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SEMESTE SEMESTE SEMESTE SEMESTE CLASSICAL, CLASSICAL, CLASSICAL, CLASSICAL, TARUN KUMAR BAND In case you face difficulty in un tbanerjee1960@gm DOWNLOAD SITE: W 1 ER I MATHEMATICS ER I MATHEMATICS ER I MATHEMATICS ER I MATHEMATICS HONOURS HONOURS HONOURS HONOURS , , , , ABSTRACT ALGEB ABSTRACT ALGEB ABSTRACT ALGEB ABSTRACT ALGEB DYOPADHYAY, DEPARTMENT OF MATHEMAT nderstanding following material , you may e-m mail.com stating your Name and R WWW.SXCCAL.E S S S S BRA BRA BRA BRA TICS mail to me at Roll No. EDU

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SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS

CLASSICAL,CLASSICAL,CLASSICAL,CLASSICAL,

TARUN KUMAR BANDYOPADHYAY, DEPARTMENT OF MATHEMATICS

In case you face difficulty in understanding following material ,

[email protected]

DOWNLOAD SITE: WWW.SXCCAL.EDU

1

SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS

HONOURSHONOURSHONOURSHONOURS

CLASSICAL,CLASSICAL,CLASSICAL,CLASSICAL, ABSTRACT ALGEBRAABSTRACT ALGEBRAABSTRACT ALGEBRAABSTRACT ALGEBRA

TARUN KUMAR BANDYOPADHYAY, DEPARTMENT OF MATHEMATICS

In case you face difficulty in understanding following material , you may e-mail to me at

[email protected] stating your Name and Roll No.

WWW.SXCCAL.EDU

SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS SEMESTER I MATHEMATICS

ABSTRACT ALGEBRAABSTRACT ALGEBRAABSTRACT ALGEBRAABSTRACT ALGEBRA

TARUN KUMAR BANDYOPADHYAY, DEPARTMENT OF MATHEMATICS

mail to me at stating your Name and Roll No.

WWW.SXCCAL.EDU

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2

SECT ION I : ABSTRACT ALGEBRA

Text Abstract Algebra—Sen, Ghosh, Mukhopadhyay

CHAPTER I

SETS AND FUNCTIONS

In Mathematics, we define a mathematical concept in terms of

more elementary concept(s).For example, the definition of

perpendicularity between two straight lines is given in terms of

the more basic concept of angle between two straight lines. The

concept of set is such a basic one that it is difficult to define this

concept in terms of more elementary concept. Accordingly, we do we do we do we do

not define ‘set’not define ‘set’not define ‘set’not define ‘set’ but to explain the concept intuitivelyintuitivelyintuitivelyintuitively we say: a set

is a collection of objects having the property that given any

abstract (the thought of getting 100%marks at the term-end

examination) or concrete (student having a particular Roll No. of

semester II mathematics general) objet, we can say without any without any without any without any

ambiguityambiguityambiguityambiguity whether that object belongs to the collection(collection

of all thoughts that came to one’s mind on a particular day or the

collection of all students of this class) or not. For example, the

collection of ‘good’ students of semester II will not be a set unless

the criteria of ‘goodness’ is made explicit! The objects of which a

set A is constituted of are called elements of the set A. If x is an

element of a set A, we write xxxx�AAAA; otherwise xxxx�AAAA. If every element

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of a set X is an element of set Y,X is a subset of Y, written as X �Y.

X is a proper subset of Y if X �Y and Y�X, written as � � �. For

two sets X = Y iff (if and only if, bi-implication) X �Y and Y�X. A

set having no element is called null set, denoted by .

Example1.1 a≠{a} (a letter inside envelope is different from a letter

without envelope!�, {a}�{a,{a}}, {a}�{a,{a}}, A(the premise

x� of the implication x� ⇒ x� � is false and so the

implication holds vacuouslyvacuouslyvacuouslyvacuously !), A�A, for every set A.

Set Operations: formation of new sets

Let X and Y be two sets. Union of X and Y, denoted by X�Y, is the

set {a| a�X or a�Y or both}. Intersection of X and Y, denoted by

X� �, is the set {a| a�X and a�Y}. The set difference of X and Y,

denoted by X-Y, is the set {a| a�X and a� Y}. The set difference U-

X is called complement of the set X, denoted by X/, where U is the

universal set. The symmetric set difference of X and Y, denoted by

X��, is the set (X-Y) U(Y-X). For any set X, the power set of X, P(X),

is the set of all subsets of X. Two sets X and Y are disjoint iff X� �

= . The Cartesian product of X and Y, denoted by X X Y, is defined

as the set {(x,y)| x�X, y�Y} [ (x,y) is called an ordered pair. Two

ordered pairs (x,y) and (u,v) are equal, written (x,y) = (u,v), iff x =

u and y = v]. If we take X = {1,2} and Y = {3}, then X X Y =

{(1,3),(2,3)} ≠{(3,1),(3,2)} = Y X X. Thus Cartesian product

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between two distinct sets are not necessarily commutative (Is

� �1� � �1�� ?).

Laws governing set operations

For sets X, Y, Z,

� Idempotent lawsIdempotent lawsIdempotent lawsIdempotent laws: X� � � �, X�X = X

� Commutative lawsCommutative lawsCommutative lawsCommutative laws: X�Y = Y�X, X�Y = Y� �

� Associative LawsAssociative LawsAssociative LawsAssociative Laws: (X�Y) �Z = X�(Y�Z), (X�Y) �Z =

X�(Y� �� � Distributive lawsDistributive lawsDistributive lawsDistributive laws: X�(Y�Z) = (X� �� �(X�Z), X�(Y�Z) =

(X� �� �(X�Z)

� Absorptive lawsAbsorptive lawsAbsorptive lawsAbsorptive laws: X�(X�Y)=X, X�(X�Y)=X

� De’ Morgan’s lawsDe’ Morgan’s lawsDe’ Morgan’s lawsDe’ Morgan’s laws: X-(Y�Z) = (X-Y) �(X-Z), X-(Y�Z) = (X-

Y) �(X-Z)

Note We may compare between usual addition and multiplication

of real numbers on one hand and union and intersection of sets on

the other. We see that the analogy is not complete e.g. union and

intersection both are distributive over the other but addition is

not distributive over multiplication though multiplication over

addition is. Also A� � � �, for all set A but a.a = a does not hold

for all real a.

Example1.2 Let A, B, C be three sets such that A�C = B�C and A�C/ =

B�C/ holds. Prove that A = B.

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» A = A�U (U stands for the universal set conerned) =

A�(C�C/)(definition of complement of a set) =

(A�C) �(A�C/)(distributivity of � over�� = (B�C) �(B�C/)(given

conditions)= B�(C�C/)(distributivity of � ���� ��=B.

Note MakeMakeMakeMake a habit of citing appropriate law at each step as far as a habit of citing appropriate law at each step as far as a habit of citing appropriate law at each step as far as a habit of citing appropriate law at each step as far as

practicablepracticablepracticablepracticable....

Example1.3 Let A, B, C be three sets such that A�B = A�C and A�B =

A� �, then prove B = C.

» B = B�(A�B) = B�(A� �) = (B�A) �(B�C) (distributivity of �

over�) = (C�A) �(B�C) = C�(A�B)= C�(A� �) = C.

Example1.4 A�� � ��� implies A = B: prove or disprove.

Note ProvingProvingProvingProving will involve consideration of arbitraryarbitraryarbitraryarbitrary sets A,B,C

satisfying the given condition, whereas disprovingdisprovingdisprovingdisproving consists

of giving counter-examples of three particularparticularparticularparticular sets A,B,C that

satisfies the hypothesis A�� � ��� but for which the conclusion

A = B is false.

» This is a true statement. We first prove A�B. Let x�A.

Case 1 x�C. Then x� (A-C) �(C-A) = A�� = ��� = (B-C) �(C-B).

Thus x� � �. Since x�C, x� �.

Case 2 x�C. x� (A-C) � A�� = ��� = (B-C) �(C-B). Since x�C,

x�C-B. Thus x�B-C. So x� �.

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Combining the two cases, we see A� �. Similarly, B�A.

Combining, A = B.

Example1.5 Prove or disprove: (A-B)/ = (B-A)/.

» This is a FALSE statement. COUNTEREXAMPLECOUNTEREXAMPLECOUNTEREXAMPLECOUNTEREXAMPLE: Let U = A = {1,2}, B =

{1}.Then (A-B)/ = {1} ≠ (B-A)/ = {1,2}.

Example1.6 Prove: [(A-B) �(A�B)] �[(B-A) �(A�B)/] =

» By distributivity,[(A-B) �(B-A)] �[(A-B) � " A� B)/] �[(A�B) � "B A� ] �[(A�B) � "� � ��% /]= �[(A-

B) �(A/�B/)]� � = .

PRACTICE SUMS

1. Prove or disprove: A�(B-C) = (A�B) – (A�C)

2. Prove or disprove: A-C = B-C iff A�C = B�C.( ‘IFF’ stands for’

if and only if’)

3. Prove :A X (B�C) = (A X B) � (A X C)

4. (A∆B) ∆C=A∆(B∆C)

NOTATION: N,Z,Q,R,C will denote set of all positive integers,

integers, rational numbers ,real numbers and the complex

numbers respectively.

BINARY RELATIONS

A binary relation R from a set A to a set B is a subset of AXB. A

binary relation (we shall often refer to as relation) from A to A is

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called a binary relation on A. If (a, b)�R, we say a is R-related to b,

written as aRb.

Example1.7 Let A={1,2,3} and R={(1,1),(1,3)}. Then 1R3 but 3R1 does

not hold.

Example1.8 Let A be a set and P(A) denote the power set of A. Given

any two subsets X and Y of A, that is, X,Y� P(A), either X�Y or

X�Y. Thus � is a binary relation on P(A).

Example1.9 R={(x,y) �R2/ x2+y2=9} is a relation on R.

DefinitionDefinitionDefinitionDefinition 1.11.11.11.1 Let R be a binary relation on a set A.

• R is reflexive iff aRa holds ∀a�A

• R is symmetric iff a,b�A and aRb imply bRa

• R is transitive iff a,b,c�A, aRb,bRc imply aRc

• R is an equivalence relation on A iff R is reflexive,symmetric

and transitive.

Example1.10 Let R be a relation defined on Z by aRb iff ab≥0. R is

reflexive, symmetric but not transitive: -5R0, 0R7 but -5R7 does

not hold.

Example1.11 Let S be a binary relation on the set R of real numbers .

xSy iff Reflexive Symmetric Transitive

y=2x X X X

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x<y X X Yes

x(y X Yes X

xy>0 X Yes yes

y( ) *2 Yes X X

x≤y Yes X yes

xy≥0 Yes Yes X

x=y Yes Yes yes

DefinitionDefinitionDefinitionDefinition 1.21.21.21.2 Let R be an equivalence relation on a set A. Let a�A.

[a]={x�A/xRa} (�A) is the equivalence class determined by

a with respect to R.

DefinitionDefinitionDefinitionDefinition 1.31.31.31.3 Let A be a nonempty set and P be a collection of

nonempty subsets of A. Then P is a partition of A iff (1) for X,Y� P,either X=Y or X∩Y= and (2)A=+ �,�- .

Theorem 1.1 Theorem 1.1 Theorem 1.1 Theorem 1.1 Let R be an equivalence relation on a set A. Then (1)

[a]( , ∀a�A, (2) b�[a] iff [b]=[a], (3) either [a]=[b]or [a] ∩[b]= ,

(4) A=+ ./%0�1 . Thus {[a]/ a�A} is a partition of A.

Proof (1) since R is reflexive, (a,a) �R, ∀a�A. Thus a�[a]. Hence

[a]( , ∀a�A.

(2) if [b]=[a], then b�[b]=[a]. conversely, let b�[a]. then aRb. For

x�[a], xRa holds and , by transitivity of R, xRb holds, that is , x�[b].

Hence [a] �[b]. similarly [b] �[a] can be proved . Hence [b]=[a].

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(3) let [a] ∩[b]( . Let x�[a] ∩[b]. Then aRx,xRb imply aRb, that

is, [a]=[b].

(4) by definition, [a] �A, a�A. Thus, + ./%0�1 �A. conversely, let

b�A. Then b�[b] �+ ./%0�1 . Thus A�+ ./%0�1 .Hence A=+ ./%0�1 .

Theorem 1.2Theorem 1.2Theorem 1.2Theorem 1.2 Let P be a partition of a given set A. Define R on A by: for

all a,b �A, aRb iff there exists B� P such that a,b �B. Then R is an

equivalence relation on A.

Example1.12 Verify whether the following relations on the set R of

real numbers are equivalence relations: (1) aRb iff |/ 3| 40, "2� aRb iff 1+ab>0, (3) aRb iff|/|≤b

» (1)R is neither reflexive nor transitive but symmetric: 1R0 and

0R1 hold but 1R1 does not hold.

(2) R is reflexive and symmetric but not transitive: 3R(-78� and (-

78�9" 6� hold but 3R(-6) does not hold.

(3)(-2)R(-2)does not hold: not reflexive. -2R5 holds but 5R-2 does

not: R not symmetric. Let pRq and qRs hold. Then |;|≤q≤|<|≤s

imply pRs hold.

Example1.13 Verify whether the following relations on the set Z of

integers are equivalence relations: (1)aRb iff |/ 3| = 3, (2) aRb

iff a-b is a multiple of 6, (3) aRb iff a2-b2 is a multiple of 7, (4) aRb

iff |/| � |3|, (5) aRb iff 2a+b=41.

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Example1.14 Let X( . Prove that the following conditions are

equivalent: (1) Ris an equivalence relation on X, (2) R is reflexive

and for all x,y,z�X, xRy and yRz imply zRx, (3) R is reflexive and

for all x,y,z�X, xRy and xRz imply yRz.

Example1.15 A relation R on the set of all nonzero complex numbers

is defined by uRv iff ?@A?BA is real. Prove that R is not an equivalence

relation .

DefinitionDefinitionDefinitionDefinition 1.41.41.41.4 A functionfunctionfunctionfunction from a set A to a set B, denoted by f: ACB,

is a relation from A to B satisfying the properties:

� For every x�A, there exists y�B such that (x,y) �f. y is called

image of x under f and denoted by f(x) .x is called a preimage

of y=f(x). A is called domain and B is called the codomaincodomaincodomaincodomain of

the correspondence. Note that we differentiate between f,

the correspondence, and f(x), the image of x under f.

� For a fixedfixedfixedfixed x�A, f(x) � � is unique. For two different

elements x and y of A, images f(x) and f(y) may be same or

may be different.

In brief, a function is a relation under which

� both existence and uniqueness of imageboth existence and uniqueness of imageboth existence and uniqueness of imageboth existence and uniqueness of image of allallallall elements of

the domain is guaranteedguaranteedguaranteedguaranteed butbutbutbut

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� neither the existence nor the uniqueness of preimageneither the existence nor the uniqueness of preimageneither the existence nor the uniqueness of preimageneither the existence nor the uniqueness of preimage of

some element of codomain is guaranteedguaranteedguaranteedguaranteed.

NOTATION Let f:ACB. For y�B, f-1({y}) = , if y has no

preimage under f and stands for the set of all preimages if y has at

least one preimage under f. For two elements y1,y2�B, f-1({y1,y2})=

f-1({y1})� f-1({y2}). For C�A, f(C) = {f(c)| c� ��. f(A) is called the

range of f.

Example1.16 Prove that f(A�B) �f(A) �f(B) ; give a counterexample to

establish that the reverse inclusion may notmay notmay notmay not hold.

» y� f(A�B)⇒y = f(x), x� A�B⇒ y = f(x), x� A and x�B⇒y�f(A)and

y�f(B) ⇒y� f(A) �f(B). Hence f(A�B) �f(A) �f(B). Consider the

counterexample: f: RCR,f(x) = x2, A = {2}, B = {-2}.

Example1.17 Let f: RCR, f(x) = 3x2-5. f(x) = 70 implies x = ±5. Thus f-

1{70} = {-5, 5}.Hence f[f-1{70}] = {f(-5), f(-5)} = {70}. Also, f-1({-

11}) = [x� f-1({-11})⇒3x2-5=-11⇒x2=-2].

Example1.18 Let g: RCR, g(x) = D

DEB7. Find g-1({2}).

PRACTICE SUMS

Prove that (1) f(A�B) =f(A) �f(B), (2) f-1(B1�B2) = f-1(B1) �f-1(B2),

(3) f-1(B1�B2) = f-1(B1) �f-1(B2).

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DefinitionDefinitionDefinitionDefinition 1.51.51.51.5 A function under which uniqueness of preimage is

guaranteed is called an injectiveinjectiveinjectiveinjective function. Thus f: A∩B is injective

iff a1,a2 �A, f(a1)=f(a2) imply a1=a2. A function under which

existence of preimage is guaranteed is called a surjectivesurjectivesurjectivesurjective function.

f is surjective iff codomain and range coincide, that is, for every

y�B, there exists x�A such that f(x)=y. A function which is both

injective and surjective is called bijectivebijectivebijectivebijective.

Note The injectivity, surjectivity and bijectivity depends very

much on the domain and codomain sets and may well change with

the variation of those sets even if expression of the function

remains unaltered e.g. f:ZCZ, f(x) = x2 is not injective though

g:NCZ, g(x) = x2 is injective.

Example1.19 f: RCR, f(x) = x2 – 3x+4. f(x1) = f(x2) implies (x1-

x2)(x1+x2-3) = 0. Thus f(1) = f(2) though 1≠2; hence f is not

injective[Note: for establishing non-injectivity, it is sufficient to

consider particular values of x]. Let y�R and x� f-1{y}. Then y =

f(x) = x2 – 3x+4. We get a quadratic equation x2 – 3x+(4-y) = 0

whose roots, considered as a quadratic in x, give preimage(s) of y.

But the quadratic will have real roots if the discriminant 4y-7F0,

that is , only when yF7/4. Thus, for example, f-1{1} =. Hence f is

not surjective.

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DefinitionDefinitionDefinitionDefinition 1.61.61.61.6 If f:A B and g:B C, we can define a function

g0f:A C, called the compositioncompositioncompositioncomposition of f and g, by (g0f)(a) = g(f(a)),

a A.

