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Algebra for JEE Main and Advanced 11th Edition Goyal
Digital Instant Download
Author(s): Goyal
ISBN(s): 9789312146880, 9312146882
Edition: 11
File Details: PDF, 66.69 MB
Year: 2018
Language: english
i I
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o oo

Dr. SK Goyal arihant

I

Skills in
Mathematics for
JEE MAIN &
ADVANCED

Algebra
With Sessionwise Theory & Exercises

*
I
I
i

I
J
Skills in
Mathematics for
JEE MAIN &
ADVANCED

Aleebra
With Sessionwise Theory & Exercises

Dr. SK Goyal

jjcarihant
ARIHANT PRAKASHAN (Series), MEERUT
Skills in Mathematics for
JEE MAIN & ADVANCED

arihant
ARIHANT PRAKASHAN (Series), MEERUT
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Skills in Mathematics for
JEE MAIN & ADVANCED

PREFACE
“THE ALGEBRAIC SUM OF ALL THE TRANSFORMATIONS OCCURRING INA CYCLICAL
PROCESS CAN ONLY BE POSITIVE, OR, AS AN EXTREME CASE EQUAL TO NOTHING"
MEANS IF YOU CONTINUOUSLY PUT YOUR EFFORTS ON AN ASPECT YOU HAVE VERY
GOOD CHANCE OF POSITIVE OUTCOME i.e. SUCCESS

It is a matter of great pride and honour for me to have received such an overwhelming response to
the previous editions of this book from the readers. In a way, this has inspired me to revise this
book thoroughly as per the changed pattern of JEE Main & Advanced. I have tried to make the
contents more relevant as per the needs of students, many topics have been re-written, a lot of new
problems of new types have been added in etcetc. All possible efforts are made to remove all the
printing errors that had crept in previous editions. The book is now in such a shape that the
students would feel at ease while going through the problems, which will in turn clear their
concepts too.

A Summary of changes that have been made in Revised & Enlarged Edition
• Theory has been completely updated so as to accommodate all the changes made in J EE Syllabus &
Pattern in recent years.
• The most important point about this new edition is, now the whole text matter of each chapter has
been divided into small sessions with exercise in each session. In this way the reader will be able to go
through the whole chapter in a systematic way.
• Just after completion of theory, Solved Examples of all JEE types have been given, providing the
students a complete understanding of all the formats of JEE questions & the level of difficulty of
questions generally asked in JEE.
• Along with exercises given with each session, a complete cumulative exercises have been given at the
end of each chapter so as to give the students complete practice for JEE along with the assessment of
knowledge that they have gained with the study of the chapter.
• Last 13 Years questions asked in JEE Main &Adv, IIT-JEE & AIEEE have been covered in all the
chapters.
However I have made the best efforts and put my all Algebra teaching experience in revising this
book. Still I am looking forward to get the valuable suggestions and criticism from my own
fraternity i.e. the fraternity of JEE teachers.
I would also like to motivate the students to send their suggestions or the changes that they want to
be incorporated in this book.
All the suggestions given by you all will be kept in prime focus at the time of next revision of
the book.

Dr. SK Goyal
Skills in Mathematics for
JEE MAIN & ADVANCED

CONTENTS
1. COMPLEX NUMBERS 1-102
LEARNING PART Session 4
Session 1 • nth Root of Unity
• Integral Powers of Iota (i) • Vector Representation of Complex Numbers
• Switch System Theory • Geometrical Representation of Algebraic
Operation on Complex Numbers
Session 2
• Rotation Theorem (Coni Method)
• Definition of Complex Number
• Shifting the Origin in Case of Complex
• Conjugate Complex Numbers
Numbers
• Representation of a Complex Number in
• Inverse Points
Various Forms
• Dot and Cross Product
Session 3 • Use of Complex Numbers in Coordinate
• amp (z)- amp (-z)=± p, According as amp (z) Geometry
is Positive or Negative
PRACTICE PART
• Square Root of a Complex Number
• JEE Type Examples
• Solution of Complex Equations
• Chapter Exercises
• De-Moivres Theorem
• Cube Roots of Unity

2. THEORY OF EQUATIONS 103-206

LEARNING PART Session 4
Session 1 • Equations of Higher Degree
• Polynomial in One Variable • Rational Algebraic Inequalities
• Identity • Roots of Equation with the
• Linear Equation Help of Graphs
• Quadratic Equations Session 5
• Standard Quadratic Equation • Irrational Equations
Session 2 • Irrational Inequations
• Transformation of Quadratic Equations • Exponential Equations
• Condition for Common Roots • Exponential Inequations
Session 3 • Logarithmic Equations
• Quadratic Expression • Logarithmic Inequations
• Wavy Curve Method PRACTICE PART
• Condition for Resolution into Linear Factors • JEE Type Examples
• Location of Roots (Interval in which Roots Lie) • Chapter Exercises
Skills in Mathematics for
JEE MAIN & ADVANCED

3. SEQUENCES AND SERIES 207-312

LEARNING PART Session 5
Session 1 • Mean
• Sequence Session 6
• Series • Arithmetico-Geometric
• Progression Series (AGS)
• Sigma (S) Notation
Session 2
• Natural Numbers
• Arithmetic Progression
Session 7
Session 3
• Application to Problems of Maxima and
• Geometric Sequence or Geometric Minima
Progression
PRACTICE PART
Session 4 • J EE Type Examples
• Harmonic Sequence or Harmonic Progression • Chapter Exercises

4. LOGARITHMS AND THEIR PROPERTIES 313-358

LEARNING PART Session 3
Session 1 • Properties of Monotonocity of Logarithm
• Definition • Graphs of Logarithmic Functions
• Characteristic and Mantissa PRACTICE PART
Session 2 • JEE Type Examples
• Principle Properties of Logarithm • Chapter Exercises

5. PERMUTATIONS AND COMBINATIONS 359-436

LEARNING PART Session 5
Session 1 • Combinations from Identical Objects
• Fundamental Principle of Counting Session 6
• Factorial Notation •. Arrangement in Groups
Session 2 • Multinomial Theorem
• Divisibility Test • Multiplying Synthetically
• Principle of Inclusion and Exclusion
Session 7
• Permutation
• Rank in a Dictionary
Session 3 • Gap Method
• Number of Permutations Under Certain [when particular objects are never together]
Conditions
PRACTICE PART
• Circular Permutations
• JEE Type Examples
• Restricted Circular Permutations
• Chapter Exercises
Session 4
• Combination
• Restricted Combinations
Skills in Mathematics for
JEE MAIN & ADVANCED

