k<j≤n

(akbj − ajbk)2.

Note that this identity implies that Cauchy-Schwarz inequality.

Proof : Consider

n

X

k=1

a2k

! n X

k=1

b2k

!

= X

1≤k,j≤n

a2kb2j =X

k=j

a2kb2j+X

k6=j

a2kb2j =

n

X

k=1

a2kb2k+X

k6=j

a2kb2j

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and

n

X

k=1

akbk

! n X

k=1

akbk

!

= X

1≤k,j≤n

akbkajbj =

n

X

k=1

a2kb2k+X

k6=j

akbkajbj

So,

n

X

k=1

akbk

!2

=

n

X

k=1

a2k

! n X

k=1

b2k

!

+X

k6=j

akbkajbj−X

k6=j

a2kb2j

=

n

X

k=1

a2k

! n X

k=1

b2k

!

+ 2 X

1≤k<j≤n

akbkajbj − X

1≤k<j≤n

a2kb2j + a2jb2k

=

n

X

k=1

a2k

! n X

k=1

b2k

!

− X

1≤k<j≤n

(akbj − ajbk)2.

Remark: (1) The reader may recall the relation with Cross Product and Inner Product, we then have a fancy formula:

kx × yk2+ |< x, y >|2 = kxk2kyk2, where x, y ∈ R3.

(2) We often write

< a, b >:=

n

X

k=1

akbk, and the Cauchy-Schwarz inequality becomes

|< x, y >| ≤ kxk kyk by Remark (1).

1.24 Prove that for arbitrary real ak, bk, ck we have

n

X

k=1

akbkck

!4

n

X

k=1

a4k

! n X

k=1

b2k

!2 n

X

k=1

c4k

! .

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Proof : Use Cauchy-Schwarz inequality twice, we then have

n

X

k=1

akbkck

!4

=

n

X

k=1

akbkck

!2

2

n

X

k=1

a2kc2k

!2 n

X

k=1

b2k

!2

n

X

k=1

a4k

!2 n

X

k=1

c4k

! n X

k=1

b2k

!2

=

n

X

k=1

a4k

! n X

k=1

b2k

!2 n

X

k=1

c4k

! .

1.25

Prove that Minkowski’s inequality:

n

X

k=1

(ak+ bk)2

!1/2

n

X

k=1

a2k

!1/2

+

n

X

k=1

b2k

!1/2

.

This is the triangle inequality ka + bk ≤ kak+kbk for n−dimensional vectors, where a = (a1, ..., an) , b = (b1, ..., bn) and

kak =

n

X

k=1

a2k

!1/2

.

Proof : Consider

n

X

k=1

(ak+ bk)2 =

n

X

k=1

a2k+

n

X

k=1

b2k+ 2

n

X

k=1

akbk

n

X

k=1

a2k+

n

X

k=1

b2k+ 2

n

X

k=1

a2k

!1/2 n

X

k=1

b2k

!1/2

by Cauchy-Schwarz inequality

=

n

X

k=1

a2k

!1/2

+

n

X

k=1

b2k

!1/2

2

.

(18)

So,

n

X

k=1

(ak+ bk)2

!1/2

n

X

k=1

a2k

!1/2

+

n

X

k=1

b2k

!1/2

.

1.26

If a1 ≥ ... ≥ an and b1 ≥ ... ≥ bn, prove that

n

X

k=1

ak

! n X

k=1

bk

!

≤ n

n

X

k=1

akbk

! .

Hint. P

1≤j≤k≤n(ak− aj) (bk− bj) ≥ 0.

Proof : Consider

0 ≤ X

1≤j≤k≤n

(ak− aj) (bk− bj) = X

1≤j≤k≤n

akbk+ ajbj − X

1≤j≤k≤n

akbj + ajbk

which implies that X

1≤j≤k≤n

akbj + ajbk≤ X

1≤j≤k≤n

akbk+ ajbj. (*)

Since X

1≤j≤k≤n

akbj + ajbk = X

1≤j<k≤n

akbj+ ajbk+ 2

n

X

k=1

akbk

= X

1≤j<k≤n

akbj+ ajbk+

n

X

k=1

akbk

! +

n

X

k=1

akbk

=

n

X

k=1

ak

! n X

k=1

bk

! +

n

X

k=1

akbk,

we then have, by (*)

n

X

k=1

ak

! n X

k=1

bk

! +

n

X

k=1

akbk ≤ X

1≤j≤k≤n

akbk+ ajbj. (**)

