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CHAPTER 1

THE REAL AND

COMPLEX NUMBER SYSTEMS

1.1 INTRODUCTION

Mathematical analysis studies concepts related in some way to real numbers, so

we begin our study of analysis with a discussion of the real-number system.

Several methods are used to introduce real numbers. One method starts with

the positive integers 1, 2, 3, ... as undefined concepts and uses them to build a

larger system, the positive rational numbers (quotients of positive integers), their

negatives, and zero. The rational numbers, in turn, are then used to construct the

irrational numbers, real numbers like -,/2 and iv which are not rational. The rational

and irrational numbers together constitute the real-number system.

Although these matters are an important part of the foundations of math-

ematics, they will not be described in detail here. As a matter of fact, in most

phases of analysis it is only the properties of real numbers that concern us, rather

than the methods used to construct them. Therefore, we shall take the real numbers

themselves as undefined objects satisfying certain axioms from which further

properties will be derived. Since the reader is probably familiar with most of the

properties of real numbers discussed in the next few pages, the presentation will

be rather brief. Its purpose is to review the important features and persuade the

reader that, if it were necessary to do so, all the properties could be traced back

to the axioms. More detailed treatments can be found in the references at the end

of this chapter.

For convenience we use some elementary set notation and terminology. Let

S denote a set (a collection of objects). The notation x e S means that the object x

is in the set S, and we write x 0 S to indicate that x is not in S.

A set S is said to be a subset of T, and we write S s T, if every object in S is

also in T. A set is called nonempty if it contains at least one object.

We assume there exists a nonempty set R of objects, called real numbers,

which satisfy the ten axioms listed below. The axioms fall in a natural way into

three groups which we refer to as the field axioms, the order axioms, and the

completeness axiom (also called the least-upper-bound axiom or the axiom of

continuity).

1.2 THE FIELD AXIOMS

Along with the-set R of real numbers we assume the existence of two operations,

called addition and multiplication, such that for every pair of real numbers x and y

2Real and Complex Numbcr Systems Ax.1

the sum x + y and the product xy are real numbers uniquely determined by x

and y satisfying the following axioms. (In the axioms that appear below, x, y,

z represent arbitrary real numbers unless something is said to the contrary.)

Axiom 1. x + y = y + x, xy = yx (commutative laws).

Axiom 2. x + (y + z) = (x + y) + z, x(yz) _ (xy)z (associative laws).

Axiom 3. x(y + z) = xy + xz (distributive law).

Axiom 4. Given any two real numbers x and y, there exists a real number z such that

x + z = y. This z is denoted by y - x; the number x - x is denoted by 0. (It

can be proved that 0 is independent of x.) We write - x for 0 - x and call - x the

negative of x.

Axiom S. There exists at least one real number x 96 0. If x and y are two real

numbers with x 0, then there exists a real number z such that xz = y. This z is

denoted by y/x; the number x/x is denoted by 1 and can be shown to be independent of

x. We write x-1 for 1/x if x 0 and call x-1 the reciprocal of x.

From these axioms all the usual laws of arithmetic can be derived; for example,

-(-x)=x,(x-1)-1=x, -(x-y)=y-x,x-y=x+(-y),etc. (For

a more detailed explanation, see Reference 1.1.)

1.3 THE ORDER AXIOMS

We also assume the existence of a relation < which establishes an ordering among

the real numbers and which satisfies the following axioms:

Axiom 6. Exactly one of the relations x = y, x < y, x > y holds.

NOTE. x > y means the same as y < x.

Axiom 7. If x < y, then for every z we have x + z < y + z.

Axiom 8. If x > 0 and y > 0, then xy > 0.

Axiom 9. If x > y and y > z, then x > z.

NOTE. A real number x is called positive if x > 0, and negative if x < 0. We

denote by R+ the set of all positive real numbers, and by R- the set of all negative

real numbers.

From these axioms we can derive the usual rules for operating with inequalities.

For example, if we have x < y, then xz < yz if z is positive, whereas xz > yz if

z is negative. Also, if x > y and z > w where both y and w are positive, then

xz > yw. (For a complete discussion of these rules see Reference 1.1.)

NOTE. The symbolism x < y is used as an abbreviation for the statement:

46 x < y or Y .

Th. 1.1 Intervals 3

Thus we have 2 < 3 since 2 < 3; and 2 < 2 since 2 = 2. The symbol >- is

similarly used. A real number x is called nonnegative if x >- 0. A pair of simul-

taneous inequalities such as x < y, y < z is usually written more briefly as

x<y<z.

The following theorem, which is a simple consequence of the foregoing axioms,

is often used in proofs in analysis.

Theorem 1.1. Given real numbers a and b such that

a < b + e for every e > 0. (1)

Then a S b.

Proof. If b < a, then inequality (1) is violated for e = (a - b)/2 because

b+e=b+a-b=a+b<a+a=a

2 2 2

Therefore, by Axiom 6 we must have a < b.

Axiom 10, the completeness axiom, will be described in Section 1.11.

1.4 GEOMETRIC REPRESENTATION OF REAL NUMBERS

The real numbers are often represented geometrically as points on a line (called

the real line or the real axis). A point is selected to represent 0 and another to

represent 1, as shown in Fig. I.I. This choice determines the scale. Under an

appropriate set of axioms for Euclidean geometry, each point on the real line

corresponds to one and only one real number and, conversely, each real number

is represented by one and only one point on the line. It is customary to refer to

the point x rather than the point representing the real number x.

01xyFigure 1.1

The order relation has a simple geometric interpretation. If x < y, the point

x lies to the left of the point y, as shown in Fig. 1.1. Positive numbers lie to the

right of 0, and negative numbers to the left of 0. If a < b, a point x satisfies the

inequalities a < x < b if and only if x is between a and b.

1.5 INTERVALS

The set of all points between a and b is called an interval. Sometimes it is important

to distinguish between intervals which include their endpoints and intervals which

do not.

NOTATION. The notation {x: x satisfies P} will be used to designate the set of

all real numbers x which satisfy property P.

4Real and Complex Number Systems Def. 1.2

Definition 1.2. Assume a < b. The open interval (a, b) is defined to be the set

(a, b) = {x : a < x < b}.

The closed interval [a, b] is the set {x: a < x < b}. The half-open intervals

(a, b] and [a, b) are similarly defined, using the inequalities a < x :!5; b and

a < x < b, respectively. Infinite intervals are defined as follows:

(a, + oo) = {x : x > a), [a, + oo) = {x : x a},

(- oo, a) = {x: x < a}, (- oo, a] = {x : x < a}.

The real line R is sometimes referred to as the open interval (- co, + co). A

single point is considered as a "degenerate" closed interval.

NOTE. The symbols + oo and - oo are used here purely for convenience in notation

and are not to be considered as being real numbers. Later we shall extend the

real-number system to include these two symbols, but until this is done, the reader

should understand that all real numbers are "finite."

1.6 INTEGERS

This section describes the integers, a special subset of R. Before we define the

integers it is convenient to introduce first the notion of an inductive set.

Definition 1.3. A set of real numbers is called an inductive set if it has the following

two properties:

a) The number 1 is in the set.

b) For every x in the set, the number x + 1 is also in the set.

For example, R is an inductive set. So is the set V. Now we shall define the

positive integers to be those real numbers which belong to every inductive set.

Definition 1.4. A real number is called a positive integer if it belongs to every

inductive set. The set of positive integers is denoted by V.

The set Z+ is itself an inductive set. It contains the number 1, the number

1 + 1 (denoted by 2), the number 2 + 1 (denoted by 3), and so on. Since Z' is a

subset of every inductive set, we refer to Z+ as the smallest inductive set. This

property of Z+ is sometimes called the principle of induction. We assume the

reader is familiar with proofs by induction which are based on this principle.

(See Reference 1.1.) Examples of such proofs are given in the next section.

The negatiyes of the positive integers are called the negative integers. The

positive integers, together with the negative integers and 0 (zero), form a set Z

which we call simply the set of integers.

1.7 THE UNIQUE FACTORIZATION THEOREM FOR INTEGERS

If n and d are integers and if n = cd for some integer c, we say d is a divisor of n,

or n is a multiple of d, and we write d In (read : d divides n). An integer n is called

Th. 1.7 Unique Factorization Theorem 5

a prime if n > 1 and if the only positive divisors of n are 1 and n. If n > 1 and n

is not prime, then n is called composite. The integer 1 is neither prime nor composite.

This section derives some elementary results on factorization of integers,

culminating in the unique factorization theorem, also called the fundamental theorem

of arithmetic.

The fundamental theorem states that (1) every integer n > 1 can be represented

as a product of prime factors, and (2) this factorization can be done in only one

way, apart from the order of the factors. It is easy to prove part (1).

Theorem 1.5. Every integer n > 1 is either a prime or a product of primes.

Proof. We use induction on n. The theorem holds trivially for n = 2. Assume

it is true for every integer k with 1 < k < n. If n is not prime it has a positive

divisor d with 1 < d < n. Hence n = cd, where 1 < c < n. Since both c and

d are <n, each is a prime or a product of primes; hence n is a product of primes.

Before proving part (2), uniqueness of the factorization, we introduce some

further concepts.

If d l a and d 1 b we say d is a common divisor of a and b. The next theorem

shows that every pair of integers a and b has a common divisor which is a linear

combination of a and b.

Theorem 1.6. Every pair of integers a and b has a common divisor d of the form

d = ax + by

where x and y are integers. Moreover, every common divisor of a and b divides

this d.

Proof. First assume that a > 0, b > 0 and use induction on n = a + b. If

n = 0 then a = b = 0, and we can take d = 0 with x = y = 0. Assume, then,

that the theorem has been proved for 0, 1, 2, ... , n - 1. By symmetry, we can

assume a >- b. If b = 0 take d = a, x = 1, y = 0. If b >- 1 we can apply the

induction hypothesis to a - b and b, since their sum is a = n - b < n - 1.

Hence there is a common divisor d of a - b and b of the form d = (a - b)x + by.

This d also divides (a - b) + b = a, so d is a common divisor of a and b and

we have d = ax + (y - x)b, a linear combination of a and b. To complete the

proof we need to show that every common divisor divides d. Since a common

divisor divides a and b, it also divides the linear combination ax + (y - x)b = d.

This completes the proof if a >- 0 and b Z 0. If one or both of a and b is negative;

apply the result just proved to lal and obi.

NOTE. If d is a common divisor of a and b of the form d = ax + by, then - d is

also a divisor of the same form, -d = a(- x) + b(-y). Of these two common

divisors, the nonnegative one is called the greatest common divisor of a and b,

and is denoted by gcd(a, b) or, simply by (a, b). If (a, b) = I then a and b are

said to be relatively prime.

Theorem 1.7(Euclid's Lemma). If albc and (a, b) = 1, then alc.