6. BINOMIAL THEOREM 437-518

LEARNING PART Session 4
Session 1 • Use of Complex Numbers in Binomial
• Binomial Theorem for Positive Integral Index Theorem
• Pascals Triangle • Multinomial Theorem
• Use of Differentiation
Session 2 • Use of Integration
• General Term • When Each Term is Summation Contains the
• Middle Terms Product of Two Binomial Coefficients or
• Greatest Term Square of Binomial Coefficients
• Trinomial Expansion • Binomial Inside Binomial
Session 3 • Sum of the Series
• Two Important Theorems PRACTICE PART
• Divisibility Problems • JEE Type Examples
• Chapter Exercises

7. DETERMINANTS 519-604
LEARNING PART the Same Order
Session 1 • System of Linear Equations
• Definition of Determinants • Cramers Rule
• Expansion of Determinant • Nature of Solutions of System of Linear
• Sarrus Rule for Expansion Equations
• Window Rule for Expansion • System of Homogeneous Linear Equations
Session 2 Session 4
• Minors and Cofactors • Differentiation of Determinant
• Use of Determinants in Coordinate Geometry • Integration of a Determinant
• Properties of Determinants • Walli s Formula
• Use of S in Determinant
Session 3
• Examples on Largest Value of a PRACTICE PART
Third Order Determinant • JEE Type Examples
• Multiplication of Two Determinants of • Chapter Exercises

8. MATRICES 605-690
LEARNING PART Session 2
Session 1 • Transpose of a Matrix
• Definition • Symmetric Matrix
• Types of Matrices • Orthogonal Matrix
• Difference Between a Matrix and a Determinant • Complex Conjugate (or Conjugate) of a Matrix
• Equal Matrices • Hermitian Matrix
• Operations of Matrices • Unitary Matrix
• Various Kinds of Matrices • Determinant of a Matrix
• Singular and Non-Singular Matrices
-x<- Skills in Mathematics for
XX
JEE MAIN & ADVANCED

Session 3 Session 4
• Adjoint of a Matrix • Solutions of Linear Simultaneous Equations
• Inverse of a Matrix Using Matrix Method
• Elementary Row Operations
PRACTICE PART
• Equivalent Matrices
• JEE Type Examples
• Matrix Polynomial • Chapter Exercises
• Use of Mathematical Induction

9. PROBABILITY 691-760
LEARNING PART Session 4
Session 1 • Binomial Theorem on Probability
• Some Basic Definitions • Poisson Distribution
• Mathematical or Priori or Classical Definition • Expectation
of Probability • Multinomial Theorem
• Odds in Favours and Odds Against the Event • Uncountable Uniform Spaces
Session 2 PRACTICE PART
• Some Important Symbols • JEE Type Examples
• Conditional Probability • Chapter Exercises
Session 3
• Total Probability Theorem
• Bayes Theorem or Inverse Probability

10. MATHEMATICAL INDUCTION 761-784

LEARNING PART PRACTICE PART
• Introduction • JEE Type Examples
• Statement • Chapter Exercises
• Mathematical Statement

11. SETS, RELATIONS AND FUNCTIONS 785-836

LEARNING PART Session 3
Session 1 • Definition of Function
• Definition of Sets • Domain, Codomain and Range
• Representation of a Set • Composition of Mapping
• Different Types of Sets • Equivalence Classes
• Laws and Theorems • Partition of Set
• Venn Diagrams (Euler-Venn Diagrams) • Congruences
Session 2 PRACTICE PART
• Ordered Pair • JEE Type Examples
• Definition of Relation • Chapter Exercises
• Ordered Relation
• Composition of Two Relations
 Skills in Mathematics for
JEE MAIN & ADVANCED

SYLLABUS FOR JEE MAIN

Unit I Sets, Relations and Functions Unit IV Permutations and Combinations
Sets and their representation, Union, intersection and Fundamental principle of counting, permutation as an
complement of sets and their algebraic properties, arrangement and combination as selection, Meaning of P(n,r)
Power set, Relation, Types of relations, equivalence and C (n,r), simple applications.
relations, functions, one-one, into and onto functions,
Unit V Mathematical Induction
composition of functions.
Principle of Mathematical Induction and its simple applications.
Unit II Complex Numbers
Unit VI Binomial Theorem and its
Complex numbers as ordered pairs of reals,
Representation of complex numbers in the form a+ib Simple Applications
and their representation in a plane, Argand diagram, Binomial theorem for a positive integral index, general term and
algebra of complex numbers, modulus and argument middle term, properties of Binomial coefficients and simple
(or amplitude) of a complex number, square root of a applications.
complex number, triangle inequality. Unit VII Sequences and Series
Unit III Matrices and Determinants Arithmetic and Geometric progressions, insertion of arithmetic,
Matrices, algebra of matrices, types of matrices, geometric means between two given numbers. Relation between
determinants and matrices of order two and three. AM and GM Sum upto n terms of special series: X n, £ n, £n3.
Properties of determinants, evaluation of deter­ Arithmetico - Geometric progression.
minants, area of triangles using determinants. Adjoint Unit VIII Probability
and evaluation of inverse of a square matrix using Probability of an event, addition and multiplication theorems of
determinants and elementary transformations, Test probability, Baye’s theorem, probability distribution of a random
of consistency and solution of simultaneous linear variate, Bernoulli and Binomial distribution.
equations in two or three variables using determinants
and matrices.

SYLLABUS FOR JEE ADVANCED

Algebra Logarithms and their Properties
Algebra of complex numbers, addition, multiplication, Permutations and combinations, Binomial theorem for a positive
conjugation, polar representation, properties of integral index, properties of binomial coefficients.
modulus and principal argument, triangle inequality,
Matrices as a rectangular array of real numbers, equality of
cube roots of unity, geometric interpretations.
matrices, addition, multiplication by a scalar and product of
Quadratic equations with real coefficients, relations matrices, transpose of a matrix, determinant of a square matrix of
’ between roots and coefficients, formation of quadratic order up to three, inverse of a square matrix of order up to three,
equations with given roots, symmetric functions of properties of these matrix operations, diagonal, symmetric and
roots. skew-symmetric matrices and their properties, solutions of
Arithmetic, geometric and harmonic progressions, simultaneous linear equations in two or three variables.
arithmetic, geometric and harmonic means, sums of Addition and multiplication rules of probability, conditional
finite arithmetic and geometric progressions, infinite probability, independence of events, computation of probability of
geometric series, sums of squares and cubes of the first events using permutations and combinations.
n natural numbers.
CHAPTER

Complex Numbers
Learning Part
Session 1
• Integral Powers of lota (i)
• Switch System Theory
Session 2
• Definition of Complex Number
• Conjugate Complex Numbers
• Representation of a Complex Number in Various Forms
Session 3
• amp (z) - amp (- z) = ± it, According as amp (z) is Positive or Negative
• Square Root of a Complex Number
• Solution of Complex Equations
• De-Moivre’s Theorem
• Cube Roots of Unity
Session 4
• nth Root of Unity
• Vector Representation of Complex Numbers
• Geometrical Representation of Algebraic Operation on Complex Numbers
• Rotation Theorem (Coni Method)
• Shifting the Origin in Case of Complex Numbers
• Inverse Points
• Dot and Cross Product
• Use of Complex Numbers in Coordinate Geometry

Practice Part
• JEE Type Examples
• Chapter Exercises

Arihant on Your Mobile!