(19)

In addition, X

1≤j≤k≤n

akbk+ ajbj

=

n

X

k=1

akbk+ na1b1+

n

X

k=2

akbk+ (n − 1) a2b2+ ... +

n

X

k=n−1

akbk+ 2an−1bn−1+X

k=n

akbk

= n

n

X

k=1

akbk+ a1b1+ a2b2+ ... + anbn

= (n + 1)

n

X

k=1

akbk

which implies that, by (**),

n

X

k=1

ak

! n X

k=1

bk

!

≤ n

n

X

k=1

akbk

! .

Complex numbers

1.27 Express the following complex numbers in the form a + bi.

(a) (1 + i)3

Solution: (1 + i)3 = 1 + 3i + 3i2 + i3 = 1 + 3i − 3 − i = −2 + 2i.

(b) (2 + 3i) / (3 − 4i)

Solution: 2+3i3−4i = (2+3i)(3+4i)

(3−4i)(3+4i) = −6+17i25 = −625 + 1725i.

(c) i5+ i16

Solution: i5+ i16= i + 1.

(d) 12(1 + i) (1 + i−8)

Solution: 12(1 + i) (1 + i−8) = 1 + i.

1.28 In each case, determine all real x and y which satisfy the given relation.

(20)

(a) x + iy = |x − iy|

Proof : Since |x − iy| ≥ 0, we have

x ≥ 0 and y = 0.

(b) x + iy = (x − iy)2

Proof : Since (x − iy)2 = x2− (2xy) i − y2, we have x = x2− y2 and y = −2xy.

We consider tow cases: (i) y = 0 and (ii) y 6= 0.

(i) As y = 0 : x = 0 or 1.

(ii) As y 6= 0 : x = −1/2, and y = ±

3 2 . (c) P100

k=0ik= x + iy Proof : Since P100

k=0ik= 1−i1−i101 = 1−i1−i = 1, we have x = 1 and y = 0.

1.29 If z = x + iy, x and y real, the complex conjugate of z is the complex number ¯z = x − iy. Prove that:

(a) Conjugate of (z1+ z2) = ¯z1+ ¯z2

Proof : Write z1 = x1+ iy1 and z2 = x2+ iy2, then z1+ z2 = (x1+ x2) + i (y1+ y2)

= (x1+ x2) − i (y1+ y2)

= (x1− iy1) + (x2− iy2)

= ¯z1+ ¯z2.

(b) z1z2 = ¯z12

Proof : Write z1 = x1+ iy1 and z2 = x2+ iy2, then z1z2 = (x1x2− y1y2) + i (x1y2+ x2y1)

= (x1x2− y1y2) − i (x1y2+ x2y1) and

¯

z12 = (x1− iy1) (x2− iy2)

= (x1x2− y1y2) − i (x1y2+ x2y1) .

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So, z1z2 = ¯z12 (c) z ¯z = |z|2

Proof : Write z = x + iy and thus

z ¯z = x2+ y2 = |z|2.

(d) z + ¯z =twice the real part of z Proof : Write z = x + iy, then

z + ¯z = 2x, twice the real part of z.

(e) (z − ¯z) /i =twice the imaginary part of z Proof : Write z = x + iy, then

z − ¯z i = 2y, twice the imaginary part of z.

1.30 Describe geometrically the set of complex numbers z which satisfies each of the following conditions:

(a) |z| = 1

Solution: The unit circle centered at zero.

(b) |z| < 1

Solution: The open unit disk centered at zero.

(c) |z| ≤ 1

Solution: The closed unit disk centered at zero.

(d) z + ¯z = 1

Solution: Write z = x + iy, then z + ¯z = 1 means that x = 1/2. So, the set is the line x = 1/2.

(e) z − ¯z = i

(22)

Proof : Write z = x + iy, then z − ¯z = i means that y = 1/2. So, the set is the line y = 1/2.

(f) z + ¯z = |z|2

Proof : Write z = x + iy, then 2x = x2+ y2 ⇔ (x − 1)2+ y2 = 1. So, the set is the unit circle centered at (1, 0) .