6Real and Complex Number Systems Th. 1.8

Proof. Since (a, b) = 1 we can write 1 = ax + by. Therefore c = acx + bcy.

But aiacx and ajbcy, so aic.

Theorem 1.8. If a prime p divides ab, then pia or pib. More generally, if a prime p,

divides a product a1 . ak, then p divides at least one of the factors.

Proof. Assume plab and that p does not divide a. If we prove that (p, a) = 1,

then Euclid's Lemma implies pib. Let d = (p, a). Then dip so d = I or d = p.

We cannot have d = p because dia but p does not divide a. Hence d = 1. To

prove the more general statement we use induction on k, the number of factors.

Details are left to the reader.

Theorem 1.9 (Unique factorization theorem). Every integer n > I can be repre-

sented as a product of prime factors in only one way, apart from the order of the

factors. .

Proof. We use induction on n. The theorem is true for n = 2. Assume, then,

that it is true for all integers greater than 1 and less than n. If n is prime there is

nothing more to prove. Therefore assume that n is composite and that n has two

factorizations into prime factors, say

n = PiP2 ' * *Ps = g1g2 "' qt (2)

We wish to show that s = t and that each p equals some q. Since pt divides the

product q 1 q 2 q, it divides at least one factor. Relabel the q's if necessary so

that p1l q 1. Then pt = q 1 since both pt and q 1 are primes. In (2) we cancel pt

on both sides to obtain

n -=P2...PS=g2"'gr

Since n .is composite, 1 < n/p1 < n; so by the induction hypothesis the two

factorizations of n/p1 are identical, apart from the order of the factors. Therefore

the same is true in (2) and the proof is complete.

1.8 RATIONAL NUMBERS

Quotients of integers alb (where b : 0) are called rational numbers. For example,

1/2, - 7/5, and 6 are rational numbers. The set of rational numbers, which we

denote by Q, contains Z as a subset. The reader should note that all the field

axioms and the order axioms are satisfied by Q.

We assume that the reader is familiar with certain elementary properties of

rational numbers. For example, if a and b are rational, their average (a + b)/2 is

also rational and lies between a and b. Therefore between any two rational numbers

there are infinitely many rational numbers, which implies that if we are given a

certain rational number we cannot speak of the "next largest" rational number.

7b. 1.11 Irrational Numbers 7

1.9 IRRATIONAL NUMBERS

Real numbers that are not rational are called irrational. For example, the numbers

-,/2, e, ir and e" are irrational.

Ordinarily it is not too easy to prove that some particular number is irrational.

There is no simple proof, for example, of the irrationality of e. However, the

irrationality of certain numbers such as and 3 is not too difficult to establish

and, in fact, we easily prove the following :

Theorem 1.10. If n is a positive integer which is not a perfect square, then i is

irrational.

Proof Suppose first that n contains no square factor > 1. We assume that is

rational and obtain a contradiction. Let n = a/b, where a and b are integers

having no factor in common. Then nb2 = a2 and, since the left side of this equation

is a multiple of n, so too is a2. However, if a2 is a multiple of n, a itself must be a

multiple of n, since n has no square factors > 1. (This is easily seen by examining

the factorization of a into its prime factors.) This means that a = cn, where c is

some integer. Then the equation nb2 = a2 becomes nb2 = c2n2, or b2 = nc2.

The same argument shows that b must also be a multiple of n. Thus a and b are

both multiples of n, which contradicts the fact that they have no factor in common.

This completes the proof if n has no square factor > 1.

If n has a square factor, we can write n = m2k, where k > 1 and k has no

square factor > 1. Then = m,/k; and if '/n were rational, the number fk-

would also be rational, contradicting that which was just proved.

A different type of argument is needed to prove that the number e is irrational.

(We assume familiarity with the exponential e" from elementary calculus and its

representation as an infinite series.)

Theorem 1.11. If ex = 1 + x + x2/2! + x3/3! + + x"/n! + , then the

number e is irrational.

Proof. We shall prove that a-1 is irrational. The series for a-1 is an alternating

series with terms which decrease steadily in absolute value. In such an alternating

series the error made by stopping at the nth term has the algebraic sign of the first

neglected term and is less in absolute value than the first neglected term. Hence,

if s" = Ek=o (- 1)k/k!, we have the inequality

-1

0 < e - s2k_1 < (2k) ! '

from which we obtain

0 < (2k - 1)! (e-1 - s2k-1) < 2k 2' (3)

for any integer -k z 1. Now (2k - D! s2k _ 1 is always an integer. If e-' were

rational, then we could choose k so large that (2k - 1)! a-1 would also be an

8Real and Complex Number Systems Def. 1.12

integer. Because of (3) the difference of these two integers would be a number

between 0 and 1, which is impossible. Thus a-1 cannot be rational, and hence e

cannot be rational.

NOTE. For a proof that iv is irrational, see Exercise 7.33.

The ancient Greeks were aware of the existence of irrational numbers as early

as 500 B.c. However, a satisfactory theory of such numbers was not developed

until late in the nineteenth century, at which time three different theories were.

introduced by Cantor, Dedekind, and Weierstrass. For an account of the theories

of Dedekind and Cantor and their equivalence, see Reference 1.6.

1.10 UPPER BOUNDS, MAXIMUM ELEMENT, LEAST UPPER BOUND

(SUPREMUM)

Irrational numbers arise in algebra when we try to solve certain quadratic equa-

tions. For example, it is desirable to have a real number x such that x2 = 2. From

the nine axioms listed above we cannot prove that such an x exists in R because

these nine axioms are also satisfied by Q and we have shown that there is no

rational number whose square is 2. The completeness axiom allows us to introduce

irrational numbers in the real-number system, and it gives the real-number system

a property of continuity that is fundamental to many theorems in analysis.

Before we describe the completeness axiom, it is convenient to introduce

additional terminology and notation.

Definition 1.12. Let S be a set of real numbers. If there is a real number b such

that x < b for every x in S, then b is called an upper bound for S and we say that

S is bounded above by b.

We say an upper bound because every number greater than b will also be an

upper bound. If an upper bound b is also a member of S, then b is called the

largest member or the maximum element of S. There can be at most one such b.

If it exists, we write b = max S.

A set with no upper bound is said to be unbounded above.

Definitions of the terms lower bound, bounded below, smallest member (or

minimum element) can be similarly formulated. If S has a minimum element we

denote it by min S.

Examples

1. The set R+ = (0, + oo) is unbounded above. It has no upper bounds and no max-

imum element. It is bounded below by 0 but has no minimum element.

2. The closed interval S = [0, 1 ] is bounded above by 1 and is bounded below by 0.

In fact, max S = 1 and min S = 0.

3. The half-open interval S = [0, 1) is bounded above by 1 but it has no maximum

element. Its minimum element is 0.

Th. 1.14 Some Properties of the Supremum 9

For sets like the one in Example 3, which are bounded above but have no

maximum element, there is a concept which takes the place of the maximum ele-

ment. It is called the least upper bound or supremum of the set and is defined as

follows :

Definition 1.13. Let S be a set of real numbers bounded above. A real number b is

called a least upper bound for S if it has the following two properties:

a) b is an upper bound for S.

b) No number less than b is an upper bound for S.

Examples. If S = [0, 1 ] the maximum element 1 is also a least upper bound for S. If

S = [0, 1) the number 1 is a least upper bound for S, even though S has no maximum

element.

It is an easy exercise to prove that a set cannot have two different least upper

bounds. Therefore, if there is a least upper bound for S, there is only one and we

can speak of the least upper bound.

It is common practice to refer to the least upper bound of a set by the more

concise term supremum, abbreviated sup. We shall adopt this convention and write

b = sup S

to indicate that b is the supremum of S. If S has a maximum element, then

max S = sup S.

The greatest lower bound, or infimum of S, denoted by inf S, is defined in an

analogous fashion.

1.11 THE COMPLETENESS AXIOM

Our final axiom for the real number system involves the notion of supremum.

Axiom 10. Every nonempty set S of real 'numbers which is bounded above has a

supremum; that is, there is a real number b such that b = sup S.

As a consequence of this axiom it follows that every nonempty set of real

numbers which is bounded below has an infimum.

1.12 SOME PROPERTIES OF THE SUPREMUM

This section discusses some fundamental properties of the supremum that will be

useful in this text. There is a corresponding set of properties of the infimum that

the reader should formulate for himself.

The first property shows that a set with a supremum contains numbers arbi-

trarily close to its supremum.

Theorem 1.14 (Approximation property). Let S be a nonempty set of real numbers

with a supremum, say b = sup S. Then for every a < b there is some x in S such

that a<x<b.

10 Real and Complex Number Systems Th. 1.15

Proof. First of all, x < b for all x in S. If we had x < a for every x in S, then a

would be an upper bound for S smaller than the least upper bound. Therefore

x > a for at least one x in S.

Theorem 1.15 (Additive property). Given nonempty subsets A and B of R, let C

denote the set C= {x + y:xc-A, yEB}.

If each of A and B has a supremum, then C has a supremum and

sup C = sup A + sup B.

Proof. Let a = sup A, b = sup B. If z e C then z = x + y, where x c- A,

y e B, so z = x + y <- a + b. Hence a + b is an upper bound for C, so C has a

supremum, say c = sup C, and c < a + b. We show next that a + b < c.

Choose any e > 0. By Theorem 1.14 there is an x in A and a y in B such that

a - E<x and b - E<y.

Adding these inequalities we find

a + b - 2E<x+y<c.

Thus, a + b < c + 2E for every e > 0 so, by Theorem 1.1, a + b < c.

The proof of the next theorem is left as an exercise for the reader.

Theorem 1.16 (Comparison property). Given nonempty subsets S and T of R such

that s <- t for every s in S and tin T. If T has a supremum then S has a supremum

and sup S < sup T.

1.13 PROPERTIES OF THE INTEGERS DEDUCED FROM THE

COMPLETENESS AXIOM

Theorem 1.17. The set Z+ of positive integers 1, 2, 3, ... is unbounded above.

Proof. If Z+ were bounded above then Z+ would have a supremum, say a =

sup V. By Theorem 1.14 we would have a - 1 < n for some n in Z+. Then

n + 1 > a for this n. Since n + 1 e Z+ this contradicts the fact that a = sup Z+.

Theorem 1.18. For every real x there is a positive integer n such that n > x.

Proof. If this were not true, some x would be an upper bound for Z+, contra-

dicting Theorem 1.17. -

1.14 THE ARCHIMEDEAN PROPERTY OF THE REAL NUMBER SYSTEM

The next theorem describes the Archimedean property of the real number system.

Geometrically, it tells us that any line segment, no matter how long, can be

Th. 1.20 Finite Decimal Approximations 11

covered by a finite number of line segments of a given positive length, no matter

how small.