Exercises with the @ symbol can be practised on your mobile. See inside cover page to activate for free.
 2 Textbook of Algebra

The square of any real number, whether positive, negative Remark

or zero, is always non-negative i.e. x2 > 0 for all x e R. 4^ = ija, where a is positive quantity. Keeping this result in
mind, the following computation is correct
Therefore, there will be no real value of x, which when J-a 7^3 = i 4a-i 4b -i2 4ab - - 4ab
squared, will give a negative number.
where, a and b are positive real numbers.
Thus, the equation x2 +1 =0 is not satisfied for any real But the computation, 44i 44b = J(-a)(-b) = 71 al |b| is wrong.
value of x. ‘Euler’ was the first Mathematician to Because the property, 4a 4b = 4ab is valid only when atleast one
introduce the symbol i (read ‘Iota’) for the square root of of a and b is non-negative,
If a and b are both negative, then 4a4b = -^|a||b|.
-1 with the property i2 = -1. The theory of complex
number was later on developed by Gauss and Hamilton. I Example 1. Is the following computation correct?
According to Hamilton, “Imaginary number is that If not, give the correct computation.
number whose square is a negative number ”. Hence, the V^2^3 = 1/(-2)(-3)=V6
equation x2 +1=0
Sol. No,
x2 = -l If a and b are both negative real numbers, then JaJb=-Jab
or x = ±V-l Here, a = - 2 and b = - 3.
(in the sense of arithmetic, J-l has no meaning). 7=2 7-3 = -7(“2) (-3) = - V6
Symbolically, V-l is denoted by i (the first letter of the
8 Example 2. A student writes the formula
word ‘Imaginary ’).
Job = JaJb. Then, he substitutes a = -1 and b = -1
Solutions of x2 +1 = 0 are x = ± i. and finds 1 = -1. Explain, where he is wrong.
Also, i is the unit of complex number, since i is present in Sol. Since, a and b are both negative, therefore Jab * Ja Jb .
every complex number. Generally, if a is positive quantity, Infact a and b are both negative, then we havejajb=- Jab.
then
J-a x V-a = V(“ 1) x a x V(-l) x a I Example 3. Explain the fallacy
I
= V-l x Va x 7-1 x Va
-1 = i x i = J^\ x V-l = 7H)x(-1) = Vi = 1.
Sol. If a and b are both negative, then
a JaJb = -JJ\b\
= i2a = -a V3TxVzi = -7|-i||-i| =-i

Session 1
Integral Powers of lota (/), Switch System Theory
Integral Powers of lota (/)
(i) If the index of i is whole number, then When, 0 < r < 3

i° =1,/’ =j,i2 =(V=1)2 =-l, ... in=i4’+''=(i4)^(Iy=(i)’.(i)r=ir

In general, i4n =1, i4n + 1 =i, i4n+2 = -l,
i3 =i.i2 =-iti4 = (i2)2 =(-i)2 =i
• 4n + 3 _ _ • £or any whoie number n.
To find the value of in (n > 4) First divide n by 4.
(ii) If the index of i is a negative integer, then
Let q be the quotient and r be the remainder.
i.e. 4) n (q i 1 = 1 = 1 == -,;,-
-i,i ’=1 = -l,
i ii 2 -1 i2
“4q
r i 1 =■.-L=i,i
1 = ,;,-. =1 = 1 = 1,etc.
i3 I*
i4 i4 1
 Chap 01 Complex Numbers 3

2 1 + i2 +2iy
I Example 4. Evaluate. Sol. ••• a2
(i) I1998 2 2 /
(ii) _(1-1 + 2i)
=i
(iii) ,ne N I 2
a1929 = a-a1928 = a-(a2)964 =a(i)',964
Sol. (i) 1998 leaves remainder 2, when it is divided by 4.
= a(i)4x241 = a-(i4)241 = a
i.e. 4) 1998(499
1996
2
§ Example 7. Dividing f(z) by z - i, where i = V-1, we
. z-1998
= i2 = -1 obtain the remainder i and dividing it by z + i, we get
the remainder 1 + i. Find the remainder upon the
Aliter
■ 2000 J
division of f(z) by z2 + 1.
p998 = -1
I2 “1 Sol. z - i = 0 => z = i

(ii) 9999 leaves remainder 3, when it is divided by 4. Remainder, when f (z) is divided by (z - i) = i
i.e. 4) 9999 (2499 i.e. j\i) = i ...(i)
9996 and remainder, when /(z) is divided by (z + 1) = 1 + i
3 i.e. /(-i) = l + i [vz + i = 0=>z = -i]...(ii)
• - 9999 1 1 i i Since, z2 + 1 is a quadratic expression, therefore remainder
•9999
=i
i3 i 1 when /(z) is divided by z2 + 1, will be in general a linear
Aliter expression. Let g(z) be the quotient and az + b (where a
and b are complex numbers) be the remainder, when /(z) is
■ - 9999 1 i _i_.
■ 9999 -10000
10000 | divided by z2 +1.