Example1.20 f:Z Z and g: Z Z by f(n) = (-1)n and g(n) = 2n. Then

g0f: Z Z, (g0f)(n)=g((-1)n) = 2(-1)n and (f0g)(n) = (-1)2n. Thus g0f ≠

f0g. Commutativity of composition of functions need not hold.

ASSOCIATIVITY OF COMPOSITION OF FUNCTIONS

Theorem 1.3 Theorem 1.3 Theorem 1.3 Theorem 1.3 Let f:A∩ B , g:B∩C and h: C∩D. Then h0(g0f)=(h0g)0f.

Proof Both h0(g0f):A∩D and (h0g)0f: A∩D. for x�A, [h0(g0f)](x) =

h[(g0f)(x)]=h[g(f(x))] and [(h0g)0f](x)=(h0g)(f(x))=h[g(f(x))].

Hence [h0(g0f)](x)= [(h0g)0f](x), ∀x�A.

Theorem 1.4Theorem 1.4Theorem 1.4Theorem 1.4 Suppose f:A ∩B and g:B∩C. Then

(1) if f and g are both injective, then g0f is injective

(2) if g0f is injective, then fis injective

(3) if f and g are both surjective, then g0f is surjective

(4) if g0f is surjective, then g is surjective

DefinitionDefinitionDefinitionDefinition 1.71.71.71.7 Let f:A∩A. f0(a)=a, f1(a)=f(a), fn+1(a)=(f0fn)(a) for all

a�A and for all natural number n.

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Lemma 1.1Lemma 1.1Lemma 1.1Lemma 1.1 Let f:A∩A be injective. Then fn: A∩A is injective for all

natural number n.

Proof If possible, let there exists positive integer n such that fn is not

injective: let k be the smallest such positive integer. Thus there

exist a,b�A, a(b such that fk(a)=fk(b) holds. Now fk (a)= fk(b) ⇒

f[fk-1(a)]=f[fk-1 (b)] ⇒ fk-1(a)= fk-1 (b) (since f is injective) ⇒a=b

(since fk-1 is injective), contradiction. Hence the proof.

Theorem 1.5Theorem 1.5Theorem 1.5Theorem 1.5 Let A be finite and f: A∩A be injective. Then f is

surjective.

Proof Let a�A. Let B={a, f(a),f2(a),…} �A. since A is finite, there exist

positive integers r and s such that r>s and fr (a)=fs(a). By

injectivity of fs, fr-s(a)=a. if r-s = 1, a� f-1{a}. If r-s>1, then fr-s-1(a) is

a preimage of a under f. hence the result.

DefinitionDefinitionDefinitionDefinition 1.8 1.8 1.8 1.8 Let f:A∩B. f is called left(right) invertible if there

exists g:B∩A (resp. h:B∩A) such that g0f=1A(resp.f0h=1B). f is

invertible iff f is both left and right invertible.

Example1.21 (1) Let f,g:N∩N , f(n)=n+1; g(1)=1, g(n)=n-1 for n>1.

Then g0f=1N. But (f0g)(1)=2; so f0g(1N. Thus g is a left but not a

right inverse of f.

(2) Let f:Z∩E (E is the set of all even nonnegative integers),

f(x)=x+|)| and g:E∩Z, g(x)=x/2. Then f0g=1E but (g0f)(-1)=0 hence

g0f(1Z. Hence g is right but not left inverse of f.

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(3) Let f,g: R∩R, f(x)=3x+4, g(x)=D@GH . Then g0f = f0g=1R. Thus g is

both left and right inverse of f.

Note If g:B∩A and h:B∩A be a left inverse and a right inverse of f:

A∩B, then g=h: for g=g01B=g0(f0h)=(g0f)0h=1A0h=h. as we shall see

below, a function may have many left (right) inverses without

having any right(left respectively) inverse.

Theorem 1.6Theorem 1.6Theorem 1.6Theorem 1.6 Let f: A∩B. Then (1) f is left invertible iff f is injective,

(2) f is right invertible iff f is surjective (3) f is invertible iff f is

bijective.

Proof (1) Let f be left invertible. Then there exists g:B ∩A such that

g0f=1A. f(a1)=f(a2), a1,a2�A ⇒g(f(a1))=g(f(a2)) ⇒(g0f)(a1)=(g0f)(a2)

⇒1A(a1)=1A(a2) ⇒a1=a2. Hence f is injective. Conversely, let f be

injective. Fix a0�A. Define g:B∩A by : g(b)=a, if a be the unique (by

injectivity of f, if preimage of b exists under f) preimage of b under

f and g(b)=a0,if b has no preimage under f. Clearly g:B∩A is a

function and (g0f)(a)=g(f(a))=a (by the definition of g)=1A(a),

∀a�A. Hence g is a left inverse of f.

Example1.22 Verify whether the following functions are injective

and/or surjective: f:RCR, f(x) = x|)|, f: (-1,1) CR, f(x) = D

7B|D|.

Example1.23 Let f:A∩B. Prove that (1) f is injective iff X=f-1(f(X)) for

all X�A, (2) f is surjective iff f(f-1(Y))=Y for all Y�B, (3) if f is

injective, then f (Y-Z) �f(Y)-f(Z) for Y,Z�A.

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Example1.24 Prove that f(X�Y)=f(X) �f(Y), f(X∩Y) �f(X) ∩f(Y); give

counterexample to show that the inclusion may be strict (f:R∩R,

f(x)=x2, X={2},Y={-2}).

DefinitionDefinitionDefinitionDefinition 1.9 1.9 1.9 1.9 Let f:ACB be a bijective function. We can

define a function f-1: BC A by f-1(y) = x iff f(x) = y. Convince

yourself that because of uniqueness and existence of preimage

under f ( since f is injective and surjective), f-1 is indeed a function.

The function f-1 is called the inverse functioninverse functioninverse functioninverse function to f. The graphs of f

and f-1 for a given f can be seen here: ..\Documents\x8.mw

Note Graph of f-1 can be obtained by reflecting the graph of f

about line x=y.

Example1.25 Let f: (0,1) C(1/2,2/3) be defined by f(x) = DB7DBI. Verify

that f is bijective.

[explanation:f(x)=f(y)⇒ DB7DBI�JB7JBI⇒x=y. Let c�(1/2,2/3). If possible, let x be a pre-image of c under f, that is, f(x)=c. then c=DB7DBI implying x=7@Ibb@7 �(0,1) since -1/2<c-1<-1/3, -1/3<1-2c<0]. f-1:

(1/2,2/3) C (0,1) is to be found. Now, let f-1(y) = x, y� (1/2,2/3) . Then f(x) = y. So

DB7DBI=y and hence x =

7@IJJ@7 = f-1(y).

BINARY OPERATIONS

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DefinitionDefinitionDefinitionDefinition 1.10 1.10 1.10 1.10 Let A≠. A binary operation ‘0’ on A is a function

from A X A to A. In other words , a binary operation ‘0’ on A is a

rule of correspondence that assigns to each ordered pair (a1,a2) �

A X A, some element of A, which we shall denote by a1 0a2. Note

that a1 0a2 need not be distinct from a1 or a2.

Example1.26 Subtraction is a binary operation on Z but not on N;

division is a binary operation on the set Q* of all nonzero rational

number but not on Z.

DefinitionDefinitionDefinitionDefinition 1.11 1.11 1.11 1.11 Let 0 be a binary operation on A≠. (A,0)is called a

mathematical system. 0 is commutativecommutativecommutativecommutative iff x0y = y0x holds ,for all

x,y�A. 0 is associativeassociativeassociativeassociative iff x0(y0z) = (x0y)0z holds for all x,y,z�A. An

element e�A is a left left left left identityidentityidentityidentity of the system (A,0) iff e0x =

x holds ∀x�A. An element e�A is a right identityright identityright identityright identity of the

system (A,0) iff x0e = x holds ∀x�A. An element in a system which

is both a left and a right identity of the system is called an

identityidentityidentityidentity of the sytem. (A,0) be a system with an identity e

and let x, y�A such that x0y = e holds. Then y(x) is called a right right right right

inverseinverseinverseinverse to x(x is a left left left left inverseinverseinverseinverse of y respectively) in (A,0). y �A is an

inverse to x�A iff x0y=y0x=e.

Example1.27 Consider the system (R,0) defined by x0y = x, ∀x,y�R (R

stands for the set of real numbers). Verify that 0 is non-

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commutative, associative binary operation and that (R,0) has no

left identity though(R,0) has infinite number of right identity.

Example1.28 Verify that subtraction is neither associative nor

commutative binary operation on Z. (Z,-) does not have any

identity.

Example1.29 Consider the system (Z,*) where the binary operation *

is defined by a*b = |/ * 3|, a,b�Z. Verify that * is commutative but

not associative [ note, for example, that {(-1) *2}*(-3)≠(-1) *{2*(-

3)}]. (Z,*) does not have an identity.

Example1.30 (R,+) is commutative, associative, possess an identity

element 0 and every element of (R,+) has an inverse in (R,+).

NoteNoteNoteNote From examples 1.12 to 1.15 it is clear that associativity and

commutativity of a binary operation are properties independent

of each other, that is, one can not be deduced from the other.

Example1.31 Let 2Z denote set of all even integers. 2Z, under usual

multiplication, form a system which is associative, commutative

but possesses no identity.

Example1.32 Let M2(Z) =de/ 3f gh | /, 3, f, g � �i . M2(Z) under usual

matrix addition forms a system which is commutative,

associative. (M2(Z),+) possesses an identity, namely the null

matrix, and every element in (M2(Z),+) has an inverse in

(M2(Z),+).

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Example1.33 Let GL(2,R) denote the set of all 2x2 real non-singular

matrices under usual matrix multiplication. The system is

associative, non-commutative, possesses an identity and every

element has an inverse in the system.

SEMIGROUP, MONOID AND GROUP

DefinitionDefinitionDefinitionDefinition 1.121.121.121.12 A nonempty set S with an associative binary

operation ‘.’defined on S forms a semigroup . A semigroup M that

contains an identity element is called a Monoid . A monoid in

which every element is invertible is called a Group . Thus (G,.) is

a group iff following conditions are satisfied: (1) . is associative,

(2) (G,.) has an identity element, generally denoted by e and (3)

every element x�G has an inverse element x-1�G. If, in addition,

(G,.) is commutative , (G,.) is an abelian group.

NoteNoteNoteNote x-1 is not to be confused with 1/x, which may be a

meaningless expression keeping the generality of the underlying

set into account. If the group operation is denoted by ‘+’, then

inverse of x is denoted by – x.

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Example of formation of negation of a mathematical statement:

Does there exist

elements a,b in G such

that a.b∉∉∉∉GGGG

YES

Ex: (N,-)

(G,.)

NOT A

GROUP

Does there exist three

elements a,b,c in G such

that (a.b).c ≠a.(b.c)

holds

NOEx: Nonzero

rationals

under

division

(G,.)

NOT A

GROUP

NO

DOES THERE NOT EXIST AN ELEMENT

e SUCH THAT e.a=a.e=a HOLDS FOR

ALL a IN G AND e IS INDEPENDENT

OF a IN G

YES Ex:REAL NUMBERS

UNDER a.b=

ba.

ba.

(G,.)

NOT A

GROUP

NO

DOES THERE EXIST TWO ELEMENTS

a,b SUCH THAT NOT BOTH a.b=e

AND b.a=e HOLD

YESNON-NEGATIVE

INTEGERS UNDER

ADDITION

(G,.)

NOT A

GROUP

(G,.) IS A GROUP

NO

YES

UNDER WHAT CIRCUMSTANCE DO WE SAY (G,.) IS NOT A GROUP?

Example1.34 Verify whether (Z,0) defined by a0b = a+b – ab (usual

operations on Z on the RHS) forms a group.

» (a0b)0c = (a+b – ab)0c = (a+b – ab)+c – (a+b-ab)c = a+b+c – ab –

bc – ca+ abc

a0(b0c) = a0(b+c – bc) = a+(b+c – bc) – a(b+c – bc) = a+b+c – ab –

bc – ca+abc.

So, (a0b)0c = a0(b0c), for a,b,c�Z. Thus (Z,0) is associative.

Clearly, a00 = 00a = a, ∀a�Z. Thus (Z,0) possesses an identity 0�Z.

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Let a�z and b�Z be an inverse to a. by definition, a0b = b0a = 0.

Thus a+b – ab = 0. Hence , for a≠1, b = a/(a-1). But , in particular,

for a = 3, b = 3/2∉Z. Hence (Z,0) does NOT form a group.

Example1.35 Verify whether (R,0) defined by a0b =|/| * |3|, a,b�R

forms a group.

» Verify that (R,0) is associative and commutative. If (R,0) has an

identity e�R, then a0e = e0a = a holds for all a �R. Thus |/| * |�| =

a. hence |�| = a - |/|<0 for a<0, contradiction. Hence (R,0) has no

identity element and hence does NOT form a group.

Example1.36 Let A( and X={f/f:A∩B}. The system (X,0) where 0 is

composition of functions, is associative but not commutative if A

contains at least three elements.

Example1.37 Let n be a natural number. Define a binary relation S on

Z by: aSb iff a-b=nk, ∃k�Z. S is an equivalence relation and the set

of equivalence classes Zn= {[0],[1],…,[n-1]} form a partition of Z,

where [r] = {nk+r/ k�Z}. define a binary operation +n on Zn by:

[r]+n[s]=[r+s], ∀[r],[s] � Zn. the definition is meaningful :If [r]=[r1]

and [s]=[s1], then r=nk+r1 and s=nm+s1 and hence

r+s=(k+m)n+(r1+s1) so that [r+s]=[r1+s1]. +n is associative:

([r]+n[s])+n[t]=[r+s]+n[t]=[(r+s)+t]=[r+(s+t)]=[r]+n([s]+n[t]). [0] is

an identity in (Zn,+n). inverse of [0] is [0] and inverse of [r] is [n-r],

0<r<n. Thus (Zn,+n) is an an abelian group.

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Example1.38 Let n be a prime integer and let Zn={[1],[2],…,[n-1]}.

Define [r].n[s]=[rs]. Definition is meaningful: [r]=[r1] and [s]=[s1]

imply r=kn+r1,s=mn+s1, ∃k,m�Z, so that rs=n(mr1+ks1+kmn)+r1s1

implying [rs]=[r1s1]. Associativity of .n follows from associativity

of . .[1] is an identity of (Zn,.n). Let [r] �Zn, 1=r<n. since n is prime,

r and n are relatively prime so that there exist integers p,q such

that 1=pr+qn. Thus [1]=[pr]=[p].n[r]=[r].n[p] implying [p] is an

inverse of [r]. hence (Zn,.n) is an abelian group.

NoteNoteNoteNote If n is not prime, (Zn,.n) is not a group: if 1<p<n,1<q<n,n=pq ,

p,q natural,then [p],[q] �Zn but [pq]=[n]=[0] ∉Zn.

Example1.39 Let X( and S(X) be the set of all bijective functions

from X onto X.(S(X),0)is a group. If X contains at least three

elements, then S(X) is not commutative. Consider f,g� S(X)

defined by f(a)=a, f(b)=c,f(c)=b; g(a)=b,g(b)=a,g(c)=c. Then

(f0g)(a)((g0f)(a). Hence f0g(g0f.

Example1.40 consider the group GL(2,R), the set of all 2x2 real

nonsingular matrices with usual multiplication of matrices.

GL(2,R) is non-commutative.

PRACTICE SUMS

Verify whether following system forms group or not:

(1) (Z,0) , a0b = a|3|, a,b�Z

(2) (Z,0) , a0b = a+b+2

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(3) (2Z+1,*), a*b = a+b-1 [2Z+1 stands for set of all odd

integers]

(4) Let S( and P(S) be the power set of S. Consider

(P(S), �).

(5) Let S( and P(S) be the power set of S. Consider

(P(S), �).

(6) Let Q[l2% = {a+√2 b| a,b�Q}. consider (Q[l2%-{0},.)

Elementary properties of Group

Theorem 1.7 Theorem 1.7 Theorem 1.7 Theorem 1.7 Let (G,.) be a group. The following properties hold:

(1) If G has a left identity e and a right identity f, then e=f.

In particular, identity element in G is unique.

(2) If an element x of G has a left inverse y and a right

inverse z, then y=z. In particular, x-1, inverse element to x, is

unique.

(3) e-1 = e, (x-1)-1 = x, ∀x�G.

(4) (Cancellation Laws) for a,b,c�G, a.c = b.c⇒a = b(right

cancellation property) c.a = c.b⇒a = b(left cancellation

property)

(5) (a.b)-1 = b-1.a-1, a,b�G.

Proof (1) Let e,f be two identities of (G,.). Then e = e.f (f is a right

identity element) = f (e is a left identity element).

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(2) y.x = e and x.z = e. Then y = y.e = y.(x.z) = (y.x).z = e.z = z. hence

x-1, inverse element for given x is unique.

(3) Since e.e= e and x.x-1= e, it follows from definition of identity

and inverse element.

(4) a.c = b.c⇒(a.c).c-1 = (b.c).c-1⇒a.(c.c-1) = b.(c.c-1) (associativity) ⇒a.e = b.e⇒a = b.

(5) (a.b).(b-1.a-1) =a.(b.b-1).a-1 = (a.e).a-1 = a.a-1 = e. Similarlyimilarlyimilarlyimilarly (need be mentioned)(b-1.a-1).(a.b) = e. Hence.

NotationNotationNotationNotation Let G be a group, a�G, n�Z. a0=e(identity), an =((

a.a)…a)( n times, n� r),an = (a-1).(a-1)…(a-1) (-n times, -n�N).

Example1.41 Let (G,.) be a group such that (a.b)-1 = a-1.b-1, 'a,b�G.

Prove that G is abelian.

»For all a,b�G, (a.b)-1 =a-1.b-1 = (b.a)-1. Hence a.b = [(a.b)-1]-1 =

[(b.a)-1]-1 = b.a, 'a,b�G.

Example1.42 Let (G,.) be a finite abelian group and G={a1,a2,…,an}. Let

x = a1.a2….an� G. Prove that x2 = e.

» using commutativity and associativity of (G,.), x2 can be

expressed as finite product of pairwise product of pair of

elements of G, each pair consisting of elements which are

mutually inverse to each other. Hence the result.