Theorem 1.19. If x > 0 and if y is an arbitrary real number, there is a positive

integer n such that nx > y.

Proof. Apply Theorem 1.18 with x replaced by y/x.

1.15 RATIONAL NUMBERS WITH FINITE DECIMAL REPRESENTATION

A real number of the form r=ao+++---+

10 102 10"

where ao is a nonnegative integer and al, ... , a are integers satisfying 0 < at < 9,

is usually written more briefly as follows:

r =

afinite decimal representation of r. For example,

1__ 5 =0.5, 1 __ 2 =0.02, 29=7+ 2+ 5 =7.25.

210 50 102 410 102

Real numbers like these are necessarily rational and, in fact, they all have the form

r = a/10", where a is an integer. However, not all rational numbers can be ex-

pressed with finite decimal representations. For example, if } could be so expressed,

then we would have I = a/10" or 3a = 10" for some integer a. But this is im-

possible since 3 does not divide any power of 10.

1.16 FINITE DECIMAL APPROXIMATIONS TO REAL NUMBERS

This section uses the completeness axiom to show that real numbers can be

approximated to any desired degree of accuracy by rational numbers with finite

decimal representations.

Theorem 1.20. Assume x > 0. Then for every integer n >_ 1 there is a finite

decimal r" = ao. ala2 -- a" such that

r"<x<r"+-. Ion

Proof. Let S be the set of all nonnegative integers <x. Then S is nonempty,

since 0 e S, and S is bounded above by x. Therefore S has a supremum, say

ao = sup S. It is easily verified that ao e S, so ao is a nonnegative integer. We

call ao the greatest integer in x, and we write ao = [x]. Clearly, we have

ao<x<ao+1.

12 Real and Complex Number Systems Th. 1.20

Now let a1 = [lox - 10ao], the greatest integer in lOx - 10ao. Since

0 < lOx - 10ao = 10(x - ao) < 10, we have 0 < a1 5 9 and

a1<lox -l0ao<a1+l.

In other words, a1 is the largest integer satisfying the inequalities

ao+io_<<x<ao+a11+01

More generally, having chosen a1, ... , an_1 with 0 <- ai 5 9, let a" be the

largest integer satisfying the inequalities

am a" < x < ao + a1 ++ a"

ao+0+...+10"o... 10" (4)

Then 0 < a" < 9 and we have

r"<x<r"+ion

where r" = ao. a1a2 a". This completes the proof. It is easy to verify that x is

actually the supremum of the set of rational numbers r1, r2, ... .

1.17 INFINITE DECIMAL REPRESENTATIONS OF REAL NUMBERS

The integers ao, a1, a2, ... obtained in the proof of Theorem 1.20 can be used to

define an infinite decimal representation of x. We write

x = ao. ala2 .. .

to mean that a" is the largest integer satisfying (4). For example, if x = } we find

ao = 0, a1 = 1, a2 = 2, a3 = 5, and a" = 0 for all n Z 4. Therefore we can

write * = 0.125000

If we interchange the inequality signs S and < in (4), we obtain a slightly

different definition of decimal expansions. The finite decimals r,, satisfy r" < x <

r" + 10-" although the digits ao, a1, a2i ... need not be the same as those in (4).

For example, if we apply this second definition to x = } we find the infinite decimal

representation * = 0.124999

The fact that a real number might have two different decimal representations is

merely a reflection of the fact that two different sets of real numbers can have the

same supremum.

1.18 ABSOLUTE VALUES AND THE TRIANGLE INEQUALITY

Calculations with inequalities arise quite frequently in analysis. They are of

particular importance in dealing with the notion of absolute value. If x is any real

Th. 1.22 Cauchy-Schwarz Inequality

number, the absolute value of x, denoted by lxj, is defined as follows:

Ixl x, ifx 0,

-x, ifx50.

A fundamental inequality concerning absolute values is given in the following:

Theorem 1.21. If a >- 0, then we have the inequality Ixl 5 a if, and only if,

-a<xSa.

Proof. From the definition of Ixl, we have the inequality - Ixl 5 x < Ixl, since

x = Ixl or x = - lxl. If we assume that lxl 5 a, then we can write -a 5 - Ixl 5

x 5 lxl 5 a and thus half of the theorem is proved. Conversely, let us assume

-a 5 x 5 a. Then ifx >_ 0, we have lxl = x 5 a, whereas ifx < 0, we have

lxl = -x 5 a. In either case we have lxi 5 a and the theorem is proved.

We can use this theorem to prove the triangle inequality.

Theorem 1.22. For arbitrary real x and y we have

Ix + yl <- Ixl + IYI (the triangle inequality).

Proof. We have - lxl -< x 5 lxl and -I yj <- y <- l yl. Addition gives us

-(Ixl + IYI) <- x + y 5 lxl + lyl, and from Theorem 1.21 we conclude that

Ix + yl 5 lxl + l yl. This proves the theorem.

The triangle inequality is often used in other forms. For example, if we take

x = a - c and y = c - b in Theorem 1.22 we find

la-bl5 la - cl+lc - bl.

Also, from Theorem 1.22 we have Ixl >- Ix + yl - IYI Taking x = a + b,

y = - b, we obtain Ia + bl z lal - lbl.

Interchanging a and b we also find J a + bl > I bI - j al = - (lal - Ibl), and

hence la + bl z hlal - IbIl.

By induction we can also prove the generalizations

Ix1 + x2 + ... + x.l 5 Ix1l + Ixzl + ... +

and Ixl + x2 + ... + x.l z Ix1l - Ix21 14-

1.19 THE CAUCHY-SCHWARZ INEQUALITY

We shall now derive another inequality which is often used in analysis.

14 Real and Complex Number Systems Tb.1.23

Theorem 1.23 (Cauchy-Schwarz inequality). If a1,.. ., an and b1i ... , bn are

arbitrary real numbers, we have

Ck = 1

akbkl2 - E a)(kr b k) .

Moreover, if some a. 0 equality holds if and only if there is a real x such that

akx + bk = 0 for each k = 1, 2,,, .. , n.

Proof. A sum of squares can never be negative. Hence we have

(akx + bk)2 > 0

k=1

for every real x, with equality if and only if each term is zero. This inequality can

be written in the form

where Ax2+2Bx+C>-0,

Aak, Bakbk,

k=1

k=11

C=Ebk. k=1

If A > 0, put x = -B/A to obtain B2 - AC < 0, which is the desired inequality.

If A = 0, the proof is trivial.

NOTE. In vector notation the Cauchy-Schwarz inequality takes the form

(a b)2 < Ila11211b112,

where a = (a1, ... , an), b = (b1i ... , bn) are two n-dimensional vectors,

a b = akbk,

k=1

is their dot product, and Ilall = (a a)''2 is the length of a.

1.20 PLUS AND MINUS INFINITY AND THE EXTENDED REAL NUMBER

SYSTEM R*

Next we extend the real number system by adjoining two "ideal points" denoted

by the symbols + oo and - oo ("plus infinity" and "minus infinity").

Definition 1.24. By the extended real number system R* we shall mean the set of

real numbers R together with two symbols + co and - oo which satisfy the following

properties:

a) If x e R, then we have

x+(+oo)= +oo, x+(-co)= -oo,

x - (+00} = -oo, x - (-00) = +00,

x/(+oo)=x/(-co)=0.

Def. 1.26 Complex Numbers

b) If x > 0, then we have

x(+ oo) = + oo,

c) If x < 0, then we have

x(+oo) = -oo,

X(- 00) = -oo.

X(- 00) = +oo.

d)-,k+ oo) + (+oo) = (+oo)(+oo) = (-oo)(-oo) = +oo,

(-co) + (-00) = (+c0)(-o0) = -oo.

e) If X E R, then we have - oo < x < + co.

NOTATION. We denote R by (- oo, + co) and R* by [ - oo, + oo]. The points in R

are called "finite" to distinguish them from the "infinite" points + oo and - oo.

The principal reason for introducing the symbols + oo and - oo is one of

convenience. For example, if we define + oo to be the sup of a set of real numbers

which is not bounded above, then every nonempty subset of R has a supremum

in R*. The sup is finite if the set is bounded above and infinite if it is not bounded

above. Similarly, we define - oo to be the inf of any set of real numbers which is

not bounded below. Then every nonempty subset of R has an inf in R*.

For some of the later work concerned with limits, it is also convenient to

introduce the following terminology.

Definition 1.25. Every open interval (a, + oo) is called a neighborhood of + 00 or

a ball with center + co. Every open interval (- oo, a) is called a neighborhood of

- o0 or a ball with center - oo.

1.21 COMPLEX NUMBERS

It follows from the axioms governing the relation < that the square of a real

number is never negative. Thus, for example, the elementary quadratic equation

2 x = - I has no solution among the real numbers. New types of numbers, called

complex numbers, have been introduced to provide solutions to such equations. It

turns out that the introduction of complex numbers provides, at the same time,

solutions to general algebraic equations of the form

where the' coefficients ao, a,, ... , a" are arbitrary real numbers. (This fact is

known as the Fundamental Theorem of Algebra.)

We shall now define complex numbers and discuss them in further detail.

Definition 1.26. By a complex number we shall mean an ordered pair of real numbers

which we denote by (x,, x2). The first member, x,, is called the real part of the

complex number; the second member, x2, is called the imaginary part. Two complex

numbers x = (x,, x2) and y = (y,, y2) are called equal, and we write x = y, if,

16 Real and Complex Number Systems Th. 1.27

and only if, x1 = y1 and x2 = Y2. We define the sum x + y and the product xy by

the equations

x + y = (x1 + Y1, x2 + Y2), xY = (x1 Y1 - x2Y2, x1 Y2 + x2Y1)

NOTE-, The set of all complex numbers will be denoted by C.

Theorem 1.27. The operations of addition and multiplication just defined satisfy

the commutative, associative, and distributive laws.

Proof. We prove only the distributive law; proofs of the others are simpler. If

x = (x1, x2), y = (Y1, y2), and z = (z1, z2), then we have

x(Y + z) = (x1, x2)(Y1 + z1, Y2 + z2)

= (x1Y1 + x121 - x2Y2 - x222, x1Y2 + x122 + x2Y1 + x221)

= (x1Y1 - x2Y2, x1Y2 + x2Y1) + (x121 - x222, x122 + x221)

= xy + xz.

Theorem 1.28.

(x1, x2) + (0, 0) = (x1, x2), (x1, x2)(0, 0) _ (0, 0),

(x1, x2)(1, 0) = (x1, x2), (x1, x2) + (-x1, -x2) = (0, 0)

Proof. The proofs here are immediate from the definition, as are the proofs of

Theorems 1.29, 1.30, 1.32, and 1.33. '

Theorem 1.29. Given two complex numbers x = (x1, x2) and y = (y1, Y2), there

exists a complex number z such that x + z = y. In fact, z = (y1 - x1, Y2 - x2).