(iii) 4n + 3 leaves remainder 3, when it is divided by 4. Then, /(z) = (z2 + l)g(z) + az + t ...(iii)
i.e., 4)4n+3(n f (i) = (i2 + 1) g (i) + ai + b = ai + b
4n or ai + b = i [from Eq. (i)]... (iv)
3 and f (“ 0 = O’2 + 0 g 0 ~ ai + b =- ai + b
■ 4n + 3 •‘-Z3 = -i
I — I
or - ai + b = l + i [from Eq. (ii)] ,..(v)
= _(t-)4n + 3
Now, (-J^T)in+3 = (-i)4n + 3 From Eqs. (iv) and (v), we get
= ~(~i) , 1 . i
b-- + i and a = -
=i 2 2
Aliter (-V-1)'|4n + 3 _ ,4n + 3 +3 Hence, required remainder = az + b
1 . 1 .
=-(i4)n-i3 = - iz + - + i
2 2
= -(l)n(-0 =i

S Example 5. Find the value of 1+/2 + l4 + i6 +...+ i2n The Sum of Four Consecutive .
where i = and n e N. Powers of / [Iota] is Zero
Sol. v 1 + i2 + i4 + i6 +... + i2n = 1-1 + 1-! + ... + (-1)" If n e I and i = ypl, then
■n 4-jn + 1 q.jn + ^
Case I If n is odd, then
1 + i2 + i4 + i6 +... + i2n= 1 -1 + 1 -1 +... + 1-1 = 0
Case II If n is even, then
1 + i2 + i4 + i6 +... + i2n = 1 - 1 + 1 - 1 +... + 1 = 1 Remark
m m-p+ 1

1. r2^= S f(r + p-1)

I Example 6. If a = where i = -^-1, then find the =p r=1
m P+1

value of a1929. 2. r=S-pz(r)= rX=1

 4 Textbook of Algebra

13
I Example 8. Find the value of y (/ n 4-iin+])
(where,/ = n-1
Switch System Theory
13 13 13
[Finding Digit in the Unit's Place]
Sol. V y (in4-in + 1)= £in4- £in+1 = (i + o) +(i2 + o) We can determine the digit in the unit’s place in
n =1 n =1 n=1
ab, where a,b& N. If last digit of a are 0,1,5 and 6, then
13 13
L
v £ in =0and y in + 1 = o digits in the unit’s place of a are 0, 1, 5 and 6
= i-l n=2 n =2
respectively, for all be N.
(three sets of four consecutive powers off)

I Example 9. Find the value of y in!

100
Powers of 2
2 1,22,2 3,2 4,2 5,2 6,2 7,2 8,2 9,...the digits in unit’s place
(where,/= ^^1). n=0
of different powers of 2 are as follows :
Sol. n! is divisible by 4, V n > 4.
100 97 2, 4, 8, 6, 2, 4, 8, 6, 2,... (period being 4)
... y j-n! _ y j(n + 3)! *p» /p>. /p* /p's
n=4 n =1
®©®@®@®®® (switch number)
= i° 4- i° 4- i° 4-... 97 times = 97 ...(i)
100 3 100
(The remainder when b is divided by 4, can be 1 or 2 or 3 or 0).
.-. £ «n! = 1/ + £'n!
n=0 n =0 n =4 Then, press the switch number and then we get the digit
in unit’s place of ab (just above the switch number) i.e.
= i0! 4- i•1!” 4- i2' + ;i3!
3! 4- 97 [from Eq. (i)]
‘press the number and get the answer’.
= i1 4- i1 4- i2 4- i6 4- 97 = i 4- i - 1 - 1 4- 97
= 95 4- 2i E Example 12. What is the digit in the unit’s place of
4n + 7
(5172)11327?
! Example 10. Find the value of y ir
i— r=1 Sol. Here, last digit of a is 2.
(where, / = ^-1). The remainder when 11327 is divided by 4, is 3. Then,
4n + 7 4n + 7 4n + 4 press switch number 3 and then we get 8.
Sol. y ir = il 4- i2 4- 13 4- y ir = i - 1 - i 4- £ ir + 3 Hence, the digit in the unit’s place of (5172)I11327
1 is 8.
r=1 r=4 r=1

= -14-0 [(n 4-1) sets of four consecutive powers of i]

= -l Powers of 3
S1^2^ 3,3 4,35,3 6,37,3 8,... the digits in unit’s place of
I Example 11. Show that the polynomial
xz'p + xz,q+1 + x4r+2 + xz,s+3 is divisible by different powers of 3 are as follows:
x3 + x2 + x +1, where p, q, r, s g N. 3, 9, 7, 1, 3, 9, 7, 1,... (period being 4)
/^
Sol. Let f(x) = x4fi 4- x4<? +1 4- x4r + 2+ x4s + 3
®©®®®@®@ ••• (switch number)
and x3+x2+x + l = (x2+l)(x + 1)
= (x + i)(x-i)(x + l), The remainder when b is divided by 4, can be 1 or 2 or 3
where i = J- 1 or 0. Now, press the switch number and get the unit’s
Now, f (i) = i4p 4-i4?+ 14-i4r +2 + i4s + 3 = 1 + i + i2+ i3 =0 place digit (just above).
[sum of four consecutive powers of i is zero]
C Example 13. What is the digit in the unit’s place of
f (-/) = (-j)4P + (-i)4’ +1 + (-i)4r + 2 + (-i)4s+ 3
(143)86 ?
= 1 + (-i)1 + (—i)2 4- (— i).33 = 1 - i - 1 + i = 0
and f (- 1) = (- I)4' + (- 1)4<? +1 + (- l)4r + 2 + (- i)4‘+ 3 Sol. Here, last digit of a is 3.
The remainder when 86 is divided by 4, is 2.
= 1-14-1-1=0
Then, press switch number 2 and then we get 9.
Hence, by division theorem, f (x) is divisible by
Hence, the digit in the unit’s place of (143)86 is 9.
x3+x2 + x + l.
 Chap 01 Complex Numbers 5

Powers of 4 Powers of 8
41,4z,43,44,45,...the digits in unit’s place of different 8 3,8 4,85,8 6,87,8 8,...the digits in unit’s place of
powers of 4 are as follows: different powers of 8 are as follows:
4, 6, 4, 6, 4, ... (period being 2) 8, 4, 2, 6, 8, 4, 2, 6,... (period being 4)
T T T ? ? w
©©(D©® (switch number)
••• (switch number)
The remainder when b is divided by 2, can be 1 or 0. Now, The remainder when b is divided by 4, can be 1 or 2 or 3
press the switch number and get the unit’s place digit or 0.
(just above the switch number).
Now, press the switch number and get the unit’s place
I Example 14. What is the digit in unit’s place of digit (just above the switch number).

(1354)22222? I Example 16. What is the digit in the unit’s place of

Sol. Here, last digit of a is 4. (1008)786 ?
The remainder when 22222 is divided by 2, is 0. Then, Sol. Here, last digit of a is 8.
press switch number 0 and then we get 6. The remainder when 786 is divided by 4, is 2. Then, press
Hence, the digit in the unit’s place of (13 54 )22222 is 6. switch number 2 and then we get 4.
Hence, the digit in the unit’s place of (1008)>786 is 4.