1.31 Given three complex numbers z1, z2, z3 such that |z1| = |z2| = |z3| = 1 and z1+ z2+ z3 = 0. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with center at the origin.

Proof : It is clear that three numbers are vertices of triangle inscribed in the unit circle with center at the origin. It remains to show that |z1− z2| =

|z2 − z3| = |z3 − z1| . In addition, it suffices to show that

|z1− z2| = |z2− z3| . Note that

|2z1+ z3| = |2z3+ z1| by z1 + z2+ z3 = 0 which is equivalent to

|2z1+ z3|2 = |2z3+ z1|2 which is equivalent to

(2z1+ z3) (2¯z1+ ¯z3) = (2z3+ z1) (2¯z3+ ¯z1) which is equivalent to

|z1| = |z3| .

1.32 If a and b are complex numbers, prove that:

(a) |a − b|2 ≤ 1 + |a|2

1 + |b|2 Proof : Consider

1 + |a|2

1 + |b|2 − |a − b|2 = (1 + ¯aa) 1 + ¯bb − (a − b) ¯a − ¯b

= (1 + ¯ab) 1 + a¯b

= |1 + ¯ab|2 ≥ 0,

(23)

so, |a − b|2 ≤ 1 + |a|2

1 + |b|2

(b) If a 6= 0, then |a + b| = |a| + |b| if, and only if, b/a is real and nonnegative.

Proof : (⇒)Since |a + b| = |a| + |b| , we have

|a + b|2 = (|a| + |b|)2 which implies that

Re (¯ab) = |a| |b| = |¯a| |b|

which implies that

¯

ab = |¯a| |b|

which implies that

b a = ab¯

¯

aa = |¯a| |b|

|a|2 ≥ 0.

(⇐) Suppose that

b

a = k, where k ≥ 0.

Then

|a + b| = |a + ka| = (1 + k) |a| = |a| + k |a| = |a| + |b| .

1.33 If a and b are complex numbers, prove that

|a − b| = |1 − ¯ab|

if, and only if, |a| = 1 or |b| = 1. For which a and b is the inequality

|a − b| < |1 − ¯ab| valid?

Proof : (⇔) Since

|a − b| = |1 − ¯ab|

⇔ ¯a − ¯b (a − b) = (1 − ¯ab) 1 − a¯b

⇔ |a|2+ |b|2 = 1 + |a|2|b|2

⇔ |a|2 − 1

|b|2− 1 = 0

⇔ |a|2 = 1 or |b|2 = 1.

(24)

By the preceding, it is easy to know that

|a − b| < |1 − ¯ab| ⇔ 0 < |a|2− 1

|b|2− 1 .

So, |a − b| < |1 − ¯ab| if, and only if, |a| > 1 and |b| > 1. (Or |a| < 1 and

|b| < 1).

1.34 If a and c are real constant, b complex, show that the equation az ¯z + b¯z + ¯bz + c = 0 (a 6= 0, z = x + iy)

represents a circle in the x − y plane.

Proof : Consider z ¯z − b

−az −¯

¯b

−az + b

−a

"

 b

−a

#

= −ac + |b|2 a2 , so, we have

z −

 b

−a



2

= −ac + |b|2 a2 .

Hence, as |b|2 − ac > 0, it is a circle. As −ac+|b|a2 2 = 0, it is a point. As

−ac+|b|2

a2 < 0, it is not a circle.

Remark: The idea is easy from the fact

|z − q| = r.

We square both sides and thus

z ¯z − q ¯z − ¯qz + ¯qq = r2.

1.35 Recall the definition of the inverse tangent: given a real number t, tan−1(t) is the unique real number θ which satisfies the two conditions

−π

2 < θ < +π

2, tan θ = t.

If z = x + iy, show that

(a) arg (z) = tan−1 yx , if x > 0

(25)

Proof : Note that in this text book, we say arg (z) is the principal argu- ment of z, denoted by θ = arg z, where −π < θ ≤ π.

So, as x > 0, arg z = tan−1 yx .