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Example1.43 Let G be a group such that a2 = e, for all a�G. Prove that

G is abelian.

» For a� s,a.a = a2=e = a.a-1⇒ a = a-1⇒a.b = a-1.b-1 = (b.a)-1 = b.a, for a,b�G. Hence.

Example1.44 Let G be a group such that for a,b,c� s , a.b = c.a implies

that b = c. Show that G is abelian.

» (a.b).a = a.(b.a), for a,b,c�G (by associativity) ⇒a.b = b.a

Example1.45 Prove that G is abelian iff (a.b)2 = a2.b2, 'a,b�G.

» Sufficiency a.(b.a).b = (a.b)2 = (a.a)(b.b) = a.(a.b).b⇒ b.a = a.b for a,b�G.

Necessity Necessity Necessity Necessity If G is abelian, then (a.b)2 = (a.b).(a.b) = a.(b.a).b = a.(a.b).b = (a.a).(b.b) = a2.b2.

Theorem 1.8Theorem 1.8Theorem 1.8Theorem 1.8 A semigroup (S,.) is a group iff (1) ∃e�S, 'a�S such that

e.a=a and (2) 'a �S, ∃b�S such that b.a=e

Proof Let S be a semigroup which satisfies conditions (1) and (2).

Let a�S. By (2), corresponding to a�S, there exists b�S such that

b.a=e. for b�S, by(2), there exists c�S such that c.b=e. now,

a=e.a=(c.b).a=c.(b.a)=c.e and a.b=(c.e).b=c.(e.b)=c.b (by(1))=e.

hence a.b=b.a=e. Also, a.e=a.(b.a)=(a.b).a=e.a=a. Thus a.e=e.a=a,

'a�S. Thus e�S is an identity and b �S is an inverse of a�S. Thus

(S,.) is a group. Converse part follows from definition of a group.

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Example1.46 Let J=de) u) uh ), u � R, x * y ( 0i. Show that J is a

semigroup under matrix multiplication . Show that J has a left

identity and each element of J has a right inverse.

Theorem 1.9Theorem 1.9Theorem 1.9Theorem 1.9 A semigroup(S,.) is a group iff ∀a,b�S, the equations

a.x=b and y.a=b have solutions for x and y in (S,.).

Proof Let a�S. The equation x.a=a has a solution ,say,u�S. Then

u.a=a.Let b�S. The equation a.x=b has solution,say,c�S. Thus

a.c=b. Now u.b=u.(a.c)=(u.a).c (associativity in (S,.))=a.c=b. Since

b�S was arbitrary, u�S is a left identity in (S,.). Again, the

equation y.a=u has solution , say d�S. Then d�S is a left inverse of

a in S. Thus (S,.) is a group, by previous theorem.Converse part is

obvious.

Theorem 1.10 Theorem 1.10 Theorem 1.10 Theorem 1.10 A finite semigroup (S,.) is a group iff (S,.) satisfies both

the cancellation laws: for a,b,c�S, a.b=a.c⇒b=c (left cancellation

law) and b.a=c.a⇒b=c(right cancellation law).

Proof Let (S,.) be a finite semigroup in which both the cancellation

laws hold. By previous theorem, it is sufficient to show that the

equations a.x=b and y.a=b have solutions in S, ∀a,b�S. Let

S={a1,…,an}. Let a,b�S. clearly A={a.a1,…,a.an}�S. Also all the

elements of A are distinct by left cancellation law. Thus A and S

contains same number of elements and A�S. Thus b

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�S=A={a.a1,…,a.an}. Hence b=a.ai, for some i,1≤i≤n. Thus a.x=b has

solution in S, ∀a,b�S. Similarly, x.a=b has solution in S, ∀a,b�S.

Hence (S,.) is a group. Converse part is trivial.

In a semigroup, cancellation laws may or may not hold. In (N,+),

cancellation laws hold. In the semigroup S of all 2x2matrices over

integers under multiplication, for A=e1 00 0h,B=e0 00 1h,C=e0 01 0h,

AB=AC holds but B(C.

DefinitionDefinitionDefinitionDefinition 1.131.131.131.13 Let (G,.) be a group, a�G, n�Z. a0=e;an=a.an-1, if n�N;

an=(a-1)-n, if -n�N.

DefinitionDefinitionDefinitionDefinition 1.141.141.141.14 Let (G,.) be a group and a�G. If there exists positive

integer n such that an=e, then the smallest such positive integer n

is the order of a, denoted by 0(a). If no such positive integer exist,

then a is of infinite order.

NoteNoteNoteNote In a group G, the only element of order 1 is identity. All

elements in a finite group must be of finite order.

Example1.47 (R,+)is an infinite group in which all elements other

than 0 are of infinite order. (P(R), ∆) is an infinite group in which

all nonidentity elements are of order 2: that is, every element is of

finite order. In (Z6,+6), elements [0],[1],[2],[3],[4],[5] have

respective orders 1,6,3,2,3,6.

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Theorem 1.11Theorem 1.11Theorem 1.11Theorem 1.11 Let G be a group and a�G such that 0(a)=n. Then (1) if

am=e for some positive integer m, then n divides m, (2) for every

positive integer t, 0(at)=w

xyz ({,w).

Proof (1) by division algorithm of integers, ∃q,r�Z such that

m=nq+r, where 0≤r<n. Now ar=am-nq=am.(an)q=e. Since n is the

smallest positive integer such that an=e and ar=e for 0≤r<n, it

follows that r=0.Thus m=nq, that is, n divides m.

(2) Let 0(at)=k. then akt=e. by (1), n divides (kt). Then there exists

r�Zsuch that kt=nr.Let gcd(t,n)=d. then ∃u,v�Z such that t=du,

n=dv and gcd(u,v)=1. Now kt=nr implies ku=rv. Thus v divides

(ku). Since gcd(u,v)=1, v divides k. thus n/d divides k. Also,

(at)n/d=/|}~ = /w? = (/w)? = �. Since 0(at)=k, k divides n/d. since

k and n/d are positive integers, k=n/d. Hence, 0(at)=w

xyz ({,w).

Example1.48 In a group of even order, there exists at least one

nonidentity element of order 2.

» Let A={g�G/g(g-1}⊆G. Then e∉ A and g� A implies g-1�A. Thus

A=+ ��, �@7���1 . Hence number of elements of A is even and so

A∪{e} contains odd number of elements. Since the number of

elements of G is even, there exist g�G such that g∉ A∪{e}, so that

,g=g-1 and g(e. thus 0(g)=2.

Example1.49 Let G be a group and a,b�G, a2=e,aba=b7. Prove b48=e.

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» b=a2ba2=ab7a=(aba)7=b49. Result follows by cancellation law.

Example1.50 Let S be a semigroup such that for all a�S, there exist

x�S such that a=axa. If S has a single idempotent element, prove

that S is a group.

» Let the single idempotent of S be denoted by e. Let a�S. Then

there exist x�S such that a=axa so that ax=axax=(ax)2. Thus ax is

an idempotent in S and by the uniqueness of idempotent in S,

ax=e. Similarly it can be shown that xa=e. Hence a=axa=ae and

a=axa=ea. Thus e is an identity in S. Let b�S. then there xeist y�S

such that b=byb. Hence by and yb are idempotent and so by=yb=e.

Hence S is a group.

Example1.51 Let G be a group and (a.b)n=an.bn holds for all a,b in G

and for three consecutive integers n. Prove that G is commutative.

» let (ab)n=anbn, (ab)n+1=an+1bn+1, (ab)n+2=an+2bn+2, 'a,b�G.

Then an+1bn+1=(ab)n+1=(anbn)(ab) ⇒abn=bna.

Again an+2bn+2=(ab)n+1(ab)=an+1bn+1ab

⇒abn+2=bn+1ab=b(bna)b=b(abn)b⇒ab=ba. Hence.

Example1.52 Let G={a1,…,an} be a finite abelian group and x=a1…an�G.

Prove that x2=e.

» using commutativity and associativity, x2=( a1…an)( a1…an)can

be expressed as finite products of expressions like (ai./��), where

ai and /�� are inverse to each other. Hence ai./��=e and thus x2=e.

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Example1.53 Let G be a group and x,y�G such that xy2=y3x, yx2=x3y.

Prove x=y=e.

» xy2=y3x⇒x=y3xy-2⇒x2=xy3xy-2=(xy2)yxy-2=(y3x)yxy-

2⇒x2y=y3xyxy-1 (1). Now yx2=x3y⇒yx2=xy3xyxy-1⇒x2=y-1xy3xyxy-

1⇒x2y=y-1xy3xyx (2). By (1) and (2), y3xyxy-1=y-

1xy3xyx⇒y4xyx=xy3xyxy⇒y4xyx=xy2yxyxy=y3 (xy)3⇒(yx)2=(xy)2

(3). Interchanging x and y in (3),we get (xy)2=(yx)3 (4). Now (3)

and (4) imply (xy)2=(yx)3=(yx)2(yx)=(xy)3(yx) ⇒e=xy2x⇒x-2=y2.

Further xy2=y3x⇒xx-2=yx-2x⇒x-1=yx-1⇒y=e. finally,

yx2=x3y⇒ex2=x3e⇒x=e.

Example1.54 Let S be a finite semigroup. Show that there is an

idempotent element e in S .

» Let x�S. Since S is finite, all the elements x,x2,x3,… cannot be

distinct. Thus there exist integers m,n, m>n, such that xm=xn. Thus

there exists integer k such that xn+k=xn. Now x2n+k=xn+k.xn=x2n. By

induction, xmn+k=xmn for any m�N. Also xmn+2k=xmn+k=xmn. By

induction, xmn+lk=xmn, for any l�N. In particular, xkn+nk=xkn, that is,

e2=e, where e=xkn�S.

Example1.55 Prove that a finite semigroup G with identity is a group

iff G contains only one idempotent. Give a counterexample to

show that if we drop the requirement of G possessing identity,

then G need not be a group.

» consider the semigroup {[0],[2]} under +4.

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Example1.56 If G is a group in which (ab)2=a2b2 holds 'a,b�G, then G

is abelian. Give example to show that the result does not hold for

semigroup.

» consider the semigroup (S,.), where S is any nonempty subset

and a.b=a, 'a,b�S. Then S is not a group though (ab)2=a2b2 holds

'a,b�S.

Example1.57 Show that if G is a finite semigroup with cross-

cancellation law, that is, xy=yz implying x=z, for all x,y,z in G, then

G is an abelian group.

» xy=xz⇒x(yx)=(xz)x⇒yx=xz⇒y=z. similarly right cancellation

law holds. Hence G is a group. Also (xy)x=x(yx) imply xy=yx, for

all x,y in G.

Example1.58 Let Sbe a semigroup and for all x,y in S, x2y=y=yx2 hold.

Prove that S is an abelian group.

Example1.59 Let G be a group and a,b�G such that ab=ba and 0(a) and

0(b) are relatively prime. Then prove that 0(ab)=0(a)0(b).

Example1.60 Find the number of elements of order 5 in (Z20,+20).

»Let 0([a])=5. Then 5 is the smallest positive integer such that

[0]=5[a]=[5a], that is, 20 divides 5a-0=5a. Thus 4 must divide

a,0≤a<20. Hence [a]=[4],[8],[12]or[16].

Example1.61 Suppose a group G contains elements a,b such that

0(a)=4,0(b)=2, a3b=ba. Find 0(ab).

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» a3b=ba⇒e=a4b2=(ab)2⇒0(ab) ≤2. If 0(ab)=1, then ab=e, a=b-1,

implying 4=0(a)=0(b)=2, contradiction.

Permutation GroupsPermutation GroupsPermutation GroupsPermutation Groups

DefinitionDefinitionDefinitionDefinition 1.151.151.151.15 Let A( . A permutation on A is a bijective mapping

of A onto itself. If A={1,2,…,n}, then the group Sn formed by the set

of all permutations on A under composition of functions as

composition is called Symmetric Group on n symbols . Order of Sn,

that is , the number of elements of Sn , is n!

N o t a t i o nN o t a t i o nN o t a t i o nN o t a t i o n If ��Sn, then we often denote � using two-row

notation: � = � 1 2 � ��"1� �"2�� �"���. In this notation, if

� � � 1 2 � ��"1� �"2�� �"��� and �

� 1 2 � ��"1� �"2�� �"���, then

�0 �=� 1 2 � ��"�"1�� �"�"2��� �"�"����.

Theorem 1.12Theorem 1.12Theorem 1.12Theorem 1.12 If n is a positive integer ,n≥3, then Sn is a

noncommutative group.

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DefinitionDefinitionDefinitionDefinition 1.161.161.161.16 A permutation � on {1,2,…,n} is a k-cycle iff there

exist distinct elements i1,…,ik�{1,2,…,n} such that �(i1)=i2,…, �(ik-

1)=ik, �(ik)=i1 and �")� � ) for all x�{1,2,…,n}-{ i1,…,ik}. we denote

� by (i1 i2 …ik). A transposition is a 2-cycle.

NoteNoteNoteNote Product of two cycles need not be a cycle: for �=(5 6) and

�=(3 2 4), �0 � is not a cycle.

DefinitionDefinitionDefinitionDefinition 1.171.171.171.17 Two cycles (i1…im) and (j1…jk) are disjoint iff {

i1,…,im}∩{ j1,…,jk}=.

Theorem 1.13Theorem 1.13Theorem 1.13Theorem 1.13 Let � and � be two disjoint cycles. Then �0 �= �0�.

Theorem 1.14Theorem 1.14Theorem 1.14Theorem 1.14 Any nonidentity permutation of Sn (n≥2) can be

expressed as a product of disjoint cycles , where each cycle is of

length ≥2.

Theorem 1.15Theorem 1.15Theorem 1.15Theorem 1.15 Any cycle of length≥2is either a transposition or can be

expressed as a product of transpositions.

Proof Note that (i1 i2…ik)=(i1 ik)0(i1 ik-1)0…0(i1 i2).

Theorem 1.16Theorem 1.16Theorem 1.16Theorem 1.16Any nonidentity permutation is either a transposition

or can be expressed as a product of transpositions.

DefinitionDefinitionDefinitionDefinition 1.181.181.181.18 A permutation � is even(odd) iff � can be expressed

as product of even (odd respectively)number of transpositions.

Theorem 1.17 Theorem 1.17 Theorem 1.17 Theorem 1.17 Any permutation in Sn is either an even permutation or

an odd permutation but never both.

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NoteNoteNoteNote Identity permutation I is even since I=(1 2)0 (2 1). If � is

even, then �-1 is even, since �0 �-1=I and I is even. Thus An, the set

of all even permutations on {1,2,…,n}, forms a group under

composition of functions. An is called the alternating group.

Theorem 1.18Theorem 1.18Theorem 1.18Theorem 1.18 Let n≥2 and �� Sn be a cycle. Then � is a k-cycle iff

0(�)=k.

Theorem 1.19Theorem 1.19Theorem 1.19Theorem 1.19 Let �� Sn, n≥2, and �= �1 0…0 �k be a product of disjoint

cycles. Let 0(�i)=ni, i=1,2,…,k. then 0(�)=lcm{n1,n2,…,nk}.

Example1.62 Show that the number of even permutations in Sn is

same as the number of odd permutations in Sn.

Proof Define f:An∩(Sn-An) by f(�)= �0 (1 2). Verify that f is bijective.

Example1.63 If ��S7 and �G=(2 1 4 3 5 6 7), then find �.

» 7=0(�G)=�(�)

xyz �G,�(�)�. Thus 0(�)=7,14,28. Now � is expressible as

product of disjoint cycles in S7. Thus 0(�)(14,28. Hence 0(�)=7.

Example1.64 If �=(1 2 3)0(1 4 5), write �88 in cycle notation.

» �� = �.Thus �7��=I and hence �88=�@7=(4 1 3 2 5).

Example1.65 In S6, let �=(1 2 3) and �=(4 5 6). Find a permutation x

in S6 such that x �x-1= �.

Example1.66 Find the number of elements of order 3 in A4.

Example1.67 Find the number of elements of order 6 in S4.

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Example1.68 Find the number of elements of order 2 in A4.

SUBGROUPSUBGROUPSUBGROUPSUBGROUP

Let (G,0) be a group and (H⊆G. H is closed under 0 iff ∀ h1,h2�H,

h1 0h2�H. If H be closed under 0, then the restriction of 0 to H X H is

a mapping from HX H into H. Thus the binary operation 0 on G

induces binary operation, also denoted by 0, on H.

DefinitionDefinitionDefinitionDefinition 1.191.191.191.19 Let (G,0) be a group and (H⊆G. (H,0) is a

subgroup of (G,0) iff H is closed under 0 and (H,0) is a group.

Every group G has at least two subgroups , namely, {e} and G.

These are called trivial subgroups. Other subgroups, if there be

any, are called nontrivial subgroups of G.

Example1.69 (2Z,+) is a subgroup of (Z,+) , where 2Z is the set of all

odd integers. Since the set 2Z+1 of all odd integers is not closed

under addition of integers, (2Z+1,+) is not a subgroup. Though N

is closed under addition, (N,+) does not form a group; hence (N,+)

is not a subgroup of (Z,+).

Example1.70 (An,0) is a subgroup of (Sn,0). {[0],[2]} is a subgroup of

(Z4,+4). {z�C/|�| = 1} is a subgroup of the multiplicative group of

all nonzero complex numbers.

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Theorem 1.20 Theorem 1.20 Theorem 1.20 Theorem 1.20 All subgroups of a group (G,0) have the same identity

element. If H be a subgroup of a group G, a�H and aG-1, aH

-1 be the

inverses of a in G and H respectively, then aG-1=aH

-1.

Proof Let eG and eH be the identities of G and H respectively. Then eH

0 eH=eH (since eH�H and eH is identity in (H,0) ) =eH0 eG (since

eH�H⊆G and eG is identity in (G,0) ) which by cancellation

property in (G,0) implies eH= eG.

Using cancellation property of G, from the equality a. aG-

1=eG=eH=a. aH-1, we conclude aG-1=aH-1.

NoteNoteNoteNote ��H: H is a subgroup of G� (

NoteNoteNoteNote A group may be non-commutative but one of its subgroup

may be commutative: H=de / 3 3 /h //g ( 0i is acommutative

subgroup of the non-commutative group GL(2,R).