Powers of 7
7l,7 2,7 3,7 4,7 5,7 6,7 7,7 8>... the digits in unit’s place of
Powers of 9
9 \ 92,9 3,9 4,95,... the digits in unit’s place of different
different powers of 7 are as follows:
powers of 9 are as follows:
7, 9, 3, 1, 7, 9, 3, 1,...(periodbeing4)
9, 1, 9, 1, 9,...(period being 2)

••• (switch number)

••• (switch number)
(The remainder when b is divided by 4, can be 1 or 2 or 3 The remainder when b is divided by 2, can be 1 or 0.
or 0). Now, press the switch number and get the unit’s Now, press the switch number and get the unit’s place
place digit (just above). digit (just above the switch number).

I Example 15. What is the digit in the unit’s place of I Example 17. What is the digit in the unit’s place of
(13057)941120579 ? (2419)111213?
Sol. Here, last digit of a is 7. Sol. Here, last digit of a is 9.
The remainder when 941120579 is divided by 4, is 3. Then, The remainder when 111213 is divided by 2, is 1. Then,
press switch number 3 and then we get 3. press switch number 1 and then we get 9.
Hence, the digit in the unit’s place of (13O57)94112057’ is 3. Hence, the digit in the unit’s place of (2419)111213 is 9.
6 Textbook of Algebra

g Exercise for Session 1 I

/
1 lf(1 + /)2n + (1-/)2n = -2:n+1
‘ (where, / =7-^)for all those n, which are
(a)even (b)odd
(c) multiple of 3 (d) None of these
2 If / = 7~1. the number of values of/n + i~n for different/? el is
(a) 1 (b)2
(c)3 (d)4
3 If a > 0 and b < 0, then 7a 4b is equal to (where, / = 7=1)

(a)-#W (b)TTJb]/
(d) None of these

4 Consider the following statements.

Sj:-6 = 2/x3/ =7F4)><7(-9) (where,/=7^1) S2:7F4)x7F9)=7(-4)x(-9)
S3:7(-4)x(-9)=V36 S4:736=6
Of these statements, the incorrect one is
(a) S, only (b) S2 only
(c) S3 only (d) None of these
50
5 The value of X /(2n+1)1 (where,/=J^1)is
n=0
(a)/ (b) 47-/
(c) 48 + / (d)0
1003
6 The value of X ir (where/ = 7~1)is
r=-3
(3)1 (b)-1
(C)/ (d)-/
7 The digit in the unit's place of (153)98 is
(a) 1 (b)3
(c)7 (d)9
8 The digit in the unit’s place of (141414)I12121
1 is
(3)4 (b)6
(c)3 (d)1
Session 2
Definition of Complex Number, Conjugate Complex
Numbers, Representation of a Complex Number in
Various Forms
Definition of Complex Number Algebraic Operations on
A number of the form a + ib, where a, b 6 R and i = 7~1, is Complex Numbers
called a complex number. It is denoted by z i.e. z = a + ib. Let two complex numbers be z j = a + ib and z2 = c + id,
A complex number may also be defined as an ordered pair where a,b,c,de R and i = -J-l.
of real numbers; and may be denoted by the symbol (a, b). 1. Addition Zj + z2 =(a + ib)+(c + id)
If we write z = (a, b), then a is called the real part and b is
= (a + c) + i(b + d)
the imaginary part of the complex number z and may be
denoted by Re (z) and Im (z), respectively i.e., a = Re (z) 2. SubtractionZ] -z2 = (a + ib)-(c + id)
and b = Im(z). ~(a-c) + i(b-d)
Two complex numbers are said to be equal, if and only if 3. Multiplication z i • z2 = (a + ib ) • (c + id)
their real parts and imaginary parts are separately equal. = ac + iad 4- ibc + i2 bd
Thus, a + ib = c + id
= ac + i (ad + be) - bd
a = c and b = d
= (ac - bd) + i (ad + be)
where, a, b, c, d 6 R and i = -/-I.
4 n • • zi (a + ib) (c-id)
i.e. zx = z2 z2 (e + id) (c - id)
Re(z,) =Re(z2) and Im (zx) =Im (z2)
[multiplying numerator and denominator by c - id
Important Properties of Complex Numbers where atleast one of c and d is non-zero]
1. The complex numbers do not possess the property of order, _ ac - iad + ibc - i2bd _ac + i (be - ad) + bd
i.e.. (a + ib) > or < (c + id) is not defined. For example, (c)2—(id)2 c2-i2d2
9 + 6/ > 3 + 2/ makes no sense.
2. Areal number a can be written as a + Z-0. Therefore, every _ (ac + bd) + i (be - ad) _ (ac + bd ) t . (be-ad)
real number can be considered as a complex number, whose \c2+d2)
c2+d2 (c2+d2)
imaginary part is zero. Thus, the set of real numbers (R) is a
■>

proper subset of the complex numbers (C) i.e. R c C. Hence,

the complex number system is A/ c IV c / c Q cR cC Remark
3. A complex number z is said to be purely real, if Im (z) = 0; and = / and -—- = - i, where i = V-i.
is said to be purely imaginary, if Re (z) = 0. The complex 1-/ 1+/
numberO =0 + i-0 is both purely real and purely imaginary.
4. In real number system, a2+ b2 = 0 => a = 0 = b. Properties of Algebraic Operations
But ifz, and z2 are complex numbers. thenz,2 + z 2 = 0 on Complex Numbers
does not imply z, = z2 = 0.
Let zx,z2 and z3 be any three complex numbers.
For example, z, = 1 + / and z 2 = 1 - /
Then, their algebraic operations satisfy the following
Here, z, #0, z2 #0
But z2 + z2 = (1 + i)2 + (1 - i)2 = 1 + /2 + 2/ + 1 + i2 - 2/ properties:
= 2 + 2/2 =2-2 = 0 Properties of Addition of Complex Numbers
However, if product of two complex numbers is zero, then (i) Closure law zx + z2 is a complex number.
atleast one of them must be zero, same as in case of real
numbers. (ii) Commutative lawZj + z2 =z2 +Zj
If z3z 2 =0, then zx = 0,z2 *0 or z, *0,z2 = 0 (iii) Associative law (zj +z2) +z3 = zx +(z2 +z3)
or z,=0,z2=0
 8 Textbook of Algebra