(b) arg (z) = tan−1 yx + π, if x < 0, y ≥ 0

Proof : As x < 0, and y ≥ 0. The point (x, y) is lying on S = {(x, y) : x < 0, y ≥ 0} . Note that −π < arg z ≤ π, so we have arg (z) = tan−1 yx + π.

(c) arg (z) = tan−1 yx − π, if x < 0, y < 0 Proof : Similarly for (b). So, we omit it.

(d) arg (z) = π2 if x = 0, y > 0; arg (z) = −π2 if x = 0, y < 0.

Proof : It is obvious.

1.36 Define the folowing ”pseudo-ordering” of the complex numbers:

we say z1 < z2 if we have either

(i) |z1| < |z2| or (ii) |z1| = |z2| and arg (z1) < arg (z2) . Which of Axioms 6,7,8,9 are satisfied by this relation?

Proof : (1) For axiom 6, we prove that it holds as follows. Given z1 = r1ei arg(z1), and r2ei arg(z2), then if z1 = z2, there is nothing to prove it. If z1 6= z2, there are two possibilities: (a) r1 6= r2, or (b) r1 = r2 and arg (z1) 6=

arg (z2) . So, it is clear that axiom 6 holds.

(2) For axiom 7, we prove that it does not hold as follows. Given z1 = 1 and z2 = −1, then it is clear that z1 < z2 since |z1| = |z2| = 1 and arg (z1) = 0 < arg (z2) = π. However, let z3 = −i, we have

z1+ z3 = 1 − i > z2+ z3 = −1 − i since

|z1+ z3| = |z2+ z3| =√ 2 and

arg (z1+ z3) = −π

4 > −3π

4 = arg (z2+ z3) .

(3) For axiom 8, we prove that it holds as follows. If z1 > 0 and z2 > 0, then |z1| > 0 and |z2| > 0. Hence, z1z2 > 0 by |z1z2| = |z1| |z2| > 0.

(4) For axiom 9, we prove that it holds as follows. If z1 > z2 and z2 > z3, we consider the following cases. Since z1 > z2, we may have (a) |z1| > |z2| or (b) |z1| = |z2| and arg (z1) < arg (z2) .

As |z1| > |z2| , it is clear that |z1| > |z3| . So, z1 > z3.

(26)

As |z1| = |z2| and arg (z1) < arg (z2) , we have arg (z1) > arg (z3) . So, z1 > z3.

1.37 Which of Axioms 6,7,8,9 are satisfied if the pseudo-ordering is defined as follows? We say (x1, y1) < (x2, y2) if we have either (i) x1 < x2 or (ii) x1 = x2 and y1 < y2.

Proof: (1) For axiom 6, we prove that it holds as follows. Given x = (x1, y1) and y = (x2, y2) . If x = y, there is nothing to prove it. We consider x 6= y : As x 6= y, we have x1 6= x2 or y1 6= y2. Both cases imply x < y or y < x.

(2) For axiom 7, we prove that it holds as follows. Given x = (x1, y1) , y = (x2, y2) and z = (z1, z3) . If x < y, then there are two possibilities: (a) x1 < x2 or (b) x1 = x2 and y1 < y2.

For case (a), it is clear that x1+ z1 < y1+ z1. So, x + z < y + z.

For case (b), it is clear that x1+ z1 = y1+ z1 and x2+ z2 < y2+ z2. So, x + z < y + z.

(3) For axiom 8, we prove that it does not hold as follows. Consider x = (1, 0) and y = (0, 1) , then it is clear that x > 0 and y > 0. However, xy = (0, 0) = 0.

(4) For axiom 9, we prove that it holds as follows. Given x = (x1, y1) , y = (x2, y2) and z = (z1, z3) . If x > y and y > z, then we consider the following cases. (a) x1 > y1, or (b) x1 = y1.

For case (a), it is clear that x1 > z1. So, x > z.

For case (b), it is clear that x2 > y2. So, x > z.

1.38 State and prove a theorem analogous to Theorem 1.48, expressing arg (z1/z2) in terms of arg (z1) and arg (z2) .

Proof : Write z1 = r1ei arg(z1) and z2 = r2ei arg(z2), then z1

z2 = r1

r2ei[arg(z1)−arg(z2)]. Hence,

arg z1 z2



= arg (z1) − arg (z2) + 2πn (z1, z2) , where

n (z1, z2) =

0 if − π < arg (z1) − arg (z2) ≤ π 1 if − 2π < arg (z1) − arg (z2) ≤ −π

−1 if π < arg (z1) − arg (z2) < 2π .