Theorem 1.21 Theorem 1.21 Theorem 1.21 Theorem 1.21 Let G be a group and (H�G. Then H is a subgroup of

G iff a.b-1 �H, ∀a,b �H.

Proof Let (H�G and let a.b-1 �H, ∀a,b �H. Since H( , let a�H. By

assumption, e=a.a-1�H. let b�H. Then b-1= e.b-1�H. also, for a,b �H,

a,b-1�H and hence a.b=a.(b-1)-1�H. Associativity, being a

hereditary property, holds in (H,.) since it holds in (G,.). Thus H is

a subgroup of G. Converse part is trivial.

Theorem 1.22 Theorem 1.22 Theorem 1.22 Theorem 1.22 Let G be a group and (H�G, H finite. Then H is a

subgroup of G iff a.b �H, ∀a,b �H.

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Proof (sufficiency)Let h�H. Then A={h,h2,h3,…} �H. Since H is finite,

there exist integers r and s, 0≤r<s and hr=hs. Hence e=hs-r �H. e=

hs-r=h.hs-r-1 implies h-1=hs-r-1�H. Let a,b�H. Then a,b-1�H. Hence by

hypothesis, a.b-1�H. thus by previous theorem, H is a subgroup of

G.

Theorem 1.23Theorem 1.23Theorem 1.23Theorem 1.23 The intersection of all subgroups of a group G is a

subgroup of G. For two subgroups H and K of a group G, H�K is a

subgroup of G iff H�K or K�H.

Proof Let {Hx}x be the collection of all subgroups of a group G. Since

e���D , ��D ( . Let a,b���D . Then a,b� Hx and since Hx is a

subgroup of G, a.b-1� Hx. Thus a.b-1���D . Hence ��D is a

subgroup of G.

Let H ,K and H�K be subgroups of G. To prove: H�K or K�H. If

possible, let H �K and K�H. Let a� H-K, b�K-H. Since H�K is a

subgroup, a.b-1� H�K. If a.b-1�K, then a=(ab-1)b�K, contradiction.

Similarly, if a.b-1�H, then b=(a.b-1)-1.a-1�H,contradiction. Hence

H�K either H�K or K�H.

DefinitionDefinitionDefinitionDefinition 1.201.201.201.20 Let G be a group. Z(G)={x�G/x.g=g.x, ∀g�G} is

called the centre of G. If G is commutative, then G=Z(G).

Example1.71 Z(G) is a subgroup of G.

» Since e� Z(G), Z(G)( . Let a,b� Z(G). Then for g�G,

(a.b).g=a.(b.g)=a.(g.b)=(a.g).b=(g.a).b=g.(a.b) so that a.b� G. Let a�

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Z(G). then a.g=g.a, 'g�G. Then a-1.(a.g)a-1=a-1.(g.a).a-1 implying g.a-

1=a-1.g; hence a-1� Z(G).

Example1.72 Prove that every subgroup of (Z,+) is of the form

Nz={na:a�Z}, where n is a nonnegative integer.

» Let H be a subgroup of Z. If H={0}, then H=0Z.

Let {0}⊊H. Let 0(a�H. Since H is a subgroup, -a�H. Thus H

contains positive integers. By well-oredring principle of N, the set

A={a�H:a�N}⊆N has a smallest element, say, n. We shall prove

H={nr:r�Z}. Since n�H and H is a subgroup, {nr:r�Z}⊆H.

Conversely, let b�H. By division algorithm for integers, there exist

p,r�Z such that b=pn+r,0≤r<n. r=b-pn�H(since H is a subgroup).

Since r<n and n is the smallest positive integer such that n�H, r=0.

Thus b=pn�nZ. Thus H⊆nZ. Combining, H=Nz.

Example1.73 Let G be a group and a�G. Let C(a)={x�G:a.x=x.a}.Show

that C(a) is a subgroup of G and Z(G) is contained in C(a).

Example1.74 Let G be a commutative group . Prove that the set H of all

elements of G of finite order is a subgroup of G.

Example1.75 In the group S3, show that the subset H={� �S3:0(�)

divides 2} is not a subgroup of S3.

Example1.76 In the symmetric group S3, show that H={I,(2 3)} and

K={I,(1 2)} are subgroups but H∪K is not a subgroup of S3.

Example1.77 If a group has finitely many subgroups, then G is finite.

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Example1.78 True or false: the multiplicative group of nonero real

numbers has no finite subgroups other than {1}.

Example1.79 True or false:Let G be a group and H be a nonempty

subset of G such that a-1�H for all a�H. Then H is a subgroup.

Example1.80 True or false:There does not exist a proper subgroup H

of (Z,+) such that H contains both 5Z and 7Z.

CYCLIC GROUPS

A group G is a cyclic group iff there exists a�G such that

G=(a)={an:n�Z}. Such an element a is called a generator of G.

Example1.81 (Z,+) is a cyclic group since Z=(1). (Zn,+n) is cyclic with

[1] as one of its generators. (R,+) is not cyclic: a rational can not

generate irrational and vice versa. The multiplicative group of all

four fourth roots of unity is cyclic with i as one of its generators.

Theorem 1.24 Theorem 1.24 Theorem 1.24 Theorem 1.24 Every cyclic group is commutative ; converse may not

hold.

Proof Let G=(a) and let b,c�G. Then b=an,c=am, there exist integers

m,n. Thus bc=an.am=an+m=am+n=am.an=cb, proving that G is abelian.

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Consider the Klein’s 4-group: G={e,a,b,c} with the binary

operation 0 defined by the property a2=b2=c2=e2=e. Then G is

abelian but not cyclic.

Theorem 1.25 Theorem 1.25 Theorem 1.25 Theorem 1.25 A finite group G is cyclic iff there exists a�G such that

0(a)=0(G) (0(G), order of G, stands for the number of elements of

G).

Proof Let G be a finite cyclic group of ordern. Hence there exists

element a�G such that G=(a)={ai:i�Z}. Since G is finite, there exist

i,j �Z with i<j such that ai=aj. Thus aj-i=e, j-i�N. Let m be the

smallest positive integer such that am=e. Then 0(a)=m and for all

integers i,jsuch that 0≤i<j<m, ai(aj or else aj-i=e, which contradicts

the minimality of m. Hence the elements of the set S={e,a,a2,…,am-

1} are distinct. Clearly, S⊆(a). Conversely, let ak�(a). Then there

exist q,r�Z such that k=qm+r,0≤r<m. Thus

ak=aqm+r=(/�)� .ar=ar�S. Hence S=(a). Since elements of S are

distinct and 0((a))=n, m=n. Hence 0(a)=n.

Conversely assume G is a finite group of order n and G has an

element a such that 0(a)=n. since o(a)=n, all the elements of

A={e,a,a2,…,an-1} are distinct ,A⊆G and 0(A)=0(G). Hence G=A=(a).

C o r o l l a r yC o r o l l a r yC o r o l l a r yC o r o l l a r y Let (a) be finite cyclic group. Then 0(a)=0((a)).

Example1.82 Klein’s 4-group is not cyclic since it has no element of

order 4.

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Theorem 1.25 Theorem 1.25 Theorem 1.25 Theorem 1.25 Let G=(a) , 0(G)=n. then for any integer k, 1≤k<n, ak is a

generator of G iff gcd(n,k)=1.

Proof If G=(ak), then 0(ak)=0(G)=n. since G=(a), we have 0(a)=n.

Thus n=0(ak)=�(0)

xyz (w,�) = wxyz (w,�). So gcd(n,k)=1.

Conversely, let gcd(n,k)=1. Then 0(ak)=�(0)

xyz (w,�)=0(a)=n=0(G).

Hence G=(ak).

Example1.82 For the group G={1,-1.i,-i}, 0(G)=4 and 1 and 3 are the

only positive integers less than 4 and relatively prime to 4.Hence i

and i3=-i are the only generators of G.

Theorem 1.26Theorem 1.26Theorem 1.26Theorem 1.26 Every subgroup of a cyclic group is cyclic.

Proof Let H be a subgroup of a cyclic group G=(a). If H={e}, then

H=(e). Suppose H({e}. Then there exists b�H , b(e. Thus there

exist nonzero integer m such that b=am. Since H is a subgroup, a-

m=b-1�H. either m or –m is a positive integer. Hence there exists

positive integer I such that ai�H. Thus A={n�N:an�H} ( . By

well-ordering principle of N, A has a least element n. we prove

that H=(an). since an�H and H is a subgroup, (an) ⊆H. Let

h�H⊆G=(a). thusthere exist integer k such that h=ak. by division

algorithm for integers, there exist integers q,r such that

k=nq+r,0≤r<n. thus ar=ak.(/w)@��H (sine H is a subgroup). Since n

is the smallest positive integer such that an�H and 0≤r<n, r=0.

Thus k=nq and so h=ak=anq=(/w)��(an). Hence H=(an).

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Example1.83 Find all the generators of (Z10,+10).

Z10=([m])=(m[1]) iff gcd(m,10)=1. Thus the generators of Z10 are

[1],[3],[7] and [9].

Example1.84 The group (Q,+) is not cyclic. Hence (R,+) is not cyclic.

If possible, let Q=(x). Clearly, x(0. Hence x=��, where p,q are prime

to each other and q(0. But �

I��Q=(��), contradiction.

COSETS AND LAGRANGE ’ S THEOREM

Let G be a group and H be a subgroup of G. Define a relation R on

G by: a,b�G, aRb iff a.b-1�H. R is an equivalence relation on G. For

a�G, [a], the equivalence class containing a, is given by:

[a]={b�G:bRa}={b�G: ba-1�H}=Ha={ha:h�H} (Ha is called right

coset of H in G generated by a). we know that equivalence classes

are nonempty, any two distinct equivalence classes are disjoint

and union of all the equivalence classes equals G.

L e m m aL e m m aL e m m aL e m m a Let H be a subgroup of a group G. If a�G, then

0(aH)=0(H).

Proof f:H∩aH, f(h)=a.h, is a bijection. Hence the result.

Theorem 1.27Theorem 1.27Theorem 1.27Theorem 1.27 (Lagrange) (Lagrange) (Lagrange) (Lagrange) Let H be a subgroup of a finite group G.

Then0(G)=[G:H]0(H), where [G:H], the index of H in G, is the

number of distinct right cosets of H in G.

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43

Proof Since G is finite, [G:H] is finite. Let [G:H]=r. Let

A={Ha1,Ha2,…,Har} be the collection of right cosets of H in G. Since

A is a partition of G, 0(G)= 0(Ha1)+…+0(Har)=r0(H)( by lemma

above). Hence the result.

Converse of Lagrange’s Theorem may not hold. It can be proved

that 6 divides 0(A4) but A4 has no subgroup of order 6. But the

converse holds for a cyclic group.

Theorem 1.28 Theorem 1.28 Theorem 1.28 Theorem 1.28 Let G =(a), 0(G)=n. if m is a positive integral factor of n,

then there exists a unique subgroup of G of order m.

Proof There exists positive integer k such that n=mk. Now

0(ak)=�(0)

xyz (�,�(0))=w�=m; hence 0((ak))=0(ak)=m.

Now suppose K is a subgroup of G of order m. K=(at), for some

t�Z. 0(at)=0((at))=m. thus amt=e. Since 0(a)=n, n divides mt, that

is, mt=nr for some integer r. By Lagrange’s Theorem, n=km, for

some natural k. Hence mt=kmr, so that t=kr. So at=(ak)r�(ak)=H.

Hence K=(at) ⊆(ak)=H. Both these subgroups have order m. hence

H=K.

Example1.85 Let G be a group of order 28. Show that G has a

nontrivial subgroup.

»If G is cyclic, G has a subgroup corresponding to every positive

integral factor of 28, say, corresponding to 4.

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If G is not cyclic, consider the cyclic subgroup (a) generated by

e(a�G. Clearly {e}⊊(a)⊊G (G is not cyclic).

Example1.86 Let G=(a) be an infinite cyclic group. Prove that (1) ar=at

only if r=t, r,t�Z , (2) G has only two generators.

» Let ar=at, r>t. then there exist natural number m such that

at+m=at, so that, am=e. Thus 0(a) is finite and 0((a)) is infinite,

contradiction.

Let (a)=G=(b), for some b�G. Since a�(b) and b�(a), there exist

integers r and t such that a=br, b=at. thus a=art, which by part (1)

implies rt=1.Thus r=1 or -1. Hence b=a or b=a -1. Thus G has

exactly two generators.

Example1.87 If a group G has only two subgroups, then G is a cyclic

group.

»G({e} since G has two subgroups. Let H=(a), e(a�G. Since H({e}

(a�H), G=H=(a).

Example1.88 Let G be a cyclic group of order 42. Find the number of

elements of order 6 and of order 7.

» Let G=(a), 0(a)=42. Let b=ar�G, o(b)=6. Then a6r=e. Thus 42

must divide 6r, that is, 7 must divide r and r<42. Thus

r=7,14,21,28,35. Thus there are five elements of G of order 6.

Example1.89 Let G be a cyclic group such that G has exactly three

subgroups: G,{e} and a subgroup of order 5. Find the order of G.

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Example1.90 Let G be a cyclic group of infinite order. Find the number

of elements of finite order.

Theorem 1.29 Theorem 1.29 Theorem 1.29 Theorem 1.29 Every group of prime order is cyclic.

Proof Let 0(G)=p, p prime. Thus p≥2. Let e(a�G. Then 0((a))

divides 0(G)=p ,0((a))=0(a)>1. Thus 0((a))=p=0(G), (a) ⊆G. Hence

G=(a).

Theorem 1.30Theorem 1.30Theorem 1.30Theorem 1.30 Let H and K be finite subgroups of a group G. Then

0(HK)=�(�)�(�)�(���) .

Example1.91 Let G be a group of order pn, p prime. Prove that G

contains an element of order p.

» Let e(a�G. Let H=(a). Then 0(a) divides 0(G)=pn. Thus

0(H)=0(a)=pm, 1≤m≤n. Now in the cyclic group (a) of order pm,

there exists a cyclic subgroup C=(c) of order p. Thus c�G and

0(c)=p.

Example1.92 Let G be a group of order pq where p and q are prime.

show that every subgroup H(( s)is cyclic.

» 0(H) divides pq=0(G). Since p,q are primes, 0(H)=1,p,q (note:

H( s). Sine every group of prime order is cyclic, result follows.

Example1.93 Find all subgroups of Klein’s 4-group.

Example1.94 Prove that every proper subgroup of S3 is cyclic though

S3 is not cyclic.

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Example1.95 Prove that every group of order 49 contains a subgroup

of order 7.

Example1.96 Let G be a group such that 0(G)<320. Suppose G has

subgroups of order 35 and 45. Find the order of G.

Example1.97 Let A and B are two subgroups of a group G. If 0(A)=p (p

prime), then show that A∩B={e} or A⊆B.

SECT ION I I : C LAS S I CAL ALGEBRA

Text (1) Classical Algebra—S.K.Mapa

CHAPTER I

COMPLEX NUMBERS: DE’ MIVRE’S THEOREM

A complex number z is an ordered pair of real numbers (a,b): a is called

Real part of z, denoted by Re z and b is called imaginary part of z, denoted

by Im z. If Re z=0, then z is called purely imaginary ; if Im z =0, then z is

called real. On the set C of all complex numbers, the relation of equality and

the operations of addition and multiplication are defined as follows:

(a,b)=(c,d) iff a=b and c=d, (a,b)+(c,d)=(a+c,b+d), (a,b).(c,d)= (ac-bd,ad+bc)

The set C of all complex numbers under the operations of addition and

multiplication as defined above satisfies following properties:

� For z1,z2,z3�C, (1) (z1+z2)+z3=z1+(z2+z3)(associativity),

(2)z1+(0,0)=z1, (3) for z=(a,b)�C, there exists –z=(-a,-b)�C such that

(-z)+z=z+(-z)=(0,0), (4)z1+z2=z2+z1.

� For z1,z2,z3�C, (1) (z1. z2).z3=z1.(z2.z3)(associativity), (2)z1.(1,0)=z1,

(3) for z=(a,b)�C,z((0,0), there exists 7� �C such that z. 7�=7�.z=1,

(4)z1.z2=z2.z1.

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47

� For z1,z2,z3�C, z1.(z2+z3)=(z1.z2)+(z1.z3).

Few ObservationsFew ObservationsFew ObservationsFew Observations

(1) Denoting the complex number (0,1) by i and identifying a real

complex number (a,0) with the real number a, we see

z=(a,b)=(a,0)+(0,b)=(a,0)+(0,1)(b,0) can be written as z=a+ib.

(2) For two real numbers a,b , a2+b2=0 implies a=0=b; same

conclusion need not follow for two complex numbers, for example,

12+i2=0 but 1∩(1,0)((0,0) ∩0 and i=(0,1) ((0,0) (∩ denotes

identification of a real complex number with the corresponding real

number).

(3) For two complex numbers z1,z2, z1z2=0 implies z1=0 or z2=0.

(4) i2=(0,1)(0,1)=(-1,0) ∩-1.

(5) Just as real numbers are represented as points on a line,

complex numbers can be represented as points on a plane:

z=(a,b)�P: (a,b). The line containing points representing the real

complex numbers (a,0), a real, is called the real axis and the line

containing points representing purely imaginary complex numbers

(0,b) ∩ib is called the imaginary axis. The plane on which the

representation is made is called Gaussian Plane or Argand Plane.

DefinitionDefinitionDefinitionDefinition 1.11.11.11.1 Let z=(a,b) ∩a+ib. The conjugate of z, denoted by �� , is (a,-b)

∩a-ib.

Geometrically, the point (representing) �� is the reflection of the point

(representing) z in the real axis. The conjugate operation satisfies the

following properties:

(1) ���=z , (2) �7 * �I��������� = �7� * �I� , (3) �7�I������ = �7� * �I� , (4)e���E

h����� = ������E��� , (4)

z+��=2 Rez, z-��=2i Im(z)

DefinitionDefinitionDefinitionDefinition 1.21.21.21.2 Let z=(a,b) ∩a+ib. The modulus of z, written as |�|, is defined

as √/I * 3I.