(iv) Additive identity z + 0 = z = 0 + z, then 0 is called

the additive identity.
Properties of Conjugate . I

(v) Additive inverse - z is called the additive inverse of Complex Numbers

Let z, Zj and z2 be complex numbers. Then,

Properties of Multiplication (i) (i)=* I

of Complex Numbers (ii) z + z =2 Re (z)
(i) Closure law z} •z2 is a complex number. (iii) z - z = 2 Im (z)
(ii) Commutative law zx • z2 = z2 ■ Z] (iv) z +z = 0 => z =-z => z is purely imaginary.
(iii) Associative law (z2 • z2) z3 =Zj (z2 • z3) (v) z - z = 0 => z=z => z is purely real.
(iv) Multiplicative identity z ■ 1 = z = 1 • z, then 1 is (vi) Zj ±z2 =z1 ±z2 Ingeneral,
called the multiplicative identity.
n = Zj ±z2 ±z3 ±...±z„
(v) Multiplicative inverse If z is a non-zero complex
(vii) Zj -z2 =Zi -z2
number, then - is called the multiplicative inverse
z
In general, z1 • z2 -z3 ...z„ =zx -z2 -z3...z„
of z i.e. z. — = 1 = — • z
z z (viii) = ^-,z2*0
(vi) Multiplication is distributive with respect to *2
addition Zj (z2 +z3) = Zj z2 +z1 z 3
(ix) z"=(z)n
(x) Zj z2 +Z] z2 = 2 Re(zj z2) = 2 Re(zj z2)
Conjugate Complex Numbers (xi) Ifz =/(zi,z2),thenz =/(z!,z2)
The complex numbers z = (a, b) = a + ib and
z = (a, - b) = a - ib, where a and b are real numbers, X-3 V-3 . , n ,
i = V-l and b * 0, are said to be complex conjugate of each
I Example 18. if ------+ ----- = i, where x, y e R and
3+i--- 3-i
other (here, the complex conjugate is obtained by just i = ^pi, find the values of x and y.
changing the sign of i). _ . x-3 y-3
Sol. ••• ------ + ------ = i
Note that, sum = (a + ib) + (a - ib) = 2a, which is real. 3+i 3-i
And product = (a + ib)(a-ib) =a2-(ib)2 (x — 3)(3 — i) + (y — 3) (3 + i) — i (3 + i) (3 — i)
=>(3x - xi - 9 + 3i) + (3y + yi - 9 - 3i) = lOi
= a -i b =a -(-l)p
=> (3x + 3y - 18) + i (y - x) = lOi
= a2 +b2, which is real. On comparing real and imaginary parts, we get
Geometrically, z is the mirror image of z along real axis on 3x + 3y - 18 = 0
■■(i)
• argand plane. x+y = 6
and y - x = 10 ...I,(ii)
Remark On solving Eqs. (i) and (ii), we get
Let z = - a - ib. a > 0, b > 0 = (- a, - b) (III quadrant) x = -2, y = 8
Imaginary axis
P(z) I Example 19. If (a+ib)5 = p + iq, where i = ypz\,
b prove that (b + ia)5 = q+ip.
0_ a □ > Real
o 0
axis Sol. (a + ib)5 = p + iq
b
(a + ib)5 = p + iq => (a - ib)5 = (p ~ iq)
0(2)
(-i2a-ib)5=(-i2p-iq) [•••i2=-l]
Then.z = -a + ib=(- a b) (II quadrant). Now,
(i) If z lies in I quadrant, then z lies in IV quadrant and => (~i)5(b + ia)3 = (-i)(q + ip)
vice-versa. => (~i)(b + ia)5 =(-i)(q + ip)
(ii) If z lies in II quadrant, then z lies in III quadrant and
vice-versa. (b+ ia)5 =(q + ip)
 Chap 01 Complex Numbers 9

I Example 20. Find the least positive integral value of I Example 23. Find real values of x and y for which

n, for which
'i-p" , where i = 7-1, is purely the complex numbers - 3+ i x 2y and x 2 + y + 4i,
<1 + ' where i = yp\, are conjugate to each other.
imaginary with positive imaginary part. Sol. Given, - 3 + zx 2y = x2 + y + 4i
2 n
"1-z 1 (1 - i 1- 1 + i2 - 2i 1 - 1 - 2i => - 3 - zx 2y = x2 + y + 4i
Sol.
<1 + i } U+i i-ij I 2 7 2 On comparing real and imaginary parts, we get
= ( i)n = Imaginary x2 +y = -3 —(i)
n = 1,3,5,... for positive imaginary part n = 3. and - x 2y = 4 .••(ii)
4
From Eq. (ii), we get x2 = -
I Example 21. If the multiplicative inverse of a
y
complex number is (73 + 4/)/19, where i = ^p\, find
4 2 4
complex number. Then, --+y=-3 putting x2 =---- in Eq. (i)
y y
Sol. Let z be the complex number.
(J~3 + 4i' y2+3y-4=0 => (y + 4)(y-l) = 0
Then, z- =1
19 . y = - 4, i
2
For y = -4,x = 1 => x = ± 1
19 X(V3~4Q ■
or For y = 1, x2 = - 4 [impossible]
(73 + 4z) (73 - 4i)
19(73 - 41) x = ± 1, y = - 4
= (73 - 41)
19 I Example 24. If x = - 5+2 yp4, find the value of
3+2/sin0 , xA+9x3 + 35x2-x + 4.
I Example 22. Find real 0, such that ------------ , where
1 — 2/sin 0 Sol. Since, x = -5 + 2 J-4 =>x+5 = 4i
i = 7-1, is => (x + 5)2 = (4i)2 => x2 + lOx + 25 = - 16
(i) purely real. (ii) purely imaginary. x2 + lOx + 41 =0 (i)
, 3 + 2i sin 0 Now, ,
Sol. Let z =-------------
1 - 2i sin 0 3
x2 + lOx + 411 x4 +9x3 +35x 2 - x + 4l x 2 - x + 4
On multiplying numerator and denominator by conjugate 2 \
x4 + 10x3 + 41x
of denominator,
(3 + 2i sin 0) (1 + 2i sin 0) (3 - 4 sin2 0) + 8i sin 0 -x3-6x2-x + 4
~ (1 -2isin0)(1 + 2isin0)~ (l + 4sin20) - x 3 -10x2 -41x
+ + +
(3- 4 sin2 9) (8 sin 0)
+i 4x2 + 40x + 4
(1 + 4 sin2 6) (1 + 4 sin2 0)
4x2 + 40X + 164
(i) For purely real, Im(z) = 0
8 sin 0 -160
=> = 0 or sin 6 = 0 x4 + 9x3 + 35x2 — x + 4
1 + 4sin2 0
= (x2 + 10x + 41)(x2 - x + 4) - 160
0 = n n, n G I
(ii) For purely imaginary, Re (z) = 0 = 0 - 160 = -160 [from Eq. (i)]
(3- 4sin20) n „ A . 2o a
----------- —- = 0 or 3 - 4sin 0=0 I Example 25. Let z be a complex number satisfying
(1 + 4sin20) the equation z 2 -(3+ i)z + X + 2i = 0, where XeR and
/ 2
i = 7^1. Suppose the equation has a real root, find the
or sin20 = — • —
= Ism 71
4 I 3 non-real root.
Sol. Let a be the real root. Then,
0 = mt ± —, n e I a2 -(3 + i)a + X+ 2i = 0
3
10 Textbook of Algebra