This z is denoted by y - x. The complex number (-x1, -x2) is denoted by -x.

Theorem 1.30. For any two complex numbers x and y, we have

(-x)Y = x(- Y) = -(xY) = (-1, 0)(xy)

Definition 1.31. If x = (x1i x2) # (0, 0) and y are complex numbers, we define

x' = [x1/(xi + xi), -x2/(xi + xz)], and y/x = yx-1.

Theorem 1.32. If x and y are complex numbers with x (0, 0), there exists a

complex number z such that xz = y, namely, z = yx-1.

Of special interest are operations with complex numbers whose imaginary

part is 0.

Theorem 1.33. (x 1, 0) + (y1, 0) = (x 1 + y 1p 0),

(x1, 0)(Y1, 0) = (x1 Y1, 0),

(x1, 0)/(Y1, 0) = (x1/Y1, 0), if Y1 # 0.

NOTE. It is evident from Theorem 1.33 that we can perform arithmetic operations

on complex numbers with zero imaginary part by performing the usual real-num-

ber operations on the real parts alone. Hence the complex numbers of the form

(x, 0) have the same arithmetic properties as the real numbers. For this reason it is

Fig. 1.3 Geometric Representation 17

convenient to think of the real number system as being a special case of the complex

number system, and we agree to identify the complex number (x, 0) and the real

number x. Therefore, we write x = (x, 0). In particular, 0 = (0, 0) and I = (1, 0).

1.22 GEOMETRIC REPRESENTATION OF COMPLEX NUMBERS

Just as real numbers are represented geometrically by points on a line, so complex

numbers are represented by points in a plane. The complex number x = (x1i x2)

can be thought of as the "point" with coordinates (x1, x2). When this is done, the

'definition of addition amounts to addition by the parallelogram law. (See Fig. 1.2.)

x+y=(x1+,1,x2+y2)

0 = (0, 0) X1 = (x1, 0) Figure 1.2

The idea of expressing complex numbers geometrically as points on a plane

was formulated by Gauss in his dissertation in 1799 and, independently, by Argand

in 1806. Gauss later coined the somewhat unfortunate phrase "complex number."

Other geometric interpretations of complex numbers are possible. Instead of

using points on a plane, we can use points on other surfaces. Riemann found the

sphere particularly convenient for this purpose. Points of the sphere are projected

from the North Pole onto the tangent plane at the South Pole and thus there

corresponds to each point of the plane a definite point of the sphere. With the

exception of the North Pole itself, each point of the sphere corresponds to exactly

one point of the plane. This correspondence is called a stereographic projection.

(See Fig. 1.3.)

18 Real and Complex Number Systems

1.23 THE IMAGINARY UNIT

Def. 1.34

It is often convenient to think of the complex number (x1, x2) as a two-dimensional

vector with components x1 and x2. Adding two complex numbers by means of

Definition 1.26 is then the same as adding two vectors component by component.

The complex number l = (1, 0) plays the same role as a unit vector in the hori-

zontal direction. The analog of a unit vector in the vertical direction will now be

introduced.

Definition 1.34. The complex number (0, 1) is denoted by i and is called the imag-

inary unit.

Theorem 1.35. Every complex number x = (x1, x2) can be represented in the form

x = x1 + 'X2-

Proof. x1 = (x1, 0), ix2 = (0, 1)(x2i 0) = (0, x2),

x1 + ix2 = (x1, 0) + (0, x2) = (x1, x2).

The next theorem tells us that the complex number i provides us with a solution

to the equation x2 = -1.

Theorem 1.36. i2 = -1.

Proof. i2 = (0, 1)(0, 1) = (-1, 0) _ -1.

1.24 ABSOLUTE VALUE OF A COMPLEX NUMBER

We now extend the concept of absolute value to the complex number system.

Definition 1.37. If x = (x1, x2), we define the modulus, or absolute value, of x to

be the nonnegative real number Ixl given by

IxI =v'x2+x2.

Theorem 1.38.

i) 1(0, 0)1 = Q, and IxI > 0 if x ; 0. ii) Ixyl = IxI IYI.

iii) 1x/y1 = Ixl/IYI, if y # 0. iv) I(xl, 0)1 = lxii.

Proof Statements (i) and (iv) are immediate. To prove (ii), we write x = x1 + ix2,

y = y1 + iy2, so that xy = x1 y1 - x2y2 + i(x1 y2 + x2 y1). Statement (ii)

follows from the relation

Ixy12 = xiyi + xiy2 + xiy2 + x2yi = (x1 + x2)(y1 + y2) = IxI21Y12

Equation (iii) can be derived from (ii) by writing it in the form IxI = IYI Ix/yl

Geometrically, IxI represents the length of the segment joining the origin to

the point x. More generally, lx - yl is the distance between the points x and y.

Using this geometric interpretation, the following theorem states that one side of

a triangle is less than the sum of the other two sides.

Def. 1.40 Complex Exponentials

Theorem 1.39. If x and y are complex numbers, then we have

Ix + y1 < Ixj + IyI (triangle inequality).

The proof is left as an exercise for the reader.

.1.25 IMPOSSIBILITY OF ORDERING THE COMPLEX NUMBERS

As yet we have not defined a relation of the form x < y if x and y are arbitrary

complex numbers, for the reason that it is impossible to give a definition of < for

complex numbers which will have all the properties in Axioms 6 through 8. To

illustrate, suppose we were able to define an order relation < satisfying Axioms

6, 7, and 8. Then, since i # 0, we must have either i > 0 or i < 0, by Axiom 6.

Let us assume i > 0. Then taking, x = y = i in Axiom 8, we get i2 > 0, or

-1 > 0. Adding 1 to both sides (Axiom 7), we get 0 > 1. On the other hand,

applying Axiom 8 to -1 > 0 we find 1 > 0. Thus we have both 0 > 1 and

1 > 0, which, by Axiom 6, is impossible. Hence the assumption i > 0 leads us

to a contradiction. [Why was the inequality -1 > 0 not already a contradiction?]

A similar argument shows that we cannot have i < 0. Hence the complex numbers

cannot be ordered in such a way that Axioms 6, 7, and 8 will be satisfied.

1.26 COMPLEX EXPONENTIALS

The exponential ex (x real) was mentioned earlier. We now wish to define eZ when

z is a complex number in such a way that the principal properties of the real

exponential function will be preserved. The main properties of ex for x real are

the law of exponents, ex,ex2 = exl+X2, and the equation e° = 1. We shall give a

definition of eZ for complex z which preserves these properties and reduces to the

ordinary exponential when z is real.

If we write z = x + iy (x, y real), then for the law of exponents to hold we

want ex+'y = exe'y. It remains, therefore, to define what we shall mean by e'y.

Definition 1.40. If z = x + iy, we define e= = ex+'y to be the complex number

e= = ex (cos y + i sin y).

This definition* agrees with the real exponential function when z is real (that

is, y = 0). We prove next that the law of exponents still holds.

* Several arguments can be given to motivate the equation e'y = cos y + i sin y. For

example, let us write e'y = f (y) + ig(y) and try to determine the real-valued functions f

and g so that the usual rules of operating with real exponentials will also apply to complex

exponentials. Formal differentiation yields e'' = g'(y) - if'(y), if we assume that

(e'y)' = ie'y. Comparing the two expressions for e'y, we see that f and g must satisfy the

equations f (y) = g'(y), f'(y) = - g(y). Elimination of g yields fly) = - f"(y). Since

we want e° = 1, we must have f (O) = 1 and f'(0) = 0. It follows that fly) = cos y and

g(y) = -f'(y) = sin y. Of course, this argument proves nothing, but it strongly suggests

that the definition e'y = cos y + i sin y is reasonable.

20 Real and Complex Number Systems Th. 1.41

Theorem 1.41. If z1 = x1 + iy1 and z2 = x2 + 1Y2 are two complex numbers,

then we have ez'ez2 = ez'+22.

Proof

ez' = ex'(cos y1 + i sin y1), ez2 = ex2(cos Y2 + i sin Y2),

ez'ez2 = ex'ex2[cos YI COS Y2 - sin y1 sin Y2

+ i(cos y1 sin Y2 + sin y1 cos Y2)].

Now ex'ex2 = ex'+x2, since xi1 and x2 are both real. Also,

cos y1 cos Y2 - sin y1 sin Y2 = cos (Y, + Y2)

and cos y1 sin Y2 + sin y1 cos Y2 = sin (y1 + Y2),

and hence

ezieZ2 = ex'+x2[cos (YI + Y2) + i sin (YI + Y2)] = ez'+Z2.

1.27 FURTHER PROPERTIES OF COMPLEX EXPONENTIALS

In the following theorems, z, z1, z2 denote complex numbers.

Theorem 1.42. ez is never zero.

Proof. eze-z = e° = 1. Hence ez cannot be zero.

Theorem 1.43. If x is real, then Ie'xI = 1.

Proof. Ie'ii2 = cost x + sin 2 x = 1, and I e'xi > 0.

Theorem 1.44. ez = 1 if, and only if, z is an integral multiple of 27ri.

Proof. If z = 2irin, where n is an integer, then

ez = cos (2irn) + i sin (21rn) = 1.

Conversely, suppose that ez = 1. This means that ex cos y = 1 and ex sin y = 0.

Since ex 0, we must have sin y = 0, y = k7r, where k is an integer. But

cos (k7r) = (- I)k. Hence ex = (-1)k, since ex cos (kir) = 1. Since ex > 0,

k must be even. Therefore ex = 1 and hence x = 0. This proves the theorem.

Theorem 1.45. ez' = eze if, and only if, z1 - z2 = 2irin (where n is an integer).

Proof. ez' = eze if, and only if, ez' `2 = 1.

1.28 THE ARGUMENT OF A COMPLEX NUMBER

If the point z -_ (x, y) = x + iy is represented by polar coordinates r and 0, we

can write x = r cos 0 and y = r sin 0, so that z = r cos 0 + it sin 0 = re'B

Def. 1.49 Integral Powers and Roots 21

The two numbers r and 0 uniquely determine z. Conversely, the positive number

r is uniquely determined by z; in fact, r = Iz I. However, z determines the angle 0

only up to multiples of 2n. There are infinitely many values of 0 which satisfy the

equations x = Iz I cos 0, y = Iz sin 0 but, of course, any two of them differ by

some multiple of 2n. Each such 0 is called an argument of z but one of these values

is singled out and is called the principal argument of z.

Definition 1.46. Let z = x + iy be a nonzero complex number. The unique real

number 0 which satisfies the conditions

x=1zIcos0, y=1zIsin0, -7r<0<-+n

is called the principal argument of z, denoted by 0 = arg (z).

The above discussion immediately yields the following theorem:

Theorem 1.47. Every complex number z 0 can be represented in the form

z = reie, where r = IzI and 0 = arg (z) + 2nn, n being any integer.