Geometrically, |�| represents the distance of the point representing z from

the origin (representing complex number (0,0) ∩0+i0). More generally,

|�7 �I| represents the distance between the points z1 and z2. The

modulus operation satisfies the following properties:

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(1) |�7 * �I| = |�7| * |�I|, (2) |�7. �I| � |�7|. |�I| (3)����E� � |��||�E| (4)

�|�7| |�I|� = |�7 �I| GEOMETRICAL REPRESENGEOMETRICAL REPRESENGEOMETRICAL REPRESENGEOMETRICAL REPRESENTATION OF COMPLEX NUTATION OF COMPLEX NUTATION OF COMPLEX NUTATION OF COMPLEX NUMBERS: THE ARGAND MBERS: THE ARGAND MBERS: THE ARGAND MBERS: THE ARGAND

PLANEPLANEPLANEPLANE

Let z=a+ib be a complex number. In the Argand plane, z is represented by

the point whose Cartesian co-ordinates is (a,b) referred to two

perpendicular lines as axes, the first co-ordinate axis is called the real axis

and the second the imaginary axis. Taking the origin as the pole and the

real axis as the initial line, let (r,�) be the polar co-ordinates of the point

(a,b). Then a=r cos �, b=r sin �. Also r=√/I * 3I=|�|. Thus

z=a+ib=|�|(cos �+isin �): this is called modulus-amplitude form of z. For a

given z(0, there exist infinitely many values of � differing from one

another by an integral multiple of 2 : the collection of all such values of �

for a given z(0 is denoted by Arg z or Amp z. The principal value of Arg z ,

denoted by arg z or amp z, is defined to be the angle � from the collection

Arg z that satisfies the condition -  c � =  . Thus Arg z={arg z+2n : n an

integer}. arg z satisfies following properties: (1)

arg(z1z2)=argz1+argz2+2k , where k is a suitable integer from the set{-

1,0,1] such that -  c argz1+argz2+2k  =  , (2) arge���Eh � argz1-argz2+2k ,

where k is a suitable integer from the set{-1,0,1] such that -  c argz1-

argz2+2k  =  .

Note Note Note Note An argument of a complex number z=a+ib is to be determined from

the relations cos �=a/|�|, sin �= b/|�| simultaneously and not from the

single relation tan �=b/a.

ExampleExampleExampleExample 1.11.11.11.1 Find arg z where z=1+i tanH¡� .

» Let 1+itanH¡� =r(cos �+i sin �). Then r2= sec2

H¡� . Thus r=- sec

H¡� >0.Thus cos

�=- cosH¡� , sin �=-sin

H¡� . Hence �=  +

H¡� . Since �>  ,arg z= �-2  =-

I¡� .

Geometrical representation of operations on complex numbers:

Addition Let P and Q represent the complex numbers z1=x1+iy1 and

z2=x2+iy2 on the Argand Plane respectively. It can be shown that the fourth

vertex R of the parallelogram OPRQ represents the sum z1+z2 ofz1 and z2.

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Product Let z1=|�7|"cos �7 * ¢ £¢� �7) and z2=|�I|"cos�I * ¢ £¢� �I) where – <�7, �I =  . Thus z1z2=|�7||�I|{cos(�7+�I)+isin(�7+�I)}. Hence the point

representing z1z2 is obtained by rotating line segment OP{ where P

represents z1} through arg z2 and then dilating the resulting line segment

by a factor of |�I|. In particular , multiplying a complex number by

i=cose¡Ih * ¢£¢� e¡Ih geometrically means rotating the line segment by ¡I.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 11 . 11 . 11 . 1 (De Moivre’s Theorem) If n is an integer and � is any real

number, then (cos �+i sin �)n= cos n �+i sin n �. If n=��, q natural, p integer,

|;|and q are realtively prime, � is any real number, then (cos �+i sin �)n has

q number of values, one of which is cos n �+i sin n �.

Proof Case 1 Let n be a positive integer.

Result holds for n=1: (cos �+i sin �)1= cos 1 �+i sin 1 �. Assume result

holds for some positive integer k: (cos �+i sin �)k= cos k �+i sin k �.Then

(cos �+i sin �)k+1=(cos �+i sin �)k(cos �+i sin �)=( cos k �+i sin k �)(

cos �+i sin �)= cos(k+1) �+isin(k+1) �. Hence result holds by mathematical

induction.

Case 2 Let n be a negative integer, say, n=-m, m natural.

(cos �+i sin �)n=(cos �+i sin �)-m=7

"b¤¥¦B�¥�w¦�§ � 7y¨©�¦B� ¥�w �¦ (by case 1) =

cos m�-isin m�=cos(-m) �+isin(-m) �= cos n �+i sin n �.

Case3 n=0: proof obvious.

Case 4 Let n=��, q natural, p integer, |;|and q are realtively prime.

Let "cos � * ¢ £¢���ª«= cos ¬+i sin ¬. Then "cos � * ¢ £¢����= (cos ¬+i

sin ¬)q. Thus cos p �+isin p �= cos q ¬+i sin q ¬. Thus q ¬=2k +p �, that is,

¬=I­¡B® ¦

� . Hence "cos � * ¢ £¢���ª«= cos(I­¡B® ¦

� )+isin(I­¡B® ¦

� �, where

k=0,1,…,q-1 are the distinct q values.

S om e A p p l i c a t i o n s o f S om e A p p l i c a t i o n s o f S om e A p p l i c a t i o n s o f S om e A p p l i c a t i o n s o f D e ’ M o i v r e ’ s T h e o r emD e ’ M o i v r e ’ s T h e o r emD e ’ M o i v r e ’ s T h e o r emD e ’ M o i v r e ’ s T h e o r em

(1) Expansion of cos n �, sin n � and tan n � where n is natural and � is real.

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50

Cos n �+i sin n �=(cos �+isin �)n=cosn �+ �7w cosn-1 �isin �+ �Iw cosn-

2 � ¢I £¢�I�+…+insinn �= (cosn �- �Iw cosn-2 � £¢�I�+…)+i( �7w cosn-

1 �sin �- �Hw cosn-3 �£¢�H�+…). Equating real and imaginary parts, cos

n �= cosn �- �Iw cosn-2 � £¢�I�+… and sin n �= �7w cosn-1 �sin �- �Hw cosn-

3 �£¢�H�+…

(2) Expansion of cosn � and sinn � in a series of multiples of �

where n is natural and � is real. Let x = cos � * ¢£¢��. Then xn=cos n �+isin n �, x-n= cos n �-isin n �.

Thus (2 cos �)n=(x+7D�n

=(xn+7D|� * �7w (xn-2+

7D|¯E)+…=2 cos n �+ �7w (2 cos(n-2) �)+…

Similarly, expansion of sinn � in terms of multiple angle can be

derived.

(3) Finding n th roots of unity

To find z satisfying zn=1=cos(2k )+isin(2k �, where k is an integer.

Thus z=[ cos(2k )+isin(2k �]1/n=cos(I�¡w � * ¢£¢� eI�¡w h, k=0,1,…,n-1;

replacing k by any integer gives rise to a complex number in the set

A={ cos(I�¡w � * ¢£¢� eI�¡w h/ k=0,1,…,n-1}. Thus A is the set of all nth

roots of unity.

ExampleExampleExampleExample 1.21.21.21.2 Solve x6+x5+x4+x3+x2+x+1=0

»We have the identity x6+x5+x4+x3+x2+x+1=D°@7D@7 . Roots of x7-1=0 are

cosI�¡± * ¢£¢� I�¡± , k=0,1,…,6. Putting k=0, we obtain root of x-1=0. Thus the

roots of given equation are cosI�¡± * ¢£¢� I�¡± , k=1,…,6.

ExampleExampleExampleExample 1.31.31.31.3 Prove that the sum of 99 th powers of all the roots of x7-

1=0 is zero.

»The roots of x7-1=0 are {1,�, �2,…, �6}, where �=cosI¡± +isin

I¡± . Thus sum of

99th powers of the roots is 1+ �99+(�I�88 *�* "�²�88 �1+ �99+(�88�I *�* "�88�²=

7@�³³.°7@�³³ =0, since �88.±=1 and �88 (1.

ExampleExampleExampleExample 1.41.41.41.4 If |�7| � |�I| and arg z1+argz2=0, then show that z1=�I� .

»Let |�7| � |�I|=r, arg z1=�, then argz2=- �. Thus z1=r(cos �+isin �)=r(cos �-isin �)= �I� .

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ExampleExampleExampleExample 1.51.51.51.5 For any complex number z, show that |�| F |´µ �|B|¶� �|√I .

»Let z=x+iy.2(x2+y2)-(x+y)2=(x-y)2F0. Thus x2+y2F "DBJ�EI and so |�|=

l)I * uI F |´µ �|B|¶� �|√I .

ExampleExampleExampleExample 1.61.61.61.6 Prove that if the ratio �@��@7 s purely imaginary, then the

point (representing )z lies on the circle whose centre is at the point 7I "1 * ¢� and radius is

7√I.

»Let z=x+iy. Then �@��@7 � DE@DBJE@J

"D@7�EBJE * ¢ 7@D@J"D@7�EBJE. By given condition ,

)I ) * uI u=0, that is, e) 7IhI * eu 7

IhI � e 7√IhI.Thus z lies on the

circle whose centre is at the point 7I "1 * ¢� and radius is

7√I.

ExampleExampleExampleExample 1.71.71.71.7 If the amplitude of the complex number �@��B7 is

¡G, show

that z lies on a circle in the Argand plane.

»Let z=x+iy. Then �@��B7=

DEBDBJE@J"DB7�EBJE * ¢ J@D@7

"DB7�EBJE. By given condition,

J@D@7DEBDBJE@J=1. On simplification, (x+1)2+(y-1)2=1. Hence z lies on the circle

centred at (-1,1) and radius 1.

ExampleExampleExampleExample 1.81.81.81.8 If z and z1 are two complex numbers such that z+z1 and

zz1 are both real, show that either z and z1 are both real or z1=��. ExampleExampleExampleExample 1.91.91.91.9 If |�7 * �I| � |�7 �I|, prove that arg z1 and argz2 differ

by ¡I or

H¡I .

»|�7 * �I|I � |�7 �I|I Thus (z1+z2)(�7� * �I� �= (z1-z2)(�7� �I� � Or, �7�I� * �7� �I � 0 (1). Let z1=r1(cos�7 * ¢£¢��7), z2=r2(cos�I * ¢£¢��I). From

(1), cos(�7-�I)=0 proving the result.

ExampleExampleExampleExample 1.11.11.11.10000 If A,B,C represent complex numbers z1,z2,z3 in the

Argand plane and z1+z2+z3=0 and |�7|=|�I| � |�H|, prove that ABC is an

equilateral triangle.

»z1+z2=-z3. Hence |�7 * �I|I � |�H|I, that is, |�7|I+|�I|I+2z1.z2=|�H|I. By

given condition, |�7||�I|cos �=|�7|I, where � is the angle between z1 and

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52

z2.Thus cos �=-7I, that is, �=1200. Hence the corresponding angle of the

triangle ABC is 600. Similarly other angles are 600.

ExampleExampleExampleExample 1.11.11.11.11111 If (x,y) represents a point lying on the line 3x+4y+5=0,

find the minimum value of |) * ¢u|. ExampleExampleExampleExample 1.11.11.11.12222 Let z and z1 be two complex numbers satisfying z=

7B��7@�� and |�7|=1. Prove that z lies on the imaginary axis.

» z1=�@7�B7. By given condition, 1 � ��@7�B7� � |�@7|

|�B7|. If z=x+iy, x=0. Hence.

ExampleExampleExampleExample 1.11.11.11.13 3 3 3 If z1, z2 are conjugates and z3,z4 are conjugates, prove

that arg���¸ � arg

�¹�E. » Since z1, z2 are conjugates, arg z1+arg z2=0. Since z3,z4 are conjugates, arg

z3+arg z4=0. Thus arg z1+arg z2=0= arg z3+arg z4. Hence arg z1-arg z4=arg z3-

arg z2; thus result holds.

ExampleExampleExampleExample 1.11.11.11.14444 complex numbers z1,z2,z3 satisfy the relation

z12+z2

2+z32-z1z2-z2z3-z3z1=0 iff |�7 �I| � |�I �H| � |�H �7|.

» 0=z12+z2

2+z32-z1z2-z2z3-z3z1=(z1+wz2+w2z3)(z1+w2z2+wz3), where w

stands for an imaginary cube roots of unity. If z1+wz2+w2z3=0, then (z1-

z2)=-w2(z3-z2); hence |�7 �I|=|ºI||�I �H|=|�I �H|.similarly other part.

Conversely, if |�7 �I| � |�I �H| � |�H �7|, then z1,z2,z3 represent

vertices of an equilateral triangle. Then z2-z1=(z3-z1)(cos 600+isin600), z1-

z2=(z3-z2)( cos 600+isin600 ); by dividing respective sides, we get the result.

ExampleExampleExampleExample 1.11.11.11.15555 Prove that |�7 * �I|I * |�7 �I|I � 2"|�7|I * |�I|I),

for two complex numbers z1,z2.

»|�7 * �I|I � "�7 * �I�"�7� * �I� �= |�7|I * |�I|I+2 z1z2; similarly |�7 �I|I=|�7|I * |�I|I-2 z1z2; Adding we get the result.

ExampleExampleExampleExample 1.11.11.11.16666 If cos�+cos �+cos »=sin�+sin �+sin »=0, then prove

that (1) cos 3�+cos3 �+cos 3 »=3cos(� * �+ »), (2) ∑f�£I�=∑£¢�I�=3/2.

»Let x=cos �+i sin�, y= cos �+i sin�, z= cos »+i sin». Then x+y+z=0. Thus

x3+y3+z3=3xyz. By De’ Moivre’s Theorem, (cos 3 �+ cos 3 �+cos 3 »)+i(sin

3 �+ sin 3 �+sin 3 »)=3[ cos(� * � * »)+isin(� * � * »�%. Equating, we get

result.

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53

Let x=cos �+i sin�, y= cos �+i sin�, z= cos »+i sin». Then x+y+z=0. Also 7D * 7

J * 7� � 0; hence xy+yz+zx=0. Thus x2+y2+z2=0. By De’ Moivre’s

Theorem, cos 2 �+cos2 �+cos2 »=0. Hence∑f�£I�=3/2. Using sin2 �=1-

cos2 �, we get other part.

ExampleExampleExampleExample 1.11.11.11.17777 Find the roots of zn=(z+1)n, where n is a positive

integer, and show that the points which represent them in the Argand

plane are collinear.

Let w=�B7� . ¾¿�� � � 7

À@7.Now zn=(z+1)n implies wn=1.Thus, w=cosI�¡w *

¢£¢� I�¡w ,k=0,…,n-1.

So z=7

y¨©E�Á| B�¥�wE�Á| , k=1,…,n-1

= 7I �

I f� �¡w . Thus all points z satisfying zn=(z+1)n lie on the line x=-7I.

Functions of a complex variableFunctions of a complex variableFunctions of a complex variableFunctions of a complex variable

Exponential function

For a complex number z=x+iy, exp(z)=ex(cos y +i sin y).

NoteNoteNoteNote

(1) If z=x+i0 is purely real, exp(z)=ex.

(2) If z = 0+iy is purely imaginary, exp(z)=exp(iy)=cos y+i sin y.

(3) Since 0 (ex= |exp �|, thus exp(z) is a non-zero complex

number for every complex number z.

(4) For every non-zero complex number u+iv=r(cos�+isin�),

exp(ln r+i�)=eln r(cos�+isin�)=u+iv. Thus exp:CCC-{0+i0}.

(5) exp(z1).exp(z2)=exp(z1+z2)

» Let z1=x1+iy1,z2=x2+iy2. exp(z1)exp(z2)

= �D�BDE(cosy1+isiny1)(cosy2+isiny2)

= �D�BDE[cos(y1+y2)+isin(y1+y2)]= exp(z1+z2).

(6) exp(z1-z2)=ÃÄ® "���ÃÄ® "�E� ; in particular exp(-z)=

7ÃÄ® "��.

(7) If n be an integer, (exp z)n=exp(nz) (follows from property(5)

and (6))

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54

(8) If n be an integer, exp(z+2n i)=exp(z).(follows from (5), since

exp(2n i)=1)

ExampleExampleExampleExample 1.11.11.11.18 8 8 8 Find all complex number z such that exp(z)=-1

» Let z=x+iy. Then ex(cos y+isiny)=-1. Thus excos y=-1, exsin y=0. Squaring,

adding gives e2x=1 whence ex=1(since ex>0); thus x=0. cosy=-1 and siny=0

gives y=(2n+1) . Thus z=(2n+1)  i, n integer.

Logarithmic function

Let z be a non-zero complex number. Then there exists a complex number

w such that exp(w)=z; further if exp(w)=z, then exp(w+2n  ¢)=z. For a

given non-zero complex number z, we define Log z={w�C/ exp(z)=w} and

call the set Logarithm of z.

Let z=r(cos �+isin �),-  < � =  . Let w=u+iv be a logarithm of z(that is, a

member of Log z). Then exp(u+iv)= r(cos �+isin �). Thus eucos v=r cos �,

eusin v=r sin �. Hence eu=r (by squaring and adding last two relations), cos

v=cos �, sinv=sin �. So u=ln r,v= �+2n , n integer. Thus Logz=ln|�|+i(arg

z+2n ).

NoteNoteNoteNote

Putting n=0 in the expression for Log z, we obtain principal value of Log z,

denoted by log z. Thus log z= ln|�|+iarg z.

(1) exp(Log z)=z for z(0; one of the values of Log(expz) is z, the

other values are z+2n i,n integer.

(2) For two distinct non-zero complex numbers z1,z2,

Log(z1z2)=Log(z1)+Log(z2); log(z1z2)(log(z1)+log(z2): take

z1=i,z2=-1.

» Let z1=r1(cos�7+isin�7), z2= r2(cos�I+isin�I). Then Log

z1=lnr1+i(�7+2n ), Log z2=lnr2+i(�I+2m ),Log(z1z2)=

z1=ln(r1r2)+i(�7+�I *2k ) where n, m, k are integers.

Log(z1)+Log(z2)=ln(r1r2)+i(�7+�I+2q �, where q=m+n. Since p,q are

arbitrary integers, Log(z1z2)=Log(z1)+Log(z2) holds.