=> (a2 -3a + A) + i(2-a) = 0 Argument of z will be 0, it - 0,n + 0 and 271-0

On comparing real and imaginary parts, we get according as the point z lies in I, II, III and IV
a2 -3a + A. = 0 ...(i) quadrants respectively, where 0 = tan -1 y
2-a =0 x
...(ii)
From Eq. (ii), a = 2
Let other root be 0.
8 Example 26. Find the arguments of z1 = 5 + 5/,
Then, a + 0 = 3 + i => 2 + 0 = 3 + i z2 = -4 + 4/, z 3 = - 3- 3i and in = 2-2i,
0=1+i where / = ^-1.
Hence, the non-real root is 1 + i. Sol. Since, z,, z 2, z 3 and z4 lies in I, II, III and IV quadrants
respectively. The arguments are given by
-i !
arg(zj) = tan = tan 1l = 7t/4
Representation of a Complex 5

Number in Various Forms arg (z2) = 7t - tan 1

4
-4
-1 71
= 7t - tan 1 = 71-----
4
3tt
4
-3
Cartesian Form arg(z3) = 7t + tan 1
-3
_i 71 571
= 7t + tan 1 = 7t 4— = —
4 4
[Geometrical Representation) -2
and arg (z4) = 2n - tan-1
Every complex number z = x + iy, where x, y E R and 2
i= can be represented by a point in the cartesian -1 71 7n
= 271 - tan 1 = 2n----
plane known as complex plane (Argand plane) by the 4 4
ordered pair (x, y).
Principal Value of the Argument
The value 0 of the argument which satisfies the inequality
Modulus and Argument of a -7t <0 <7i is called the principal value of the argument.
Complex Number If z = x + iy = (x, y), V x, y G R and i = -1, then
Let z = x + iy =(x,y) for all x,y g P and i = t/-T.
arg(z) = tan-1 — always gives the principal value. It
Imaginary axis
xJ
P(x.y) depends on the quadrant in which the point (x, y) lies.

r Imaginary
axis
i

9 P(*.y)
n +- Real axis
0 x

The length OP is called modulus of the complex number z

denoted by |z|,
X' +
i.e. OP = r = |z|=7(^22 +/) o x
Real axis
and if (x, y) # (0,0), then 0 is called the argument or
amplitude of z, (i) (x, y) G first quadrant x > 0, y > 0.

i.e. 0 = tan -1 — [angle made by OP with positive X-axis] The principal value of arg (z) = 0 = tan 1

or arg(z) = tan-1 (y / x) It is an acute angle and positive.

(ii) (x, y) G second quadrant x < 0, y > 0.
Also, argument of a complex number is not unique, since
if 0 is a value of the argument, so also is 2n7t + 0, where The principal value of arg (z) = 0
/ \
n G I. But usually, we take only that value for which
0 <0 < 271. Any two arguments of a complex number differ = 7t - tan 1 _y_
by 2nn. IM 7
 Chap 01 Complex Numbers 11

y
m Imaginary or tan 1 1, it - tan"11, - it + tan 1 1, - tan" 1 1
(x.y) axis 71 It 7t It It 371 371 It
or —, it------,-7C + —, — or —,—,------- ,
4 4 444444
e Hence, the principal values of the arguments of zb z2, z 3
, it 371 3ti it
x+ ->X and z4 are —, —,----- ,----- , respectively.
X 0
Real axis 4 4 4 4

y' Remark
1. Unless otherwise stated, amp z implies principal value of the
It is an obtuse angle and positive. argument.
(iii) (x, y) G third quadrant x < 0, y < 0. 2. Argument of the complex number 0 is not defined.
3. If z, = z2 <=> | 2i| = |z2| and arg (Zi) = arg (z2).
The principal value of arg (z) = 0 = - 7t + tan"1
4. If arg (Z) = jc/2 or — it/2, z is purely imaginary.
y 5. If arg (z) =0 or n, z is purely real.
Imaginary
axis 6 Example 28. Find the argument and the principal
value of the argument of the complex number
x 0
X+ ->X 2+ /
Real axis Z =--------------- 7 , where / =
e 4/ + (1 + /)2
2+i 2+i 2+i
Sol. Since, z =
(x.y) 4i + (1 + i)z 4i+l + i2 + 2i 6i 6 3
/
:. z lies in IV quadrant.
It is an obtuse angle and negative.
1
(iv) (x, y) G fourth quadrant x > 0,y < 0.
Here, 6 = tan 3 .= tan 1 2
The principal value of arg (z) = 0 1
/ X 6
= -tan"1
arg (z) = 2it - 0 = 2n - tan"1 2
x
\ 7
Hence, principal value of arg (z) = - 0 = - tan"1 2.
yi
Imaginary
axis Properties of Modulus
x
(i) | Z | > 0 => | Z | = 0, iff Z = 0 and j z | > 0, iff z A 0
x'^ o X
(ii) -1 z | < Re (Z) < | z | and -1 z | < Im (Z) < | z |
Real axis
(iu) |z|=|z|=j-z|=|-z|
(iv) zz = | z |2
(x.y)
(v)|Zi Z2| = |zj |Z2|
It is an acute angle and negative. In general, zx z2 z3...Z„| = |Z111z2 ||z3 |...|Z„|

I Example 27. Find the principal values of the i = N-Z2^0)

(vi)iL=pl(
arguments of Z1 =2+2/, z 2 = - 3+ 31, z3 = -4 -4/
and z4 = 5 - 5i, where i = 7-1.
*2 lzd'
*2

(vii)|Z1 ±z21< Zi|+|Z2|

Sol. Since, z ;, z 2, z 3 and z4 lies in I, H, IH and IV quadrants In general, Z1 + z2 ±z3 ± ±*n|—1*1 + |*2 |
respectively. The principal values of the arguments are
+ z3 | + ... + |z„|
given by
2
/
3
\
-4 ]
(viii)|Z1 ±z2 |>||Z1 |-|z21|
tan , it - tan - it + tan
2 H’ <~ J 4’ (ix) |Z"| = |Z|n
- tan Ml (x) ||Z1|-|Z2||<|Z1 + z21<|zx| + |Zz|
5 7
12 Textbook of Algebra

Thus, | Zi | +1z21 is the greatest possible value of sin0. GO,-,

Zi +z2
and-1 ■■■'--
11 z! | -1 z211 is the■ least possible value of L 2-
Z] + z2.
i.e. 0 < sin0( < -
(xi)|zi ±z2|2 =(zj ±z2)(zj ±z2)=|z1|2 +|z2|2 2
Inequality Eq. (i) becomes,
1 | |3 1 | |2 1
2sl|zf + -|z I1 + -
or|Zj|2 +1z2|2 ±2Re(z! z2) 2 1 +;l*l + ;lzl ' 2 2

(xii) zxz2 +zyz2 = 21 Zj 11 z21 cos(0! -02), where

3<|z|'+|z|
’dzNzl
0i = arg (zj) and 02 =arg(z2) 3S|z| + |z|2+|z| 3 + |z|4<|z| + |z|2 .