NOTE. This method of representing complex numbers is particularly useful in

connection with multiplication and division, since we have

tiO2 i(81+92) and e= rl ei(91-82)

(rleiB)(rie ) = r1r2e and r2efe2 r2

Theorem 1.48. If z1z2 0, then arg (z1z2) = arg (z1) + arg (z2) + 2irn(z1, z2),

where 0, f -n < arg (z1) + arg (z2) < +ir,

n(z1, z2) = + 1, if - 2n < arg (z1) + arg (z2) 5 - 7r,

-1, f it < arg (z1) + arg (z2) < 27r.

Proof. Write z1 = Iz11e`B', Z2 = I z21 e`02, where 01 = arg (z1) and 02 = arg (z2).

Then z1z2 = Iziz21ei(01+02) Since -n < 01 < +7t and -7r < 02 < +n, we

have -27t < 01 + 02 < 2n. Hence there is an integer n such that -7r < 01 +

02 + 2nn < it. This n is the same as the integer n(z1, z2) given in the theorem,

and for this n we have arg (z1z2) = 01 + 02 + 27rn. This proves the theorem.

1.29 INTEGRAL POWERS AND ROOTS OF COMPLEX NUMBERS

Definition 1.49. Given a complex number z and an integer n, we define the nth power

of z as follows:

z0 = 1, z n + 1 = ZnZ, if n 0,

z-"=(z-1)", ifz#Oand n>0.

Theorem 1.50, which states that the usual laws of exponents hold, can be proved

by mathematical induction. The proof is left as an exercise.

22 Real and Complex Number Systems Th. 1.50

Theorem 1.50. Given two integers m and n, we have, for z 0 0,

ZnZm = Zn+m and (Z1Z2)" = ZIZZ.

Theorem 1.51. If z 0, and if n is a positive integer, then there are exactly n

distinct complex numbers zo, z1, ... , zn_1 (called the nth roots of z), such that

zk = Z, for each k = 0, 1, 2, ... , n - 1.

Furthermore, these roots are given by the formulas

zk = Re'mk, where R = 1Z I 1 In,

and arg (z) 2nk

k= + (k=0, 1,2,...,n-1).

NOTE. The n nth roots of z are equally spaced on the circle of radius R = IzI',",

center at the origin.

Proof. The n complex numbers Re'4k, 0 < k < n - 1, are distinct and each is

an nth root of z, since (Re'mk)n = Rnei"4k = IzIe'[arg(z)+2,rk] = Z.

We must now show that there are no other nth roots of z. Suppose w = Ae" is

a complex number such that w" = z. Then I wI" = Iz I, and hence A" = Izi,

A = Iz I' I". Therefore, w" = z can be written e'"" = e'larg (0], which implies

na - arg (z) = 2nk for some integer k.

Hence a = [arg (z) + 27rk]/n. But when k runs through all integral values, w

takes only the distinct values z 0 ,... , zn _ 1. (See Fig. 1.4.)

Figure 1.4

1.30 COMPLEX LOGARITHMS

By Theorem 1.42, ez is never zero. It is natural to ask if there are other values

that ez cannot assume. The next theorem shows that zero is the only exceptional

value.

Def. 135 Complex Powers 23

Theorem 1.52. If z is a complex number # 0, then there exist complex numbers w

such that ew = z. One such w is the complex number

log IzI + i arg (z),

and any other such w must have the form

log IzI + i arg (z) + 2mri,

where n is an integer.

Proof. Since elog Izl +i arg (z) = elog IzI ei arg (z) = Iz Iei arg (z) = Z, we see that w =

log IzI + i arg (z) is a solution of the equation ew = z. But if wl is any other

solution, then ew = ewi and hence w - wl = 2niri.

Definition 1.53. Let z 0 be a given complex number. If w is a complex number

such that ew = z, then w is called a logarithm of z. The particular value of w given

by w = log IzI + i arg (z)

is called the principal logarithm of z, and for this w we write

w = Log z.

Examples

1. Since i l = 1 and arg (i) it/2, Log (f) = iir/2.

2. Since I - i l = 1 and arg (- i) _ - x/2, Log (- i) in/2.

3. Since -1 = 1 and arg (-1) = n, Log (-1) = in.

4. If x > 0, Log (x) = log x, since IxI = x and arg (x) = 0.

5. Since 11 + i I = I2 and arg (1 + i) = n/4, Log (1 + i) = log ,l2 + in/4.

Theorem 1.54. If zxz2 # 0, then

Log (zlz2) = Log zl + Log z2 + 27rin(zl, z2),

where n(zl, z2) is the integer defined in Theorem 1.48.

Proof.

Log (zlz2) = log IZ1z2I + i arg (ztz2)

= log IZ1I + log IZ2I + i [arg (z1) + arg (Z2) + 2irn(zl, Z2)]

1.31 COMPLEX POWERS

Using complex logarithms, we can now give a definition of complex powers of

complex numbers.

Definition 1.55. If z # 0 and if w is any complex number, we define

Zw = ewLogz

24 Real and Complex Number Systems Th. 1.56

Examples

1. i' = eiLogi = eUin/2) = e-rz/2

2. (-1)i = eiLog(-1) = ei(in) = e-n.

3. If n is an integer, then zi+1 = e(n+l)Logz = enLogzeLogz = znz, so Definition 1.55 does

not conflict with Definition 1.49.

The next two theorems give rules for calculating with complex powers:

Theorem 1.56. zwt ZW2 = Zwt+w2 if z 0.

Proof. Zwt+w2 = e(wt+w2)Logz = ew1Logzew2Logz = ZWIZW2.

Theorem 1.57. If z1z2 0, then

(zlz2)w = z1 2 w we 2aiwn(z1,z2)

where n(zl, z2) is the integer defined in Theorem 1.48.

Proof. (z1Z2)w = ewLog(ztz2) = ew[Logzt+Logz2+2ain(zt,z2)]

1.32 COMPLEX SINES AND COSINES

Definition 1.58. Given a complex number z, we define

eiz + e-'z e'z - e- 1Z

cos z = - --- , 2sin z = 2i

NOTE. When z is real, these equations agree with Definition 1.40.

Theorem 1.59. If z = x + iy, then we have

cos z = cos x cosh y - i sin x sinh y,

sin z = sin x cosh y + i cos x sinh y.

Proof. 2 cos z = e'z + e-1z

= e-'(cos x + i sin x) + ey(cos x - i sin x)

= cos x(ey + e-') - i sin x(ey - e-y)

= 2 cos x cosh y - 2i sin x sinh y.

The proof for sin z is similar.

Further properties of sines and cosines are given in the exercises.

1.33 INFINITY AND THE EXTENDED COMPLEX PLANE C*

Next we extend the complex number system by adjoining an ideal point denoted by

the symbol oo.

Definition 1.60. By the extended complex number system C* we shall mean the

complex plane C along with a symbol oo which satisfies the following properties:

a) If z e C, then we have z + oo = z - oo = oo, z/oo = 0.

Def. 1.61 Exercises 25

b) If z e C, but z 0, then z(oo) = oo and z/0 = oo.

c) co + oo = (co)(oo) = oo.

Definition 1.61. Every set in C of the form {z : Iz I > r > 0) is called a neighbor-

hood of oo, or a ball with center at oo.

The reader may wonder why two symbols, + oo and - oo, are adjoined to R

but only one symbol, oo, is adjoined to C. The answer lies in the fact that there is

an ordering relation < among the real numbers, but no such relation occurs

among the complex numbers. In order that certain properties of real numbers

involving the relation < hold without exception, we need two symbols, + oo and

- oo, as defined above. We have already mentioned that in R* every nonempty

set has a sup, for example.

In C it turns out to be more convenient to have just one ideal point. By way

of illustration, let us recall the stereographic projection which establishes a one-

to-one correspondence between the points of the complex plane and those points

on the surface of the sphere distinct from the North Pole. The apparent exception

at the North Pole can be removed by regarding it as the geometric representative

of the ideal point cc. We then get a one-to-one correspondence between the

extended complex plane C* and the total surface of the sphere. It is geometrically

evident that if the South Pole is placed on the origin of the complex plane, the

exterior of a "large" circle in the plane will correspond, by stereographic projection,

to a "small" spherical cap about the North Pole. This illustrates vividly why we

have defined a.neighborhood of cc by an inequality of the form Iz I > r.

EXERCISES

Integers

1.1 Prove that there is no largest prime. (A proof was known to Euclid.)

1.2 If n is a positive integer, prove the algebraic identity

n-1

an - b" = (a - b) E akbn-l-k.

k=0

1.3 If 2" - I is prime, prove that n is prime. A prime of the form 2° - 1, where p is

prime, is called a Mersenne prime.

1.4 If 2" + 1 is prime, prove that n is a power of 2. A prime of the form 22"' + 1 is

called a Fermat prime. Hint. Use Exercise 1.2.

1.5 The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13.... are defined by the recursion formula

xn+1 = x" + x"_1, with x1 = x2 = 1. Prove that (x", xn+1) = I and that x" _

(a" - bn)/(a - b), where a and bare the roots of the quadratic equation x2 - x - I = 0.

1.6 Prove that-every nonempty set of positive integers contains a smallest member.

This is called the well-ordering principle.

26 Real and Complex Number Systems

Rational and irrational numbers

1.7 Find the rational number whose decimal expansion is 0.3344444.. .

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 2"5'", where m and n are non-

negative integers.

1.9 Prove that s2 + J3 is irrational.

1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b)l(cx + d) is usually

irrational. When do exceptions occur?

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

1.12 If alb < c/d with b > 0, d > 0, prove that (a + c)l(b + d) lies between alb

and c/d.

1.13 Let a and b be positive integers. Prove that V2 always lies between the two fractions

alb and (a + 2b)l(a + b). Which fraction is closer to 2?

1.14 Prove that I n - 1 + jn + I is irrational for every integer n >- 1.

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 jkx - hi < 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.

1.16 If x is irrational prove that there are infinitely many rational numbers h/k with

k > 0 such that Ix - h/kI < 1/k2. Hint. Assume there are only a finite number

h1/k1,..., hr/kr and obtain a contradiction by applying Exercise 1.15 with N > 1/6,

where 6 is the smallest of the numbers Ix - hi/kid.

1.17 Let x be a positive rational number of the form

X= E

ak a nonnegative integer with ak 5 k - I for k >- 2 and a" > 0. Let [x]

denote the greatest integer in x. Prove that al = [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. Con-

versely, show that every positive rational number x can be expressed in this form in one

and only one way.

Upper bounds

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

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-', where p, q, and r take on all positive

integer values.

b) S = [x: 3x2 - lOx + 3 < 0}.

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

1.20 Prove the comparison property for suprema (Theorem 1.16).

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 e A and y e B.

Prove that ab = sup C.