(3) For two distinct non-zero complex numbers z1,z2,

Loge���Eh �Log(z1)-Log(z2). loge���Eh (log(z1)-log(z2): take z1=-1,z2=-i.

» Proof similar.

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55

(4) If z(0 and m be a positive integer, Log zm(mLogz: take

z=i,m=2.

ExampleExampleExampleExample 1.11.11.11.19 9 9 9 Verify Log(-i)1/2=1/2 Log(-i)

» -i=cose¡Ih+isine¡Ih. 7ILog(-i)= 7I[(2n  ¡

I�¢]=( n  ¡G)I, where n is an

integer.

Two values of (-i)1/2 are f�£ e ¡Gh+isine ¡

Gh and f�£ eH¡G h+isineH¡G h. Now

Log[f�£ e ¡Gh+isine ¡

Gh]=(2m -¡G)i and Log[f�£ eH¡G h+isineH¡G h]= (2p  *

H¡G )i=[(2p+1)   ¡

G]I, where m,p are integers.Thus the values of Log(-i)1/2

can be expressed as (n  - ¡G)I, where n is an integer. Hence the result.

ExampleExampleExampleExample 1.20 1.20 1.20 1.20 Prove that sin[ilog0@�Å0B�Å%= I0Å

0EBÅE. »Let a+ib=r(cos �+isin �),-  c � =  . Then a=r cos �, b=r sin �. Thus

z=cos(-2 �)+isin(-2 �). Thus arg z=-2 �+2k  , where k is an integer such

that -  <-2 �+2k   =  . So log z=(-2 �+2k  )i. hence sin(ilog z)=sin(2 �-

2k  )=sin2 �=I0Å0EBÅE.

Exponent function

If w be a nonzero complex number and z be any complex number, then

wz=exp(z Log w) . Since Log is many-valued, exponent is many-valued;

principal value of wz is defined as exp(z log w).

Note Note Note Note If z1,z2,w are complex numbers, w(0, then º��B�E ( º��º�E but

p.v. º��B�E=(p.v. º��)(p.v. º�E).

ExampleExampleExampleExample 1.21 1.21 1.21 1.21 Find values of ii.

» ii=exp[iLogi]=exp[i(2n  * ¡I�i]=�@"GwB7�ÁE , n any integer.

ExampleExampleExampleExample 1.22 1.22 1.22 1.22 If a,b are the imaginary cube roots of unity, prove

that (1) p.v. of aa+p.v. of bb=�@¡ √HÆ , (2) p.v. of ab+p.v. of ba =�¡ √HÆ .

p.v. of aa=exp[a log a]=exp[a{ln1+iI¡H ]], if a=-

7I* ¢ √HI

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56

=exp[-¡√H ¢ ¡²]= �@¡ √HÆ

[cos¡² ¢£¢� ¡²]. Similarly, p.v. of bb=�@¡ √HÆ [cos

¡² *¢£¢� ¡²].Hence (1) is proved. Similarly (2) can be proved.

ExampleExampleExampleExample 1.23 1.23 1.23 1.23 If iz=i, show that z is real and the general values of

z are given by z=G�B7GwB7 , m,n are integers.

ExampleExampleExampleExample 1.24 1.24 1.24 1.24 Find all the values of √2√H and show that the values

lie on a circle in the Argand plane.

Trigonometric function

From definition of exponential function, for real y, exp(iy)=cos y+isin y,

exp(-iy)=cosy-isiny. Thus for real y, cos y=ÃÄ®"�J�BÃÄ® "@�J�

I , sin

y=ÃÄ®"�J�@ÃÄ® "@�J�

I� . Backedup by theseresults, for complex z, we define cos

z=ÃÄ®"���BÃÄ® "@���

I , sin z=ÃÄ®"���@ÃÄ® "@���

I� .

(1) cos2z+sin2z=1

(2) sin(z1+z2)=sinz1cosz2+cosz1sinz2, cos(z1+z2)=cosz1cosz2-

sinz1sinz2.

(3) sin(z+2  )=sin z, sin(z+  �=-sin z

(4) if x,y are real, sin(x+iy)=sin x cosh y+icos xsinh y,

cos(x+iy)=cosx cosh y-isin xsinh y, where cosh x=ÃÄ®"D�BÃÄ® "@D�

I and

sinh x=ÃÄ®"D�@ÃÄ® "@D�

I . Thus |sin ") * ¢u�|2=sin2x+sinh2y and

|cos ") * ¢u�|2=cos2x+sinh2y: since sinh y is an unbounded function,

sin z and cos z are unbounded in absolute value. But if x is real, sin x

and cos x are bounded functions.

(5) cos(iz)=cosh z, sin(iz)=isinh z, cosh(iz)=cos z, sinh(iz)=i sin z,

where cosh z=ÃÄ®"��BÃÄ® "@��

I and sinh z=ÃÄ®"��@ÃÄ® "@��

I .

ExampleExampleExampleExample 1.25 1.25 1.25 1.25 Find all values of z such that cos z=0.

»Let z=x+iy. cos z=0 implies cos xcosh y=0, sin xsinh y=0. Since cosh y(0,

cos x=0. Thus x=(2n+1)¡I, n integer. Thus sin x=sin[(2n+1)

¡I] (0. Hence sinh

y=0. Thus y=0. Hence z==(2n+1)¡I, n integer.

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57

ExampleExampleExampleExample 1.261.261.261.26 If x=log tan(¡G * ¦

I�, where � is real, prove that �=-

iLog tane¡G * ¢ DIh.

»Since � is real, x real. ex=tan(¡G * ¦

I�=7B{0wÇE7@{0wÇE. Therefore, Â/� ¦I=

µÈ@7µÈB7. Thus

{@{¯�{B{¯�=i

µÈE@µ¯ÈEµÈEBµ¯ÈE, where t=expe�¦I h. Hence

{E@7{EB7=

� ¥�wÉÈEb¤¥ÉÈE =¥�wÊÈEb¤¥ÊÈE=tan

�DI . Thus

t2=7B{0wÊÈE7@{0wÊÈE . So exp(i�)=tane¡G * �D

Ih. �=-iLog tane¡G * ¢ DIh.

Inverse Trigonometric functions

Let z be a given complex number and w be a complex number such that sin

w=z. Then cos w=Ë√1 �I. Thus exp(iw)=izË√1 �I or, w=-

iLog(izË√1 �I). Since Log is a multi-valued function, Sin-1z=w=-

iLog(izË√1 �I) is a multiple valued function of z. The principal value of

Sin-1z is obtained by choosing cos w=+√1 �I and by taking the principal

value of the logarithm and is denoted by sin-1z.

ExampleExampleExampleExample 1.271.271.271.27 Find Cos-1(2), cos-1(2).

»Let Cos-1(2)=z. Then cos z=2. sin z=Ë√3i. Thus exp(iz)=cos z+I sin

z=2=Ë√3. When exp(iz)=2+√3, z=-iLog(2+√3)=-i[log(2+√3�+2n i]=2n -i

log(2+√3). When exp(iz)=2-√3, z=-iLog(2-√3)=-i[log(2-√3�+2n i]=2n -i

log(2-√3) =2n +i log(2+√3�. Thus Cos-1(2)= 2n  Ëi log(2+√3�. Hence cos-

1(2)=ilog(2+√3).

ExampleExampleExampleExample 1.28 1.28 1.28 1.28 If tan-1(x+iy)=a+ib, where x,y,a,b are real and

(x,y)((0, Ë1), prove that (1) x2+y2+2xcot 2a=1, (2) x2+y2+1-2ycoth 2b=0.

»tan (a+ib)=x+iy; thus tan(a-ib)=x-iy. Thus tan 2a=tan[(a+ib)+(a-

ib)]=ID

7@DE@JE. Hence the first part. Again tan(2ib)=tan[(a+ib)-(a-

ib)]= I�J7BDEBJE. But tan(2ib)=

©ÌÍ "I�Å�y¨© "I�Å� � �¥�wÉ IÅ

y¨©Î IÅ=i tanh 2b. hence the second

part.

CHAPTER II

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THEORY OF EQUATIONS

An expression of the form a0xn+a1xn-1+…+an-1x+an, where a0,a1,…,an are real

or complex constants, n is a nonnegative integer and x is a variable (over

real or complex numbers) is a polynomial in x. If a0(0, the polynomial is of

degree n and a0xn is the leading term of the polynomial. A non-zero

constant a0 is a polynomial of degree 0 while a polynomial in which the

coefficients of each term is zero is said to be a zero polynomial and no

degree is assigned to a zero polynomial.

Equality two polynomials a0xn+a1xn-1+…+an-1x+an and b0xn+b1xn-1+…+bn-

1x+bn are equal iff a0=b0,a1=b1,…,an=bn.

Addition Let f(x)= a0xn+a1xn-1+…+an-1x+an, g(x)= b0xn+b1xn-1+…+bn-1x+bn.

the sum of the polynomials f(x) and g(x) is given by

f(x)+g(x)= a0xn+…+an-m-1xm+1+(an-m+b0)xm+…+(an+bm), if m<n

= (a0+b0)xn+…+(an+bn), if m=n

= b0xm+…+bm-n-1xn+1+(bm-n+a0)xn+…+(bm+an), if m>n.

Multiplication Let f(x)= a0xn+a1xn-1+…+an-1x+an, g(x)= b0xn+b1xn-1+…+bn-

1x+bn. the product of the polynomials f(x) and g(x) is given by

f(x)g(x)=c0xm+n+c1xm+n-1+…+cm+n, where ci=a0bi+a1bi-1+…+aib0. c0=a0b0(0;

hence degree of f(x)g(x) is m+n.

Division Algorithm Let f(x) and g(x) be two polynomials of degree n and

m respectively and nFm. Then there exist two uniquely determined

polynomials q(x) and r(x) satisfying f(x)=g(x)q(x)+r(x), where the degree

of q(x) is n-m and r(x) is either a zero polynomial or the degree of r(x) is

less than m. In particular, if degree of g(x) is 1, then r(x) is a constant,

identically zero or non-zero.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 2 (1 . 2 (1 . 2 (1 . 2 ( Remainder Theorem) If a polynomial f(x) is divided by

x-a, then the remainder is f(a).

»Let q(x) be the quotient and r (constant)be the remainder when f is

divided by x-a. then f(x)=(x-a)q(x)+r is an identity. Thus f(a)=r.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 3 (1 . 3 (1 . 3 (1 . 3 ( Factor Theorem) If f is a polynomial, then x-a is a

factor of f iff f(a)=0.

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59

»By Remainder theorem, f(a) is the remainder when f is divided by x-a;

hence, if f(a)=0, then x-a is a factor of f. Conversely, if x-a is a factor of f,

then f(x)=(x-a)g(x) and hence f(a)=0.

ExampleExampleExampleExample 1.29 1.29 1.29 1.29 Find the remainder when f(x)=4x5+3x3+6x2+5 is

divided by 2x+1.

»The remainder on dividing f(x) by x-(-7I�=x+

7I is f(-

7I)=6. If q(x) be the

quotient, then f(x)=q(x)(x+7I)+6=

�"D�I (2x+1)+6. Hence 6 is the remainder

when f is divided by 2x+1.

Synthetic division

Synthetic division is a method of obtaining the quotient and remainder

when a polynomial is divided by a first degree polynomial or by a finite

product of first degree polynomials. Let q(x)=b0xn-1+b1xn-2+…+bn-1 be the

quotient and R be the remainder when f(x)=a0xn+a1xn-1+…+an is divided by

x-c. Then

a0xn+a1xn-1+…+an=( b0xn-1+b1xn-2+…+bn-1)(x-c)+R.

Above is an identity; equating coefficients of like powers of x, a0=b0, a1=b1-

cb0, a2=b2-cb1,…,an-1=bn-1-cbn-2,an=R-cbn-1. Hence

b0=a0,b1=a1+cb0,b2=a2+cb1,…,bn-1=an-1+cbn-2,R=an+cbn-1.

The calculation of b0,b1,…,bn-1,R can be performed as follows:

a0 a1 a2 … an-1 an

cb0 cb1 … cbn-2 cbn-1

b0 b1 b2 …. bn-1 R

ExampleExampleExampleExample 1.301.301.301.30 Find the quotient and remainder when 2x3-x2+1 is

divided by 2x+1.

» -7I 2 -1 0 1

-1 1 -½

……………………………..

2 -2 1 | ½

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60

Thus 2x3-x2+1=(x+7I)(2x2-2x+1)+ 7I=(2x+1)(x2-x+

7I)+ 7I; hence the quotient is

x2-x+7I and the remainder is

7I.

ExampleExampleExampleExample 1.311.311.311.31 Find the quotient and remainder when x4-

3x3+2x2+x-1 is divided by x2-4x+3.

»x2-4x+3=(x-1)(x-3). We divide x4-3x3+2x2+x-1 by x-1 and then the

obtained quotient again by x-3 by method of synthetic division. We get x4-

3x3+2x2+x-1=(x-1)[x3-2x2+1]+0=(x-1)[(x-3){x2+x+3}+10]= (x2-

4x+3)(x2+x+3)+10(x-1); hence the quotient is x2+x+3 and the remainder is

10(x-1).

Applications of the method

(1) To express a polynomial f(x)=a0xn+a1xn-1+…+an as a polynomial

in x-c.

Let f(x)=A0(x-c)n+A1(x-c)n-1+…+An. Then f(x)=(x-c)[A0(x-c)n-1+A1(x-

c)n-2+…+An-1]+An. Thus on dividing f(x) by x-c, the remainder is An and

the remainder is q(x)= A0(x-c)n-1+A1(x-c)n-2+…+An-1. Similarly if q(x) is

divided by x-c, the remainder is An-1. Repeating the process n times,

the successive remainders give the unknowns An,…,A1 and A0=a0.

(2) Let f(x) be a polynomial in x. to express f(x+c) as a polynomial

in x.

Let f(x)=a0xn+a1xn-1+…+an=A0(x-c)n+A1(x-c)n-1+…+An; by method

explained above , the unknown coefficients can be found out in terms

of the known coefficients a0,…,an. Now f(x+c)=A0xn+…+An.

ExampleExampleExampleExample 1.32 1.32 1.32 1.32 Express f(x)=x3-6x2+12x-16=0 as a polynomial in x-

2 and hence solve the equation f(x)=0.

» Using method of synthetic division repeatedly, we have

2 | 1 -6 12 -16 Hence f(x)=(x-2)3-8=0.

2 -8 8 Thus x-2=2,2w,2w2 (w is an

imaginary cube roots of unity)

---------------------------- Hence x=4,2+2w,2+2w2.

1 -4 4 | -8

2 -4

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61

-----------------------

1 -2 | 0

2

------------

1 | 0

ExampleExampleExampleExample 1.33 1.33 1.33 1.33 If f(x) is a polynomial of degree F2 in x and a,b are

unequal, show that the remainder on dividing f(x) by (x-a)(x-b)

is(D@Å)Ï(0)@(D@0)Ï(Å)

0@Å .

» By division algorithm, let f(x)=(x-a)(x-b)q(x)+rx+s, where rx+s is the

remainder. Replacing x by a and by b in turn in this identity, f(a)=ra+s,

f(b)=rb+s; solving for r,s and substituting in the expression rx+s,we get

required expression for remainder.

ExampleExampleExampleExample 1.34 1.34 1.34 1.34 If x2+px+1 be a factor of ax3+bx+c, prove that a2-

c2=ab. Show that in this case x2+px+1 is also a factor of cx3+bx2+a.

»Let ax3+bx+c=( x2+px+1)(ax+d) (*) (taking into account the coefficient of

x3). Comparing coefficients, d+ap=0,a+pd=b,d=c whence the result a2-c2=ab

follows. Replacing x by 1/x in the identity (*), we get cx3+bx2+a=(

x2+px+1)(dx+a): this proves the second part.

If f(x) is a polynomial of degree n, then f(x)=0 is called a polynomial

equation of degree n. If b is a real or complex number such that f(b)=0,

then b is a root of the polynomial equation f(x)=0 or is a zero of the

polynomial f(x). If (x-b)r is a factor of f(x) but (x-b)r+1 is not a factor of f(x),

then b is a root of f(x) of multiplicity r: a root of multiplicity 1 is called a

simple root. Thus 2 is a simple root of x3-8=0 but 2 is a root of multiplicity 3

of (x-2)3(x+3)=0.

ExampleExampleExampleExample 1.351.351.351.35 Show that 1-D7! * D(D@7)

I! *�* ( 1�w D"D@7�…"D@wB7�w! �"@7�|w! ") 1�… ") ��.

» 1,…,n are all zeros of the polynomial on the left; by factor theorem, each of

(x-1),…,(x-n) are factors of the polynomial and hence (x-1)…(x-n) is a factor

of the n th degree polynomial on the left; equating coefficient of xn from

both side, we get the constant of proportionality "@7�|w! on the right.

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T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 4 (1 . 4 (1 . 4 (1 . 4 ( Fundamental Theorem of Classical AlgebraFundamental Theorem of Classical AlgebraFundamental Theorem of Classical AlgebraFundamental Theorem of Classical Algebra))))

Every polynomial equation of degreeF1 has a root, real or complex.

Corollary Corollary Corollary Corollary A polynomial equation of degree n has exactly n roots,

multiplicity of each root being taken into account.

CorollaryCorollaryCorollaryCorollary If a polynomial f(x) of degree n vanishes for more than n distinct

values of x, then f(x) =0 for all values of x.

ExampleExampleExampleExample 1.361.361.361.36 x2-4=(x+2)(x-2) is an identity since it is satisfied by

more then two values of x; in contrast (x-1)(x-2)=0 is an equality and not

an identity.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 51 . 51 . 51 . 5 If b is a multiple root of the polynomial equation f(x)=0 of

multiplicity r, then b is a multiple root of f(1)(x)=0 of multiplicity r-1. Thus

to find the multiple roots of a polynomial equation f(x)=0, we find the h.c.f.

g(x) of the polynomials f(x) and f(1)(x). The roots of g(x)=0 are the multiple

roots of f(x)=0.

ExampleExampleExampleExample 1.37 1.37 1.37 1.37 Find the multiple roots of the equation

x5+2x4+2x3+4x2+x+2=0.