(27)

1.39 State and prove a theorem analogous to Theorem 1.54, expressing Log (z1/z2) in terms of Log (z1) and Log (z2) .

Proof : Write z1 = r1ei arg(z1) and z2 = r2ei arg(z2), then z1

z2 = r1

r2ei[arg(z1)−arg(z2)]. Hence,

Log (z1/z2) = log

z1 z2

+ i arg z1 z2



= log |z1| − log |z2| + i [arg (z1) − arg (z2) + 2πn (z1, z2)] by xercise 1.38

= Log (z1) − Log (z2) + i2πn (z1, z2) .

1.40

Prove that the nth roots of 1 (also called the nth roots of unity) are given by α, α2, ..., αn, where α = e2πi/n, and show that the roots 6= 1 satisfy the equation

1 + x + x2+ ... + xn−1 = 0.

Proof : By Theorem 1.51, we know that the roots of 1 are given by α, α2, ..., αn, where α = e2πi/n. In addition, since

xn = 1 ⇒ (x − 1) 1 + x + x2+ ... + xn−1 = 0 which implies that

1 + x + x2+ ... + xn−1 = 0 if x 6= 1.

So, all roots except 1 satisfy the equation

1 + x + x2+ ... + xn−1 = 0.

1.41

(a) Prove that |zi| < eπ for all complex z 6= 0.

Proof : Since

zi = eiLog(z) = e− arg(z)+i log|z|

,

(28)

we have

zi

= e− arg(z) < eπ by −π < arg (z) ≤ π.

(b) Prove that there is no constant M > 0 such that |cos z| < M for all complex z.

Proof : Write z = x + iy and thus,

cos z = cos x cosh y − i sin x sinh y which implies that

|cos x cosh y| ≤ |cos z| . Let x = 0 and y be real, then

ey 2 ≤ 1

2

ey + e−y

≤ |cos z| .

So, there is no constant M > 0 such that |cos z| < M for all complex z.

Remark: There is an important theorem related with this exercise. We state it as a reference. (Liouville’s Theorem) A bounded entire function is constant. The reader can see the book, Complex Analysis by Joseph Bak, and Donald J. Newman, pp 62-63. Liouville’s Theorem can be used to prove the much important theorem, Fundamental Theorem of Algebra.

1.42 If w = u + iv (u, v real), show that zw = eu log|z|−v arg(z)

ei[v log|z|+u arg(z)]

.

Proof : Write zw = ewLog(z), and thus

wLog (z) = (u + iv) (log |z| + i arg (z))

= [u log |z| − v arg (z)] + i [v log |z| + u arg (z)] . So,

zw = eu log|z|−v arg(z)ei[v log|z|+u arg(z)].

(29)

1.43 (a) Prove that Log (zw) = wLog z +2πin.

Proof : Write w = u + iv, where u and v are real. Then Log (zw) = log |zw| + i arg (zw)

= logeu log|z|−v arg(z) + i [v log |z| + u arg (z)] + 2πin by Exercise1.42

= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin.

On the other hand,

wLogz + 2πin = (u + iv) (log |z| + i arg (z)) + 2πin

= u log |z| − v arg (z) + i [v log |z| + u arg (z)] + 2πin.

Hence, Log (zw) = wLog z +2πin.

Remark: There is another proof by considering eLog(zw)= zw = ewLog(z) which implies that

Log (zw) = wLogz + 2πin for some n ∈ Z.

(b)

Prove that (zw)α= ze2πinα, where n is an integer.

Proof : By (a), we have

(zw)α = eαLog(zw)= eα(wLogz+2πin) = eαwLogze2πinα = zαwe2πinα, where n is an integer.

1.44

(i) If θ and a are real numbers, −π < θ ≤ π, prove that (cos θ + i sin θ)a = cos (aθ) + i sin (aθ) .

Proof : Write cos θ + i sin θ = z, we then have

(cos θ + i sin θ)a = za = eaLogz = ea[log|e|+i arg(e)] = eiaθ

= cos (aθ) + i sin (aθ) .

參考文獻

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