Exercises 27

1.22 Given x > 0 and an integer k >- 2. Let ao denote the largest integer <- x and,

assuming that ao, a1, ... , an_ 1 have been defined, let an denote the largest integer such

that ao+A1+ a2+ ... +Q" <x.

kk2 k"

a) Prove that 0 5 al <- k - 1 for each i = 1, 2, .. .

b) Let r" = ao + a1k-' + a2k-2 + + a"k-" and show that x is the sup of the

set of rational numbers r1, r2, .. .

NOTE. When k = 10 the integers ao, a1, a2,... are the digits in a decimal representation

of x. For general k they provide a representation in the scale of k.

Inequalities

1.23 Prove Lagrange's identity for real numbers:

k=1 =1

akbk)

=k ak) (k1 bk) l sk sn (Rkb; - a;bk)2.

Note that this identity implies the Cauchy-Schwarz inequality.

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

(akbkCk) <

k=1

1.25 Prove Minkowski's inequality: k=1

Uo (ak + bk)2)1/2 <

ak)1/2 + \k bk)1/2

This is the triangle inequality Ira + bll <- IIall + IIbIj for n-dimensional vectors, where

a = (a1,..., an), b = (b1, ... , bn) and

IIaHH = ( ak

n1/2

)

k=1

1.26 If a1 >- a2 >- >- an and b1 >- b2 >_ >_ bn, prove that

k=1 ak)(k=1 bk)

Hint. X15Jsk5n (ak - aj)(bk - bj) >- 0.

5 n F akbk.

k=1

Complex numbers

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

a) (1 + i)3, b) (2 + 3i)/(3 - 4i),

c) is + i16, d) 4(1 + i)(1 + i-8).

bk)2(E Ck)

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

100

a) x + iy = Ix - iyJ, b) x + iy = (x - iy)2, c) E ik = x + jy.

k=0

28 Real and Complex Number Systems

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) z1 + z2 = Z1 + z2, b) F, -z2 = f, Z2, C) ZZ = 1z12,

d) z + z = twice the real part of z,

e) (z - 27)/i = twice the imaginary part of z.

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

following conditions:

a)IzI=1, b)IzI<1, C)IzI_<1,

d)z+z= 1, e)z - z=i, f) z+z= IzI2.

1.31 Given three complex numbers z1, z2, z3 such that Iz, I = IZ21 = Iz31 = 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.

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

a) Ia - b12 < (1 + Ia12)(1 + Ib12).

b) If a # 0, then Ia + bI = IaI + IbI if, and only if, b/a is real and nonnegative.

1.33 If a and b are complex numbers, prove that

Ia-6I=11-abI

if, and only if, Ial = I or IbI = 1. For which a and b is the inequality Ia - bI < I 1 - abI

valid?

1.34 If a and c are real constants, b complex, show that the equation

azz+bz+bz+c=0 (a:0,z=x+iy)

represents a circle in the xy-plane.

1.35 Recall the definition of the inverse tangent: given a real number t, tan-1 (t) is the

unique real number 0 which satisfies the two conditions

tan0= t. - 2 < 0 < + 2

If z = x + iy, show that

a) arg (z) = tan-1 (!) , if x > 0,

b) arg (z) = tan-1 (!) + n, if x < 0, y >- 0,

c) arg (z) = tan-1 (y) - 7r,

xif x < 0, y < 0,

d)arg(z)= 2ifx=0,y>0;arg(z)= -2ifx=0,y<0.

Exercises 29

1.36 Define the following "pseudo-ordering" of the complex numbers: we say z1 < z2

if we have either

i) IZl I < Iz2I or ii) IZl I = Iz21 and arg (Z1) < arg (z2).

Which of Axioms 6, 7, 8, 9 are satisfied by this relation?

1.37 Which of Axioms 6, 7, 8, 9 are satisfied if the pseudo-ordering is defined as follows?

We say (x1, yl) < (x2, y2) if we have either

i) x1 < x2 or ii) x1 = x2 and y1 < y2.

1.38 State and prove a theorem analogous to Theorem 1.48, expressing arg (zl/z2) in

terms of arg (z1) and arg (z2).

1.39 State and prove a theorem analogous to Theorem 1.54, expressing Log (z,42) in

terms of Log (z1) and Log (Z2)-

1.40 Prove that the nth roots of 1 (also called the nth roots of unity) are given by a,

a2_., a", where a = e2'ri", and show that the roots :1 satisfy the equation

1 + x + x2 + + x"-1 = 0.

1.41 a) Prove that Izrl < e' for all complex z : 0.

b) Prove that there is no constant M > 0 such that Icos z I < M for all complex z.

1.42 If w = u + iv (u, v real), show that

zw = euloglzl-varg(z)eitvlogjzI+uarg(z)1

1.43 a) Prove that Log (zw) = w Log z + 2irin, where n is an integer.

b) Prove that (zw)°` = zwae2nt"a, where n is an integer.

1.44 i) If 0 and a are real numbers, - 7r < 0 < + ir, prove that

(cos 0 + i sin O)" = cos (a9) + i sin (aO).

ii) Show that, in general, the restriction - n < 0 < + it is necessary in (i) by taking

0 = - ir, a = 1.

iii) If a is an integer, show that the formula in (i) holds without any restriction on 0.

In this case it is known as DeMoivre's theorem.

1.45 Use DeMoivre's theorem (Exercise 1.44) to derive the trigonometric identities

sin 30 = 3 cos' 0 sin 0 - sin3 0,

cos 30 = cos3 0 - 3 cos 0 sine 0,

valid for real 0. Are these valid when 0 is complex?

1.46 Define tan z = (sin z)/(cos z) and show that for z = x + iy, we have

tan z = sin 2x + i sinh 2y

cos 2x + cosh 2y

1.47 Let w be a given complex number. If w 96 ± 1, show that there exist two values of

z = x + iy satisfying the conditions cos z = w and - it < x < + it. Find these values

when w = i and- when w = 2.

30 Real and Complex Number Systems

1.48 Prove Lagrange's identity for complex numbers:

k=1

IakI2 Ibkl2 Iak5J - aj5kl2.

k=1 k=1 15k<Jsn

Use this to deduce a Cauchy-Schwarz inequality for complex numbers.

1.49 a) By equating imaginary parts in DeMoivre's formula prove that

sin nO = sin" 0 {(n cotn-1 0 - (1?) coin-3 0 + (n) cot"-5 0 -

I) 3

b) If 0 < 0 < n/2, prove that

sin (2m + 1)0 = sin2,n+10Pm(cot2 0)

where Pm is the polynomial of degree m given by

Pm(X) (2m+ 11 (2m+ 1) x"`-1 + (2m+ 11 X"`_2 - +

=1 /fx"' - 3 5 J...

Use this to show that P. has zeros at them distinct points Xk = cot2 {nk/(2m + 1)}

fork = 1, 2, ... , m.

c) Show that the sum of the zeros of P. is given by

"` 2 irk m(2m - 1)

E cot =

2m + 1 3

and that the sum of their squares is given by

mcot4 irk = m(2m - 1)(4m2 + lOm - 9)

k=1 2m + 1 45

NOTE. These identities can be used to prove that E 1 n-2 = n2/6 and 1 n-4 = n4/90.

(See Exercises 8.46 and 8.47.)

1.50 Prove that z" - I = Iik=1 (z - e2nikln) for all complex z. Use this to derive the

formula

sin kn = n for n >- 2.

n2"-1

k=1

SUGGESTED REFERENCES FOR FURTHER STUDY

1.1 Apostol, T. M., Calculus, Vol. 1, 2nd ed. Xerox, Waltham, 1967.

1.2 Birkhoff, G., and MacLane, S., A Survey of Modern Algebra, 3rd ed. Macmillan,

New York, 1965.

1.3 Cohen, L., and Ehrlich, G., The Structure of the Real-Number System. Van Nos-

trand, Princeton, 1963.

1.4 Gleason, A., Fundamentals of Abstract Analysis. Addison-Wesley, Reading, 1966.

1.5 Hardy, G.- H., A Course of Pure Mathematics, 10th ed. Cambridge University Press,

1952.

References 31

1.6 Hobson, E. W., The Theory of Functions of a Real Variable and the Theory of Fourier's

Series, Vol. 1, 3rd ed. Cambridge University Press, 1927.

1.7 Landau, E., Foundations of Analysis, 2nd ed. Chelsea, New York, 1960.

1.8 Robinson, A., Non-standard Analysis. North-Holland, Amsterdam, 1966.

1.9 Thurston, H. A., The Number System. Blackie, London, 1956.

1.10 Wilder, R. L., Introduction to the Foundations of Mathematics, 2nd ed. Wiley, New

York, 1965.

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

k=0

ak bn−1−k

Proof : It suﬃces to show that

xn −1 = ( x−1)

n− 1

k=0

xk .

Consider the right hand side, we have

(x− 1)

n− 1

k=0

xk =

n− 1

k=0

xk+1 −

n− 1

k=0

k=1

xk −

n− 1

k=0

=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

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 nis a power of 2.A prime of the

form 22 m+ 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, nmust be a

power of 2.

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

x2n−1 + 1 = ( x+ 1)

2n− 2

k=0

(−1)k xk .

Proof : Consider

(x + 1)

2n− 2

k=0

(−1)k xk =

2n− 2

k=0

(−1)k xk+1 +

2n− 2

k=0

(−1)k x k

2n− 1

k=1

(−1)k+1 xk +

2n− 2

k=0

(−1)k x k

=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 x n+1 =xn +x n−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, aand b.

If we let

Fn = ( an − bn ) / ( a− b) ,

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.

Then S = N.

Proof : (A⇒ B ): If S 6= N, then N −S 6= φ. So, by ( A), there exists

the smallest integer wsuch 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∈ Swhich 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

1−1

10 

=33

102 + 4

900

=301

900 .

1.8 Prove that the decimal expansion of xwill end in zeros (or in nines)

if, and only if, x is a rational number whose denominator is of the form 2n 5m ,

where m and n are nonnegative integers.

Proof: (⇐ )Suppose that x = k

2n5m ,if n≥ m, we have

k5n−m

2n5n = 5 n−m k

10n .

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

(⇒ )Suppose that the decimal expansion of xwill end in zeros (or in

nines).

For case x =a0 .a1 a2 ···an . Then

x=Pn

k=0 10 n−ka k

10n = P n

k=0 10 n−k a k

2n5n .

For case x =a0 .a1 a2 ···an 999999 ···. Then

x=Pn

k=0 10 n−ka k

2n5n + 9

10n+1 + ... + 9

10n+m + ...

=Pn

k=0 10 n−ka k

2n5n + 9

10n+1

∞

j=0

10−j

=Pn

k=0 10 n−ka k

2n5n + 1

10n

=1 + Pn

k=0 10 n−ka k

2n5n .