Let f(x)= x5+2x4+2x3+4x2+x+2. Then f(1)(x)=5x4+8x3+6x2+8x+1. The h.c.f. of

f and f(1) are obtained by method of repeated division(at each stage, we can

multiply by a constant of proportionality to our convenience; that does not

affect the outcome):

5 8 6 8 1 | 1 2 2 4 1 2

(x 5) 5 10 10 20 5 10

…5…………8…………6………8………1………….

2 4 12 4 10

(x5) 10 20 60 20 50

10 16 12 16 2

…………………………………………………..

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63

4 48 4 48

(x1/4) 1 12 1 12

1 12 1 12 | 5 8 6 8 1

5 60 5 60

………………………………………….

-52 1 -52 1

-52 -624 -52 -624

……………………………………………

625 0 625

1 0 1 | 1 12 1 12

1 0 1

………………………….

12 0 12

12 0 12

……………………………..

Thus the h.c.f. is x2-1. Thus f(x)=0 has two double roots( that is, multiple

roots of multiplicity 2) i and –i.

Polynomial equations with Real CoefficientsPolynomial equations with Real CoefficientsPolynomial equations with Real CoefficientsPolynomial equations with Real Coefficients

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 61 . 61 . 61 . 6 If a+ib is a root of multiplicity r of the polynomial

equation f(x)=0 with real coefficients, then a-ib is a root of multiplicity r of

f(x)=0.

N o t eN o t eN o t eN o t e 1+i is a root of x2-(1+i)x=0 but not so is 1-i.

ExampleExampleExampleExample 1.38 1.38 1.38 1.38 Prove that the roots of 7

D@7 * ID@I * H

D@H = 7D are all

real.

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64

» The given equation is 7

D@7+G

D@I+8

D@H=-5 (*). Let a+ib be a root of the

polynomial equation (*) with real coefficients. Then a-ib is also a root of

(*).Thus 7

(0@7)B�Å+G

(0@I)B�Å+8

(0@H)B�Å=-5 and 7

(0@7)@�Å+G

(0@I)@�Å+8

(0@H)@�Å=-5.

Subtracting,-2ib[7

(0@7)EBÅE * G(0@I)EBÅE * 8

(0@H)EBÅE]=0 which gives b=0.

Hence all roots of given equation must be real.

ExampleExampleExampleExample 1.39 1.39 1.39 1.39 Prove that the roots of 7

DB0�* � * 7

DB0|= 7

DBÅ are all

real where a1,…,an,b are all positive real numbers and b>ai for all i.

ExampleExampleExampleExample 1.40 1.40 1.40 1.40 Solve the equation f(x)=x4+x2-2x+6=0 , given that

1+i is a root.

» Since f(x)=0 is a polynomial equation with real coefficients, 1-i is also a

root of f(x)=0. By factor theorem,(x-1-i)(x-1+i)=x2-2x+2 is a factor of f(x).

by division, f(x)=( x2-2x+2)(x2+2x+3). Roots of x2+2x+3=0 are -1Ë√2i.

Hence the roots of f(x)=0 are 1Ëi, -1Ë√2i.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 71 . 71 . 71 . 7 If a+√3 is a root of multiplicity r of the polynomial

equation f(x)=0 with rational coefficients, then a-√3 is a root of multiplicity

r of f(x)=0 where a,b are rational and b is not a perfect square of a rational

number.

Since every polynomial with real coefficients is a continuous function from

R to R, we have

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 81 . 81 . 81 . 8 (Intermediate Value Property) Let f(x) be a polynomial

with real coefficients and a,b are distinct real numbers such that f(a) and

f(b) are of opposite signs. Then f(x)=0 has an odd number of roots between

a and b. If f(a) and f(b) are of same sign, then there is an even number of

roots of f(x)=0 between a and b.

ExampleExampleExampleExample 1.411.411.411.41 Show that for all real values of a, the equation

(x+3)(x+1)(x-2)(x-4)+a(x+2)(x-1)(x-3)=0 has all its roots real and simple.

»Let f(x)= (x+3)(x+1)(x-2)(x-4)+a(x+2)(x-1)(x-3). Then limDC@Ñ Ò")�=∞,

f(-2)<0, f(1)>0, f(3)<0, limDCÑ Ò")�=∞. Thus each of the intervals ( ∞,-

2),(-2,1),(1,3),(3, ∞) contains a real root of f(x)=0. Since the equation is of

degree 4, all its roots are real and simple.

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65

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 91 . 91 . 91 . 9 (Rolle’s Theorem) Let f(x) be a polynomial with real

coefficients . Between two distinct real roots of f(x)=0 ,there is at least one

real root of f(1)(x)=0.

N o tN o tN o tN o t eeee

(1) Between two consecutive real roots of f(1)(x)=0, there is at

most one real root of f(x)=0.

(2) If all the roots of f(x)=0 be real and distinct, then all the roots of

f(1)(x)=0 are also real and distinct.

ExampleExampleExampleExample 1.421.421.421.42 Show that the equation f(x)=(x-a)3+(x-b)3+(x-c)3+(x-

d)3=0, where a,b,c,d are not all equal , has only one real root.

» Since f(x)=0 is a cubic polynomial equation with real coefficients, f(x)=0

has either one or three real roots. If � be a real multiple root of f(x)=0 with

multiplicity 3, then � is also a real root of f(1)(x)=3[(x-a)2+(x-b)2+(x-c)2+(x-

d)2]=0, and hence �=a=b=c=d (since �,a,b,c,d are real), contradiction. If

f(x)=0 has two distinct real roots, then in between should lie a real root of

f(1)(x)=0, contradiction since not all of a,b,c,d are equal. Hence f(x)=0 has

only one real root.

ExampleExampleExampleExample 1.43 1.43 1.43 1.43 Find the range of values of k for which the equation

f(x)=x4+4x3-2x2-12x+k=0 has four real and unequal roots.

» Roots of f(1)(x)=0 are -3,-1,1. Since all the roots of f(x)=0 are to be real and

distinct, they will be separated by the roots of f(1)(x)=0. Now limDC@Ñ Ò")�=∞,f(-3)=-9+k,f(-1)=7+k,f(1)=-9+k, limDCÑ Ò")�=∞. Since f(-

3)<0, f(-1)>0 and f(1)<0, -7<k<9.

ExampleExampleExampleExample 1.44 1.44 1.44 1.44 If c1,c2,…,cn be the roots of xn+nax+b=0, prove that

(c1-c2)(c1-c3)…(c1-cn)=n(c1n-1+a).

» By factor theorem, xn+nax+b=(x-c1)(x-c2)…(x-cn).Differentiating w.r.t. x,

n(xn-1+a)= (x-c2)…(x-cn)+(x-c1)(x-c3)…(x-cn)+…+ (x-c2)(x-c3)…(x-

cn).replacing x by c1 in this identity, we obtain the result.

ExampleExampleExampleExample 1.451.451.451.45 If a is a double root of f(x)=xn+p1xn-1+…+pn=0, prove

that a is also a root of p1xn-1+2p2xn-2+…+npn=0.

» Since a is a double root of f(x)=0, both f(a)=0 and f(1)(a)=0 hold. Thus

an+p1an-1+…+pn=0 (1) and nan-1+(n-1)p1an-2+…+pn-1=0(2). Multiplying both

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66

side of (1) by n and both side of (2) by a and subtracting, we get p1an-

1+2p2an-2+…+npn=0.Hence the result.

ExampleExampleExampleExample 1.461.461.461.46 Prove that the equation f(x)=1+x+DEI!+…+

D|w!=0 cannot

have a multiple root.

» If a is a multiple root of f(x)=0, then 1+a+0EI!+…+

0|w!=0 and

1+a+0EI!+…+

0|¯�"w@7�!=0;it thus follows that

0|w!=0, so that a=0; but 0 is not a root

of given equation. Hence no multiple root.

Descartes’ Rule of signsDescartes’ Rule of signsDescartes’ Rule of signsDescartes’ Rule of signs

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 1 01 . 1 01 . 1 01 . 1 0 The number of positive roots of an equation f(x)=0 with

real coefficients does not exceed the number of variations of signs in the

sequence of the coefficients of f(x) and if less, it is less by an even number.

The number of negative roots of an equation f(x)=0 with real coefficients

does not exceed the number of variations of signs in the sequence of the

coefficients of f(-x) and if less, it is less by an even number.

ExampleExampleExampleExample 1.47 1.47 1.47 1.47 If f(x)=2x3+7x2-2x-3 , express f(x-1) as a polynomial

in x. Apply Descartes’ rule of signs to both the equations f(x)=0 and f(-x)=0

to determine the exact number of positive and negative roots of f(x)=0.

» By using method of synthetic division, f(x)=2(x+1)3+(x+1)2-10(x+1)+4.

Let g(x)=f(x-1)=2x3+x2-10x+4. By Descartes’ Rule, g(x)=0 has exactly one

negative root, say, c. Thus g(c)=f(c-1)=0; hence c-1(<0) is a negative root of

f(x)=0. Since there are 2 variations of signs in the sequence of coefficients

of f(-x)and since c-1 is a negative root of f(x)=0, f(x)=0 has two negative

roots. Also, f(x)=0 has exactly one positive root ,by Descartes’ rule.

Sturm’s Method of loSturm’s Method of loSturm’s Method of loSturm’s Method of location of real roots of a polynomial equation with cation of real roots of a polynomial equation with cation of real roots of a polynomial equation with cation of real roots of a polynomial equation with

real coefficientsreal coefficientsreal coefficientsreal coefficients

Let f be a polynomial with real coefficients and f1 be its first derivative. Let

the operation of finding the h.c.f. of f and f1 be performed with the

following modifications: The sign of each remainder is to be changed before

it is used as the next divisor and the sign of the last remainder is also to be

changed. Let the modified remainders be denoted f2,…,fr. f,f1,f2,…,fr are

called Sturm’s functions. During the process of finding Sturm’s functions, at

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67

any step, we can multiply by positive constant but not by a negative

constant.

Sturm’s Theorem

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 1 1 1 . 1 1 1 . 1 1 1 . 1 1 Let f be a polynomial with real coefficients and a,b be

real numbers, a<b. The number of real roots of f(x)=0 lying between a and

b (a multiple root, if there be any, being counted only once) is equal to the

excess of the number of changes of signs in the sequence of Sturm’s

functions f,f1,…,fr when x=a over the number of changes of signs in the

sequence when x=b.

ExampleExampleExampleExample 1.481.481.481.48 Find the number and position of the real roots of the

equation x3-3x+1=0.

Let f(x)=x3-3x+1. f(1)(x)=3(x2-1). f1(x)=x2-1.The remainder on dividing f by

f1 is -2x+1. Thus f2=2x-1. Dividing 2f1 by f2 (and multiplying by 2 at an

intermediate step), the remainder is -3; hence f3 is 1.

f f1 f2 f3 No.of changes of sign

-∞ - + - + 3

0 + - - + 2

∞ + + + + 0

Hence the equation has 3 real roots,(3-2=)1 negative root and (2-0=)2

positive roots.

Location of roots

f f1 f2 f3 No.of changes of sign

-2 - + - + 3

-1 + 0 - + 2(0 to be

treated as continuation)

0 + - - + 2

1 - 0 + + 1

2 + + + + 0

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68

Thus one root lies between -2 and -1; one lies between 0 and 1 and one lies

between 1 and 2.

Note If at any stage of finding out the sequence of Sturm’s sequence, we

obtain a function all of whose roots are complex, then the h.c.f. process

need not be continued further and the determination and location of real

roots will be possible from the set of functions f,f1,…,fs. This is because fs

retains same sign for all values of x and no alteration in the number of

changes of sign can take place in the sequence of functions beyond fs.

ExampleExampleExampleExample 1.491.491.491.49 Find the number and position of the real roots of the

equation x4+4x3-x2-2x-5=0.

f(x)= x4+4x3-x2-2x-5, f1(x)=2x3+6x2-x-1, f2(x)=7x2+2x+9. We can verify all

the roots of f2=0 are complex and leading coefficient of f2 is positive; hence

f2(x)>0 for all real x. Hence the remaining Sturm functions need not be

calculated.

f f1 f2 No. of changes of sign

-∞ + - + 2

0 - - + 1

∞ + + + 0

The equation has two real roots, one positive and one negative.

Relations between roots and coefficientsRelations between roots and coefficientsRelations between roots and coefficientsRelations between roots and coefficients

Let c1,…,cn be the roots of the equation a0xn+a1xn-1+…+an-1x+an=0. By factor

theorem,

a0xn+a1xn-1+…+an-1x+an=a0(x-c1)(x-c2)…(x-cn).

Equating coefficients of like powers of x,a1=a0( ∑ f7), a2=a0∑ f7 fI ,….,an=a0

(-1)nc1c2…cn. Hence

∑ f7=-0�0Ô

, ∑ f7 fI=0E0Ô,…, c1c2…cn=(-1)n

0|0Ô.

ExampleExampleExampleExample 1.50 1.50 1.50 1.50 Solve the equation 2x3-x2-18x+9=0 if two of the roots

are equal in magnitude but opposite in signs.

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69

» Let the roots be -a, a, b .Using relations between roots and coefficients,

b=(-a)+a+b=7I and –a2b=-

8I. Hence a2=9, that is, a=Ë3. Hence the roots are 3,-

3, 7I.

ExampleExampleExampleExample 1.51 1.51 1.51 1.51 Solve x3+6x2+11x+6=0 given that the roots are in

A.P.

Symmetric functions of roots

A function f of two or more variables is symmetric if f remains unaltered by

an interchange of any two of the variables of which f is a function. A

symmetric function of the roots of a plolynomial equation which is sum of a

number of terms of the same type is represented by any one of its terms

with a sigma notation before it: for example, if a,b,c be the roots of a cubic

polynomial, then ∑ /I will stand for a2+b2+c2.

ExampleExampleExampleExample 1.521.521.521.52 If a,b,c be the roots of x3+px2+qx+r=0, find the value

of (1) ∑ /I, (2) ∑ /I 3, (3) ∑/H,(4) ∑ /I 3I,(5) ∑ 70 ,(6) ∑ 7

0Å,(7) ∑ 70E.

» (1)∑/I=(∑/)I 2∑/3=p2-2q, (2) ∑/I 3=∑/∑/ 3 -3abc=-pq+3r,

(3) ∑ /H=∑/I∑/-∑/I 3, (4) ∑/I 3I="∑/3�I-2abc∑/, (5) ∑ 70=∑0Å0Åb , (6)

∑ 70Å=

∑00Åb, (7) ∑ 7

0E=e∑ 70hI 2∑ 7

0Å.

T h e o r e mT h e o r e mT h e o r e mT h e o r e m 1 . 1 1 (1 . 1 1 (1 . 1 1 (1 . 1 1 ( Newton) Let a1,…,an be the roots of xn+p1xn-1+p2xn-

2+…+pn=0, sr=a1r+…+an

r where r is a non-negative integer. Then

(1) sr+p1sr-1+p2sr-2+….+pr-1s1+rpr=0, if 1= � = �

(2) sr+p1sr-1+p2sr-2+….+pnsr-n=0, if rFn

ExampleExampleExampleExample 1.53 1.53 1.53 1.53 If a1,a2,a3,a4 be the roots of the equation

x4+p2x2+p3x+p4=0, find the value of (1)∑/7H, (2) ∑ /7G,(3) ∑ /7².

» (1)By Newton’s Theorem, s3+p2s1+3p3=0. Here s1=0.Thus s3=-3p3.

(2)By Newton’s Theorem, s4+p2s2+p3s1+4p4=0. Here s1=0 and s2=-2p2. Thus

s4=2(p22-2p4)

(3)s6+p2s4+p3s3+p4s2=0. Hence s6=6p4p2+3p32-2p2

3.

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70

Transformation of equations

When a polynomial equation is given, it may be possible , without knowing

the individual roots, to obtain a new equation whose roots are connected

with those of the given equation by some assigned relation. The method of

finding the new equation is said to be a transformation. Study of the

transformed equation may throw some light on the nature of roots of the

original equation.

(1) Let c1,…,cn be the roots of a0xn+a1xn-1+…+an-1x+an=0; to obtain

the equation whose roots are mc1,mc2,…,mcn. (m=-1 is an interesting

case)

» Let d1=mc1. Since c1 is a root of a0xn+a1xn-1+…+an-1x+an=0, we have

a0c1n+a1c1

n-1+…+an-1c1+an=0. Replacing c1 by d1/m, we get

a0d1n+ma1d1

n-1+m2a2d1n-2+…+mn-1an-1d1+mnan=0. Thus the required

equation is a0xn+ma1xn-1+…+mn-1an-1x+mnan=0.

(2) Let c1,…,cn be the roots of a0xn+a1xn-1+…+an-1x+an=0 and let

c1c2…cn(0; to obtain the equation whose roots are 7b�

, … , 7b|

.

» Let d1=7b�

. So c1=7

Õ�. Substituting in a0c1

n+a1c1n-1+…+an-1c1+an=0, we

get a0+a1d1+a2d12+…+an-1d1

n-1+an d1n=0. Thus

7b�

, … , 7b|

are the roots of

anxn+an-1xn-1+…+a1x+a0=0.

(3) Find the equation whose roots are the roots of f(x)=x4-

8x2+8x+6=0, each diminished by 2.

» f(x)=(x-2)4+8(x-2)3+16(x-2)2+8(x-2)+6=0(by method of synthetic

division). Undertaking the transformation y=x-2, the required

equation is y4+8y3+16y2+8y+6=0

ExampleExampleExampleExample 1.541.541.541.54 If a,b,c be the roots of the equation x3+qx+r=0, find

the equation whose roots are (1) a(b+c),b(c+a),c(a+b), (2)

a2+b2,b2+c2,c2+a2, (3) b+c-2a,c+a-2b,a+b-2c.

(1) »a(b+c)=∑ /3 3f=q-0Åb

0 =q+Ö0. Thus the transformation is

y=q+ÖD. sustituting x=

ÖJ@� in x3+qx+r=0 and simplifying, we obtain the

required equation.

(2) »a2+b2=∑/I-c2=-2∑ /3-c2=-2q-c2; hence the transformation is

y=-2q-x2, or, x2=-(y+2q). the given equation can be written as

x2(x2+q)2=r2; thus the transformed equation is (y+2q)(y+q)2=- r2.

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71

(3) »b+c-2a=∑ /-3a=-3a; the transformation is y=-3x.

ExampleExampleExampleExample 1.551.551.551.55 Obtain the equation whose roots exceed the roots of

x4+3x2+8x+3=0 by 1. Use Descartes’ Rule of signs to both the equations to

find the exact number of real and complex roots of the given equation.

» Let f(x)= x4+3x2+8x+3=(x+1)4-4(x+1)3+9(x+1)2-2(x+1)-1 (by method of

synthetic division). By Descartes’ rule, f(-x) has two variations of signs in its

coefficients and hence f(x)=0 has either two negative roots or no negative

roots; also f(x)=0 has no positive root(since there is no variation of signs in

the sequence of coefficients of f). Undertaking the transformation y=x+1,

f(x)=0 transforms to g(y)=y4-4y3+9y2-2y-1; considering g(-y), by Descrtes’

Rule, g(y)=0 has a negeative root, say,a. Then f(x)=0 has a-1 as a negative

root; the conclusion is f(x)=0 has two negative root, no positive root, does

not have 0 as one of its roots and consequently exactly two complex

conjugate roots (since coefficients of f are all real).

ExampleExampleExampleExample 1.561.561.561.56 Find the equation whose roots are squares of the

roots of the equation x4-x3+2x2-x+1=0 and use Descartes’ rule of signs to

the resulting equation to deduce that the given equation has no real root.

»The given equation, after squaring, can be written as

(x4+2x2+1)2=x2(x2+1)2 ; undertaking the transformation y=x2, the

transformed equation is (y2+2y+1)2=y(y+1)2, that is, y4+3y3+4y2+3y+1=0.

The transformed equation, whose roots are squares of the roots of the

original equation, has no nonnegative root by Descartes’ Rule of signs;

hence the original equation has no real root.

ExampleExampleExampleExample 1.571.571.571.57 The roots of the equation x3+px2+qx+r=0 are a,b,c.

Find the equation whose roots are a+b-2c,b+c-2a,c+a-2b. Deduce the

condition that the roots of the given equation may be in A.P.

»a+b-2c=∑ / 3f=-p-3c. We undertake a transformation y=-p-3x,

or,x= JB�H . Thus the equation whose roots are a+b-2c,b+c-2a,c+a-2b is

(y+p)3-3p(y+p)2+9q(y+p)-27r=0. If the roots of given equation are in A.P.,

at least one root of transformed equation is zero, that is , the product of all

the roots of the transformed equation is 0. Hence the condition is 2p3-

9pq+27r=0.

ExampleExampleExampleExample 1.581.581.581.58 Find the equation whose roots are cube of the roots

of the equation x3+4x2+1=0.

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72

»Let the roots of the given equation be a,b,c. Then x3+4x2+1=(x-a)(x-b)(x-c).

In this identity, we replace x by xw and xw2, where w is the imaginary cube

roots of unity, to obtain

x3+1+4x2=(x-a)(x-b)(x-c)

x3+1+4x2w2=(xw-a)(xw-b)(xw-c)=(x-aw2)(x-bw2)(x-cw2)

x3+1+4x2w=(xw2-a)(xw2-b)(xw2-c)=(x-aw)(x-bw)(x-cw)

Multiplying respective sides,

(x3+1)3+(4x2)3=(x3-a3)(x3-b3)(x3-c3).

Undertaking the transformation y=x3, (y+1)3+4y2=(y-a3)(y-b3)(y-c3).

Thus the equation whose roots are a3,b3,c3 is (y+1)3+4y2=0, or,

y3+7y2+3y+1=0.

Cardan’s Method of solving a cubic equation

ExampleExampleExampleExample 1.591.591.591.59 Solve the equation:x3-15x2-33x+847=0.

Step 1 To transform the equation into one which lacks the second degree

term.

Let x=y+h. The transformed equation is y3+(3h-15)y2+(3h2-30h-33)y+(h3-

15h2-33h+847)=0. Equating coefficient of y2 to zero, h=5. Thus the

transformed equation is y3-108y+432=0 (*)

Step 2 Cardan’s Method

Let a=u+v be a solution of (*). Then a3-108a+432=0 . also

a3=u3+v3+3uv(u+v)= u3+v3+3uva; so a3-3uva-(u3+v3)=0. Comparing, uv=36

and u3+v3=-432. Hence u3 and v3 are the roots of t2+432t+363=0. Hence u3=-

216=v3. The three values of u are -6,-6w and -6w2, where w is an

imaginarycube roots of unite. Since uv=36, the corresponding values of v

are -6,-6w2,-6w. Thus the roots of (*) are -12,6,6 and thus the roots of the

given equation are (using x=y+5) -7,11,11.

ExampleExampleExampleExample 1.601.601.601.60 Solve the equation:x3+3x+1=0.

FERRARI’S METHOD OF SOLVING A BIQUADRATIC EQUATION

We try to write down the given biquadratic equation

ax4+4bx3+6cx2+4dx+e=0 in the form (ax2+2bx+p)2-(mx+n)2=0; equating

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73

coefficients of like powers of x, we get p,m,n; in the process we express the

given biquadratic expression as product of two quadratic expressions and

hence solve the given equation.

ExampleExampleExampleExample 1.61 1.61 1.61 1.61 Solve the equation x4-10x3+35x2-50 x+24=0.

»We try to find p,m,n such that the given equation can be written in the

form (x2-5x+p)2-(mx+n)2=0. Equating coefficients of like powers of x, we

get 35=25+2p-m2,-50=-10p-2mn and 24=p2-n2. Eliminating m and n,

(p2-24)(2p-10)=m2n2=(5p-25)2 (*)

p=5 is a solution of (*), hence m=0,n=Ë1. Thus the given equation is (x2-

5x+5)2-1=0, that is, (x2-5x+6)(x2-5x+4)=0. Hence the roots of given

equation are 1,2,3,4.

ExampleExampleExampleExample 1.62 1.62 1.62 1.62 Solve the equation x4+12x-5=0.

»We try to find p,m,n such that the given equation can be written in the

form (x2+p)2-(mx+n)2=0; equating coefficients of like powers of x, we get

p2-n2=-5, 2mn=-12,2p-m2=0. Eliminating m,n, we get 36=m2n2=(p2+5)2p;

hence p=2. Thus m=Ë2, n=Ë3. Since mn=-6<0, we take m=2, n=-3. Thus the

given equation can be written in the form (x2+2)2-(2x-3)2=0, that is, x2+2x-

5=0 and x2-2x+5=0. Hence the roots of given equation are -1Ë√2, 1Ë2i.

CHAPTER III

INEQUALITIES

• I1(Weierstrass) If a1,…,an are all positive real numbers less than 1

and sn=a1+…+an, then 1-sn<(1-a1)…(1-an)<7

7B¥| and

1+sn<(1+a1)…(1+an)<7

7@¥|.

• I2(Cauchy-Schwarz) If a1,…,an and b1,…,bn be real numbers , then

(a12+…+an

2)(b12+…+bn

2)F(a1b1+…+anbn)2, equality occurs when ai=kbi

for all i for some non-zero real k or when ai=0 for all I or bi=0 for all i.

• I3If a1,…,an be n positive real numbers, not all equal, and p,q are

rational numbers, then 0�ª×«B0Eª×«B�B0|ª×«w > or

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74

<0�ªB0EªB�B0|ªw . 0�«B0E«B�B0|«w according as p and q have the same or

opposite signs.

• I4(A.M.,G.M.,H.M.)If a1,…,an be n positive real numbers, then 0�B�B0|w F l/7…/w| F w�Ø�B�B �Ø|.

• I5(Weighted A.M.,G.M.,H.M.)If a1,…,an be n positive real numbers and

p1,…,pn be n positive rational numbers, then ��0�B�B�|0|��B�B�| F

"/7��/I�E …/w�|� �ª�×�ת| F ��B�B�|ª�Ø�B�Bª|Ø|.

• I6If a>0,a(1, and x,y are positive rational numbers, then 0È@7D 4 0Ù@7

J

if x>y.

• I7If a >0,a(1 and m be a rational number, then am-1> or <m(a-1)

according as m does not or does lie between 0 and 1.

• I8If a1,…,an be n positive real numbers, not all equal, and m be a

rational number, then 0�§B�B0|§w 4 �� c e0�B�B0|w h�according as m

does not or does lie between 0 and 1.

• I9(Holder) If a1,…,an; b1,…,bn;….;l1,…,ln be m sets of positive real

numbers each containing n members and if �7, … , ��be positive

rational numbers such that �7 *�* ��=1, then /7��37�E …Ú7�§ *�* /w��3w�E … Úw�§ = "/7 *�* /w��� …"Ú7 *�* Úw��| .

• I10(Jensen) Let a1,…,an be n positive real numbers and r,s are

positive rational numbers. Then (a1r+…+an

r)1/r>(a1s+…+an

s)1/s, if r<s.

• I11(Minkowski)Let a1,…,an and b1,…,bn be all positive real numbers

and r is a positive rational number. Then

(a1r+…+an

r)1/r+(b1r+…+bn

r)1/rF[(a1+b1)r+…+(an+bn)r]1/r, when r>1.

(a1r+…+an

r)1/r+(b1r+…+bn

r)1/r=[(a1+b1)r+…+(an+bn)r]1/r, when 0<r<1.

EXAMPLES

1. If a,b,c be all real numbers, prove that a2+b2+c2F ab+bc+ca

» a2+b2+c2 " ab+bc+ca)=1/2[(a-b)2+(b-c)2+(c-a)2] F0.

2. If a,b,c be positive real numbers, prove that 0EBÅE0BÅ * ÅEBbE

ÅBb * bEB0EbB0 F

/ * 3 * f.

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75

»2(a2+b2)=(a+b)2+(a-b)2F(a+b)2. Hence 0EBÅE0BÅ F 0BÅ

I . Similarly

ÅEBbEÅBb F ÅBb

I and 0EBbE0Bb F 0Bb

I . Adding respective sides, we get the

result.

3. If a,b,c be all positive real numbers and n be a positive rational

number, prove that an(a-b)(a-c)+bn(b-a)(b-c)+cn(c-a)(c-b) F0.

» If a=b=c, nothing remains to prove. If a=b(c, LHS equals cn(c-a)2>0.

Next, let no two of a,b,c are equal. Since the expression on the left is

symmetric in a,b,c, we can assume without loss of generality a>b>c.

Then an(a-b)(a-c)+bn(b-a)(b-c)=(a-b)[an(a-c)-bn(b-c)]>0 (since a>b

and n positive). Also cn(c-a)(c-b)>0. Hence an(a-b)(a-c)+bn(b-a)(b-

c)+cn(c-a)(c-b) F0.

4. If a,b,c,d be positive real numbers, not all equal, prove that

(a+b+c+d)(70 * 7

Å * 7b * 7

Õ�>16.

» By I3, 0BÅBbBÕ

G . 0¯�Bů�Bb¯�BÕ¯�G 4 0ÔBÅÔBbÔBÕÔG � 1. Hence.

5. If n be a positive integer, prove that 7.H.±…"I|@7�I.G.Û…I| c I|

I|�@7.

» By Weierstrass inequality, e1 7Ih e1 7

IEh… e1 7I|h c7

7B�EB�B �E|=

I|I|�@7. Hence the result.

6. If n>1, prove (1+2+…+n)(1+7I *�* 7

w� 4 �I. » By I3,

7BIB�BÍw .

7B�EB�B�|w >7.7BI.�EB�Bw.�|w =1. Hence.

7. If a1,…,an be n positive real numbers in ascending order of magnitude,

prove that a1<0�EB�B0|E0�B�B0|<an.

» a1=0�EB0�0EB�B0�0|0�B�B0| c 0�EB�B0|E0�B�B0| c 0�0|B0E0|B�B0|E0�B�B0| � /w.

8. If a,b,c be real numbers, prove that (a+b-c)2+(b+c-a)2+(c+a-

b)2Fab+bc+ca.

» LHS= 3(a2+b2+c2)-2(ab+bc+ca)F3(ab+bc+ca)-

2(ab+bc+ca)=ab+bc+ca.

9. If a,b,c,x,y,z are real numbers and a2+b2+c2=1=x2+y2+z2, prove that -

1=ax+by+cz=1.

» By Cauchy-Schwarz inequality, (ax+by+cz)2=( a2+b2+c2)(

x2+y2+z2)=1. Hence.

10. If a,b,c be positive real numbers, prove e0BÅBbH hH F / eÅBbI hI.

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76

» 7.0BI.Ü×ÝE7BI F Þ/. eÅBbI hIß

7 HÆ. Hence the result.

11. If a,b,c be positive real numbers, prove 0Å * Å

b * b0 F3.

»

ØÜBÜÝBÝØH F à0Å Åb b0¹ � 1. Hence.

12. If a,b,c be positive real numbers, 8

0BÅBb = I0BÅ * I

ÅBb * IbB0 = 7

0 *7Å * 7

b. » EØ×ÜB EÜ×ÝB EÝ×ØH F HØ×ÜE BÜ×ÝE BÝ×ØE �

H0BÅBb. Hence

80BÅBb = I

0BÅ * IÅBb * I

bB0.

Again,

�ØB�ÜI F I0BÅ ,

�ÝB�ÜI F IbBÅ, �ØB�ÝI F I

0Bb. Hence I0BÅ * I

ÅBb * IbB0 = 7

0 *7Å * 7

b. 13. If a,b,c be positive real numbers,

0ÅBb * Å

bB0 * b0BÅ>

HI, unless

a=b=c.

» eÜ×ÝØ h¯�BeÝ×ØÜ h¯�BeØ×ÜÝ h¯�H 4 áÜ×ÝØ BÝ×ØÜ BØ×ÜÝH â@7 � ãH@e�ØB�ÜB�ÝhH ä@7=å1 7H e70 * 7

Å * 7bhæ@7>1. Hence

0ÅBb * Å

bB0 * b0BÅ>3>

HI.

14. If a,b,c be positive real numbers and a+b+c=1, prove 77@0 * 7

7@Å * 77@b F 8

I. » ��¯ØB ��¯ÜB ��¯ÝH F H

H@"0BÅBb� � HI. Hence the result.

15. If a,b,c be positive real numbers and a+b+c=1, prove that "/ * 70�I * "3 * 7

�I * "f * 7b�I F 33 7H.

» 0EBÅEBbEH >e0BÅBbH hI(by I8, since m=2 does not lie between 0 and

1)=78. 0¯EBůEBb¯EH >e0BÅBbH h@I=9. Hence the result.

16. If a,b,c,d are positive real numbers, prove that

a6+b6+c6+d6Fabcd(a2+b2+c2+d2).

» 0çBÅçBbçBÕçG F 0¸BŸBb¸BÕ¸G . 0EBÅEBbEBÕEG F

7G √/G3GfGgG¸(a2+b2+c2+d2). Hence.

Page 77: Sem I,Mtma,Algebra1

77

17. If a,b,c,d are positive real numbers, prove that

(1+a4)(1+b4)(1+c4)(1+d4)F(1+abcd)4.

»By Holder’s inequality, [1.1.1.1+"/G��̧"3G��̧"fG��̧"gG��̧]="1 * /G��̧"1 * 3G��̧"1 * fG��̧"1 * gG��̧. Hence.

18. If a,b,c,d are four positive real numbers,

4(a4+b4+c4+d4)F(a+b+c+d)(a3+b3+c3+d3)F16abcd.

» ∑0¸G F ∑0¹G

∑0G . Thus 4∑/G F ∑/H∑/ F 4"/3fg�¹̧. 4"/3fg��̧ �

16/3fg. 19. If a,b,c,d be positive and s=a+b+c+d, then 81abcd=(s-a)(s-b)(s-

c)(s-d)= Û7I�²s4.

»."£ /�"£ 3�"£ f�"£ g�%�̧ = G¥@"0BÅBbBÕ�G =

H¥G . Hence (s-a)(s-

b)(s-c)(s-d)= Û7I�²s4.

Again, ¥@0H � ÅBbBÕ

H F √3fg¹. Similarly others. Hence 81abcd=(s-a)(s-

b)(s-c)(s-d).

20. If a,b,c,d be positive and s=a+b+c+d, then 7²¥ = H

¥@0 * H¥@Å * H

¥@b *H¥@Õ = 7

0 * 7Å * 7

b * 7Õ.

» �ØB�ÜB�ÝH F H0BÅBb � H

¥@Õ. Adding similar terms, 70 * 7

Å * 7b * 7

Õ F H¥@0 *

H¥@Å * H

¥@b * H¥@Õ. Again,

�é¯ØB�B �é¯~G F GG¥@"0BÅBbBÕ� � G

H¥. Hence

7²¥ = H

¥@0 * H¥@Å * H

¥@b * H¥@Õ.

21. If n be a positive integer >1, then nn>1.3.5…(2n-1)

»n=7B"Iw@7�

I 4 l1. "2� 1�, n=

HB"Iw@H�I F l3. "2� 3�,…,n=

"Iw@H�BHI F l"2� 3�. 3,n=

"Iw@7�B7I 4

l"2� 1�. 1. Multiplying respective sides, we get the result.

22. If n be a positive integer >1, prove that 7.H.�…."Iw@7�I.G.²…."Iw� >

7I√w.

» wB"w@7�I F l�"� 1�. Thus Iw@7Iw F àw@7w . Hence

7I F 7

I , HG F à7I , �² FàIH,…, Iw@7Iw F àw@7w . Multiplying respective sides, we get the result.

23. If n be a positive integer, prove 7wB7* 7

wBI *�* 7HwB7 4 1.

» �|×�B�B �¹|×�IwB7 4 IwB7"wB7�B�B"HwB7� � 7

IwB7. Hence the result.

Page 78: Sem I,Mtma,Algebra1

24. If n be a positive integer, prove

Let a=1- , m=

25. If a1,…,an be positive and s=a

.

= =

26. If a,b,c be positive rational, prove

.

27. If a,b,c are positive rationals,(a+b)

, unless a=b=c.

78

If n be a positive integer, prove

from am-1 m(a-1), we get

Hence .

be positive and s=a1+…+an, prove that

. Hence.

If a,b,c be positive rational, prove

= . Hence.

If a,b,c are positive rationals,(a+b)c(b+c)a(c+a)b<

, unless a=b=c.

.

+

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79