So, in both case, we prove that xis a rational number whose denominator is

of the form 2n 5m , 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 = a

b,where g.c.d. (a, b ) = 1.Then

b2 p= a2 ⇒ p  a2 ⇒p |a (*)

Write a = pq. So,

b2 p= p2 q2 ⇒ 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 pis a

prime.

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 =k 2

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 xis 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

dif 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

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

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+b

2and r= 1

√2 a+ 1− 1

√2 b.

(ii) (a∈ Q, b ∈ Qc ) Choose r =a+b

2and let c= 1

2n < 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 .b1 b2 ··· bn · ··

Choose n large enough so that

a < q = b0 .b1 b2 ···bn < b.

(It works since b− q = 0. 000..000b n+1 ... ≤ 1

10n )

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  √ 2 b−a

a+ b≥0.

In addition,

√ 2−a

b −  a+ 2b

a+ b−√ 2 = 2√ 2− a

b+ a+ 2b

a+ b

= 2√ 2−a 2 + 2ab + 2b 2

ab + b 2

ab + b2 h 2 √ 2−2 ab + 2√ 2− 2 b2 − a2 i

≥1

ab + b2 " 2 √ 2−2 a a

√2+  2 √ 2− 2 a

√2  2

−a2 #

= 0.

So, a+2b

a+ bis closer to √ 2.

Similarly, we also have if a/b > √ 2, then a+2b

a+ b<√ 2. Also, a+2b

a+ bis closer

to √ 2 in this case.

Remark: Note that

b< √ 2 < a+ 2b

a+ b< 2b

aby Exercise 12 and 13.

And we know that a+2b

a+ bis closer to √ 2. We can use it to approximate √ 2.

Similarly for the case 2b

a< a+ 2b

a+ b< √ 2 < a

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

2and k=1

which is absurb. So, √ n− 1 + √ n + 1 is irrational for every integer n≥ 1.

1.15 Given a real xand an integer N > 1, prove that there exist integers

hand kwith 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= minr

i=1 |x− h i /k i |>0 since xis irrational. Choose N > 1/δ, then by

Exercise 1.15, we have

N< δ ≤    

x− h

k   

which implies that 1

N< 1

which is impossible. So, there are inﬁnitely many rational numbers h/k with

k > 0 such that | x− h/k| < 1/k 2.

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.

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 |< 1

k.Consider (x− δ, x +δ ) := I, where δ > 0, and thus choos-

ing k0 large enough so that 1/k0 < δ. Deﬁne L = |k0 r − 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

k=1

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

xcan be expressed in this form in one and only one way.

Proof : ( ⇒)First,

[x ] = " a1 +

k=2

k!#

=a1 +" n

k=2

k!# since a 1 ∈ N

=a1 since

k=2

k!≤

k=2

k−1

k!=

k=2

(k− 1)! − 1

k!= 1 − 1

n! <1.

Second, ﬁxed kand consider

k! x= k!

j=1

j!= k!

k−1

j=1

j!+ a k + k!

j= k+1

and

(k− 1)!x = (k− 1)!

j=1

j!= ( k−1)!

k−1

j=1

j!+ ( k−1)!

j= k

j! .

So,

[k !x ] = " k!

k−1

j=1

j!+ a k + k!

j= k+1

j!#

=k !

k−1

j=1

j!+ a k since k!

j= k+1

j! <1

and

k[( k−1)! x] = k" ( k−1)!

k−1

j=1

j!+ ( k−1)!

j= k

j! . #

=k (k− 1)!

k−1

j=1

j!since ( k−1)!

j= k

j! <1

=k !

k−1

j=1

which implies that

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

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

integer. It is clear that

n! x= n!

k=1

k!∈ Z.

In addition,

(n− 1)!x = (n− 1)!

k=1

= (n− 1)!

n− 1

k=1

k!+ a n

/ ∈Zsince a n

n/ ∈Z.

So, we have proved it.

(⇐ )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−b

2,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 = 1

2+ 1

3+ 1

5,and inf S = 0.

(b) S = {x : 3x2 − 10x + 3 < 0}

Proof: Since 3x2 − 10x + 3 = (x− 3) (3x− 1) , we know that S = 1

3,3 .

Hence, sup S = 3 and inf S = 1

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 Sexists. 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 Sexists

and sup S≤ sup T.

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 nlarge 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 ) < a0 b0 since a− ε0 > 0 and b− ε0 > 0

which implies that

ab − ε

n( a+b )< a0 b0 := c

which implies that

ab − ε < c.

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

≤xand, assumeing that a0 , a1 , ..., an−1 have been deﬁned, let an denote the

largest integer such that

a0 + a 1

k+ a 2

k2 + ... + a n

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, ...

Proof : Choose a0 = [x ], and thus consider

[kx − ka0 ] := a 1

then

0≤k (x− a0 ) < k ⇒ 0≤a1 ≤k − 1

and

a0 + a 1

k≤ x≤a0 + a 1

k+1

Continue the process, we then have

0≤ai ≤k − 1 for each i = 1, 2, ...

and

a0 + a 1

k+ a 2

k2 + ... + a n

kn ≤ x<a 0+a 1

k+ a 2

k2 + ... + a n

kn +1

kn .(*)

(b) Let rn =a0 +a1 k−1 +a2 k−2 + ... +an k−n and show that xis 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:

k=1

ak bk !2

= n

k=1

k! n

k=1

k!− X

1≤k<j≤n

(ak bj − aj bk )2 .

Note that this identity implies that Cauchy-Schwarz inequality.

Proof : Consider

k=1

k! n

k=1

k!= X

1≤k,j ≤n

kb 2

j=X

k= j

kb 2

j+X

k6= j

kb 2

k=1

kb 2

k+X

k6= j

kb 2

and

k=1

ak bk ! n

k=1

ak bk ! =X

1≤k,j ≤n

ak bk aj bj =

k=1

kb 2

k+X

k6= j

ak bk aj bj

So,

k=1

ak bk !2

= n

k=1

k! n

k=1

k!+ X

k6= j

ak bk aj bj −X

k6= j

kb 2

= n

k=1

k! n

k=1

k!+ 2 X

1≤k<j ≤n

ak bk aj bj −X

1≤k<j ≤n

kb 2

j+a 2

jb 2

= n

k=1

k! n

k=1

k!− X

1≤k<j≤n

(ak bj − aj bk )2 .

Remark: (1) The reader may recall the relation with Cross Product

and Inner Product , we then have a fancy formula:

kx ×y k2 + |< x, y >|2 = kxk2ky k2 ,

where x, y ∈ R3 .

(2) We often write

< a, b >:=

k=1

ak bk ,

and the Cauchy-Schwarz inequality becomes

|< x, y > | ≤ kx k ky k by Remark (1).

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

k=1

ak bk ck !4

≤ n

k=1

k! n

k=1

k! 2 n

k=1

k!.

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

k=1

ak bk ck !4

=

 n

k=1

ak bk ck ! 2 



≤ n

k=1

kc 2

k! 2 n

k=1

k! 2

≤ n

k=1

k! 2 n

k=1

k! n

k=1

k! 2

= n

k=1

k! n

k=1

k! 2 n

k=1

k!.

1.25 Prove that Minkowski's inequality:

k=1

(ak +bk )2 ! 1/2

≤ n

k=1

k! 1/2

+ n

k=1

k! 1 /2

This is the triangle inequality ka +b k ≤ ka k +kb k for n − dimensional vectors,

where a = (a1 , ..., an ) , b = (b1 , ..., bn ) and

ka k= n

k=1

k! 1 /2

Proof : Consider

k=1

(ak +bk )2=

k=1

k+ 2

k=1

ak bk

≤

k=1

k+ 2 n

k=1

k! 1 /2 n

k=1

k! 1/2

by Cauchy-Schwarz inequality

=

 n

k=1

k! 1 /2

+ n

k=1

k! 1 /2



So,

k=1

(ak +bk )2 ! 1/2

≤ n

k=1

k! 1/2

+ n

k=1

k! 1 /2

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

k=1

ak ! n

k=1

bk ! ≤ n n

k=1

ak bk ! .

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

ak bk + aj bj − X

1≤j ≤k ≤n

ak bj + aj bk

which implies that

1≤j ≤k≤n

ak bj + aj bk ≤X

1≤j ≤k ≤n

ak bk + aj bj .(*)

Since

1≤j ≤k ≤n

ak bj + aj bk =X

1≤j<k ≤n

ak bj + aj bk + 2

k=1

ak bk

= X

1≤j<k ≤n

ak bj + aj bk +

k=1

ak bk ! +

k=1

ak bk

= n

k=1

ak ! n

k=1

bk ! +

k=1

ak bk ,

we then have, by (*)

k=1

ak ! n

k=1

bk ! +

k=1

ak bk ≤ X

1≤j ≤k ≤n

ak bk + aj bj .(**)

In addition,

1≤j ≤k ≤n

ak bk + aj bj

k=1

ak bk + na1 b1 +

k=2

ak bk + ( n−1) a2 b2 + ... +

k= n− 1

ak bk + 2 an−1 bn−1 +X

k= n

ak b k

k=1

ak bk + a1 b1 + a2 b2 + ... + an bn

= (n + 1)

k=1

ak bk

which implies that, by (**),

k=1

ak ! n

k=1

bk ! ≤ n n

k=1

ak bk ! .

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+2 i.

(b) (2 + 3i )/ (3 − 4i)

Solution: 2+3i

3−4i= (2+3i)(3+4i)

(3− 4 i)(3+4i )= − 6+17i

25 = − 6

25 + 17

25 i.

Solution:i5 +i 16 =i + 1.

(d) 1

2(1 + i ) (1 + i − 8 )

Solution: 1

2(1 + i ) (1 + i − 8 ) = 1 + i.

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

relation.

(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

k=0 i k =x+ iy

Proof : Since P 100

k=0 i k = 1−i101

1−i= 1−i

1−i= 1, we have x = 1 and y= 0.

1.29 If z =x + iy, x and y real, the complex conjugate of zis 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 = ¯ z1 ¯ z2

Proof : Write z1 =x1 + iy1 and z2 =x2 + iy2 , then

z1z2 = ( x1x2 − y1y2 ) + i ( x1 y2 + x2 y1 )

= (x1x2 − y1y2 )−i (x1 y2 +x2 y1 )

and

¯ z1 ¯ z2 = ( x1 − iy1 ) ( x2 − iy2 )

= (x1 x2 − y1y2 )−i (x1 y2 +x2 y1 ) .

So, z1z2 = ¯ z1 ¯ z2

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= 2 y,

twice the imaginary part of z.

1.30 Describe geometrically the set of complex numbers zwhich 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.

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

Proof : Write z =x + iy, then z− ¯ z= imeans 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 ,

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

a=¯ ab

¯ aa = | ¯ a|| b|

|a|2 ≥ 0.

(⇐ ) Suppose that

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.

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, bcomplex, 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

−a ¯ z−¯

−az+ b

−a" b

−a# = −ac + |b|2

a2 ,

so, we have

   

z− b

−a    

=−ac + |b|2

a2 .

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

a2 = 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 θwhich satisﬁes the two conditions

−π

2< θ < + π

2, tan θ= t.

If z =x + iy, show that

(a) arg (z ) = tan−1  y

x,if x > 0

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  y

x.

(b) arg (z ) = tan−1  y

x+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  y

x+π.

x−π, if x < 0 , y < 0

Proof : Similarly for (b). So, we omit it.

(d) arg (z ) = π

2if x = 0, y > 0; arg (z ) = − π

2if 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 =

r1 eiarg(z1 ) , and r2 eiarg(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.

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

x6= y: As x6= 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 =r1 eiarg(z1 ) and z2 =r2 eiarg(z2 ) , then

=r 1

ei[arg(z1 )− arg(z2 )] .

Hence,

arg  z 1

z2  = arg ( z 1 )−arg ( z 2 )+2 πn (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

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 =r1 eiarg(z1 ) and z2 =r2 eiarg(z2 ) , then

=r 1

ei[arg(z1 )− arg(z2 )] .

Hence,

Log (z1/z2 ) = log    

z2    

+i arg  z 1

z2 

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

=Log (z1 )− Log (z2 ) + i 2πn (z1 , z2 ) .

1.40 Prove that the n th 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 + ... +x n−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 + ... +x n−1 = 0 if x 6= 1.

So, all roots except 1 satisfy the equation

1 + x +x2 + ... +x n−1 = 0.

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

Proof : Since

zi = eiLog(z ) =e−arg(z)+ilog|z | ,

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

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

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 )

= log  eulog|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 + 2 πin = ( 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 + 2 πin

for some n∈ Z.

(b) Prove that (zw )α =z wα e2πinα , where n is an integer.

Proof : By (a), we have

(zw )α = eαLog(zw ) = eα(wLogz+2πin ) = eαwLogze2πinα =z αw e2πinα ,

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 iθ | +iarg ( e iθ )] =eiaθ

= cos (aθ ) + i sin (aθ ) .

Remark: Compare with the Exercise 1.43-(b).

(ii) Show that, in general, the restriction −π < θ ≤π is necessary in (i)

by taking θ = −π, a = 1

Proof : As θ = −π, and a= 1

2,we have

(−1) 1

2=e 1

2Log(− 1) =e π

2i =i6=− i= cos  −π

2 +isin  −π

2 .

(iii) If a is an integer, show that the formula in (i) holds without any

restriction on θ. In this case it is known as DeMorvre's theorem.

Proof : By Exercise 1.43, as a is an integer we have

(zw )a =z wa ,

where zw = eiθ . Then

e iθa =eiθa = cos (aθ ) + i sin (aθ).

1.45 Use DeMorvre's theorem (Exercise 1.44) to derive the trigino-

metric identities

sin 3θ= 3 cos2 θ sin θ− sin3 θ

cos 3θ = cos3 θ− 3 cos θ sin2 θ,

valid for real θ. Are these valid when θ is complex?

Proof : By Exercise 1.44-(iii), we have for any real θ,

(cos θ +i sin θ )3 = cos (3θ ) + i sin (3θ ) .

By Binomial Theorem , we have

sin 3θ= 3 cos2 θ sin θ− sin3 θ

and

cos 3θ = cos3 θ− 3 cos θ sin2 θ.

For complex θ, we show that it holds as follows. Note that sin z = e iz −e−iz

and cos z = e iz +e−iz

2,we have

3 cos2 z sin z− sin3 z = 3  e iz + e −iz

2 2  e iz −e−iz

2i −  e iz − e −iz

2i 3

= 3  e 2zi + e −2zi + 2

4 e iz −e−iz

2i + e 3zi − 3eiz + 3e−iz −e−3zi

8i  3  e 2zi + e −2zi + 2  e z i − e −zi  +  e 3zi − 3eiz + 3e−iz −e−3zi 

8i  3e3zi + 3eiz − 3e−iz − 3e−3zi  + e3zi − 3eiz + 3 e−iz −e−3zi 

8i e 3zi − e −3zi 

2i e 3zi − e −3zi 

= sin 3z.

Similarly, we also have

cos3 z− 3 cos z sin2 z = cos 3 z.

1.46 Deﬁne tan z = sin z/ cos z and show that for z =x + iy, we have

tan z = sin 2x+ i sinh 2y

cos 2x + cosh 2y.

Proof : Since

tan z = sin z

cos z = sin (x+ iy)

cos (x + iy )= sin x cosh y +i cos x sinh y

cos x cosh y− i sin x sinh y

=(sin x cosh y +i cos x sinh y ) (cos x cosh y +i sin x sinh y)

(cos x cosh y− i sin x sinh y ) (cos x cosh y +i sin x sinh y)

= sin x cos x cosh2 y− sin x cos x sinh2 y +i sin2 x cosh y sinh y + cos2 x cosh y sinh y 

(cos x cosh y )2 −(i sin x sinh y )2

=sin x cos x cosh2 y− sinh2 y +i (cosh y sinh y)

cos2 xcosh2 y + sin2 x sinh2 y since sin 2 x + cos 2 x = 1

=(sin x cos x ) + i (cosh y sinh y)

cos2 x + sinh2 y since cosh 2 y = 1 + sinh 2 y

2sin 2x+ i

2sinh 2y

cos2 x + sinh2 y since 2 cosh y sinh y = sinh 2yand 2 sin x cos x = sin 2x

=sin 2x+ i sinh 2y

2 cos2 x+ 2 sinh2 y

=sin 2x+ i sinh 2y

2 cos2 x− 1 + 2 sinh2 y+ 1

=sin 2x+ i sinh 2y

cos 2x + cosh 2y since cos 2x= 2 cos2 x− 1 and 2 sinh2 y + 1 = cosh 2y.

1.47 Let wbe a given complex number. If w 6 = ±1, show that there exists

two values of z =x + iy satisfying the conditions cos z =w and −π < x ≤ π.

Find these values when w =i and when w = 2.

Proof : Since cos z = e iz +e−iz

2,if we let e iz =u, then cos z =w implies

that

w= u 2 + 1

2u⇒ u 2 − 2wu + 1 = 0

which implies that

(u− w )2 =w2 − 1 6 = 0 since w 6 = ±1.

So, by Theorem 1.51,

eiz = u= w+  w2 −1 

1/ 2e iφ k ,where φ k =arg (w 2 −1)

2+ 2πk

2, k = 0, 1.

=w± w2 −1 

1/ 2e

i arg ( w2 −1 )

So,

ix− y= i ( x+ iy) = iz = log     

w± w2 −1 

1/ 2e i arg ( w2 −1 )

2    

+i arg 

w±  w2 −1 

1/ 2e

i arg ( w2 −1 )

2! 



Hence, there exists two values of z =x + iy satisfying the conditions cos z = w

and

−π < x = arg 

w±  w2 −1 

1/ 2e

i arg ( w2 −1 )

2! 

≤π.

For w = i, we have

iz = log   1±√ 2 i   +i arg  1±√ 2 i 

which implies that

z= arg  1±√ 2 i − ilog   1±√ 2 i   .

For w = 2, we have

iz = log   2±√ 3  +i arg  2±√ 3

which implies that

z= arg  2±√ 3 − ilog   2±√ 3  .

1.48 Prove Lagrange's identity for complex numbers:

    

k=1

ak bk     

k=1 |a k | 2

k=1 |b k | 2 −X

1≤k<j ≤na k ¯

bj −¯ aj bk  2 .

Use this to deduce a Cauchy-Schwarz ineqality for complex numbers.

Proof : It is the same as the Exercise 1.23; we omit the details.

1.49 (a) By eqating imaginary parts in DeMoivre's formula prove that

sin nθ = sinn θ (n

1) cot n−1θ−( n

3) cot n−3θ+ ( n

5) cot n−5θ−+...

Proof : By Exercise 1.44 (i), we have

sin nθ =

[n+1

k=1  n

2k− 1sin 2k−1θcos n− (2 k−1) θ

= sinn θ 









[n+1

k=1  n

2k− 1cot n−(2k − 1) θ









= sinn θ (n

1) cot n−1θ−( n

3) cot n−3θ+ ( n

5) cot n−5θ−+....

(b) If 0 < θ < π/ 2, prove that

sin (2m + 1) θ = sin2m+1 θPm  cot2 θ 

where Pm is the polynomial of degree m given by

Pm ( x) =  2m+1

1x m − 2m+1

3x m−1+ 2 m+1

5x m−2−+...

Use this to show that Pm has zeros at the m distinct points xk = cot2 {πk/ (2m + 1) }

for k = 1, 2, ..., m.

Proof : By (a),

sin (2m + 1) θ

= sin2m+1 θ n 2m+1

1cot 2 θ m − 2m+1

3cot 2 θ m−1+ 2 m+1

5cot 2 θ m−2−+...o

= sin2m+1 θPm  cot2 θ , where Pm (x ) =

m+1

k=1  2 m+1

2k− 1x m+1−k.(*)

In addition, by (*), sin (2m + 1) θ = 0 if, and only if, Pm (cot2 θ ) = 0. Hence,

Pm has zeros at the mdistinct points xk = cot2 { πk/ (2 m + 1)} for k =

1,2, ..., m.

k=1

cot2 πk

2m + 1 = m (2m− 1)

and the sum of their squares is given by

k=1

cot4 πk

2m + 1 = m (2m− 1) (4m2 + 10m− 9)

45 .

Note. There identities can be used to prove that P ∞

n=1 n − 2 =π 2 /6 and

P∞

n=1 n − 4 =π 4 /90. (See Exercises 8.46 and 8.47.)

Proof : By (b), we know that sum of the zeros of Pm is given by

k=1

xk =

k=1

cot2 πk

2m + 1 =− −  2m+1

3

2m+1

1!=m(2m− 1)

And the sum of their squares is given by

k=1

cot4 πk

2m + 1

= m

k=1

xk !2

−2 X

1≤i<j≤n

xixj !

= m (2m− 1)

3 2

−2  2m+1

5

2m+1

1!

=m (2m− 1) (4m2 + 10m− 9)

45 .

1.50 Prove that zn − 1 = Q n

k=1 z−e 2 πik/n for all complex z. Use this

to derive the formula n−1

k=1

sin kπ

n= n

2n−1 .

Proof : Since zn = 1 has exactly n distinct roots e2πik/n , where k =

0, ..., n − 1 by Theorem 1.51. Hence, zn − 1 = Q n

k=1 z−e 2 πik/n .It

implies that

zn−1 + ... + 1 =

n− 1

k=1 z−e 2 πik/n .

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