(xiii) |zj + z2|2 =|zi|2 + |z2|2 <=> — is purely imaginary.

*2
+ l»r+l*l +...+ 00
4

3<|z| + |z|2+|z| ’+|z|‘ +...+ °°

(xiv) |zj +z2 |2 +|zi ~z2 |2=2{|z,|2+|z2|2}
3<±L [here, |z| < 1]
(xv) |aZ1 - bz2^+^bzj + az2 |2=(fi2+b2)(|z1|2+|z2|2), . ,i-hl
where a, b e R 3-3| z | < | z | => 3 < 4 | z |
(xvi) Unimodular i.e., unit modulus kl>-
1 1 4
If z is unimodular, then | z | = 1. In case of unimodular,
letz = cos0 + i sin 0,0 6 R and i = V-l. Hence, -<|z |ci [•/|Z|<1]
4 1 1

Remark § Example 30.If | z- 2+ /1<2, find the greatest and

1. If f(z) is unimodular, then \f (z) | = 1 and let
f (z) = cos 0 + i sine, 0 e R and i =
least values of |z|, where i = 7=1.
2. — is always a unimodular complex number, if z * 0.
Sol. Given that, | z - 2 + i | < 2 •••(0
| z - 2 + i | > 11 z | -1 2 - z 11 [by property (x)J
(xvii) The multiplicative inverse of a non-zero complex | z-2+ i |>||z|-Vs| ...(ii)
number z is same as its reciprocal and is given by
1 _ z _ z From Eqs. (i) and (ii), we get
z zz |z|2’ | |z| - Vs| < | z-2 + i | <2

||z|-^|<2
I Example 29. If 0,- e[0,7t/6],/ = 1,2,3,4,5 and
-2<|z|-V5<2
sin0, z" +sin02 z3+sin03 z2 + sin04 z
=> Vs-2<|z|<V5 + 2
+ sin05 =2, show that - < |z | < 1.
Hence, greatest value of | z | is Vs + 2 and least value of | z |
So/. Given that, is V5-2.
sin 0j z4 + sin02 z,3 + sin03 z 2 + sin04 z + sin05 = 2
or2 = |sin0] z4 +sin0 2z 3 + sin03 z2 + sin04 z + sin05|
I Example 31. If z is any complex number such that
2 < |sin 0jz4| + Jsin02 z 3 | + |sin03 z 2 I | z+41 < 3, find the greatest value of | z +11.
+1 sin04 z | +1 sin051 [by property (vii)] So/, v | z + 11 = | (z + 4) - 3 |
=>2 < | sin 0j | J z4 | +1 sin02 || z3 | + | sin03| | z2 | = |(z + 4) + (-3)|<|z + 4| + |-3|
+1 sin04 || z | +1 sin05| [by property (v)] =|z+4 |+3
=> 2 < | sin 0j || z |4 +1 sin02 || z |3 + | sin03|| z |2 <3+3=6 [■••|z + 4|<3]
+1 sin04 j| z | +1 sin05| [by property (ix)] ...(!) | z + 11 < 6
But given, 0( 6 [0, Jt/6]
Hence, the greatest value of | z + 11 is 6.
 Chap 01 Complex Numbers 13

I Example 32.lf |z, | = 1,|z2 | = 2,|z31 = 3and I Ii22 I — I2

=> I zt — 2z2 | = | 2 - Z1Z2 |
9x^2 + 4z3z, + z2z3 | = 6, find the value of => (zj -2z2)(z1 -2z2) = (2-z1z2)(2-z1z2)
z1+z2 + z3l.
[by property (iv)]
Sol. v k.|=* =» | Zi |2 = 1 (zi -2z2)(zi -2z2) = (2-z1z2)(2-z1z2)
1 _ => ZjZj - 2zjZ2 - 2z2Zj + 4z2z2
Z1 Z1 = 1 — = 21
21 = 4 - 2ZjZ2 - 2zjZ2 + ZjZtz2z2

I Z2 I = 2 => | z2 |2 = 4 => z2 z2 = 4 | Zi |2 + 4 | z2 |2 = 4 +1 Zj |21 z2 |2

|Z112 - 1^1 |2-|z2 |2 +4|z2 |2 -4=0

— = z2 and I z 3 I = 3 => |z 312 = 9
z2
- n 9 -
(l zi P “ 4 0
=> z3 z3 = 9 => — = z3
23 But | Z2 | * 1 [given]
and given | 9z xz 2 + 4z 3z j + z 2 z 3 | = 6 i z« r = 4
Hence, |Z1| = 2
I Zi z2 z3 I — + — + — = 6
z3 z2 Zi

| 2111 22 I |z3 ||z3 +Z2 + Z11 = 6 Properties of Arguments

1 _ 4 _ , 9 _ (i) arg(z1z2) = arg(z1) + arg(z2)+2fcjt, Ice I
•/ — = z p — — Z2 and — — z 3
zi z2 z3 In general, arg(zj z2 z3...zn)
=> 1-2-3 | zx +z2 + z3 1 = 6 = arg(zj) + arg (z2) + arg (z3) +... + arg (zn) + 2/cTt,
ke I.
I Zi +z2 +z3 | = 1 [v|z| = |z|]
z
(ii) arg — =arg(zj-arg(z2) + 2kit, ke I
\Z2j
I Example 33. Prove that
I z ]
|zi| + |z2|= -(Z1+Z2) + #1^2 + i(Zj +z2)-7zTzT. (iii) arg — = 2 arg (z) + 2Att, k e I
(z J

Sol. RHS = (Zj + z2) + 7^2 + 1/ x I-----

-(Zi +Z2)" ^2
(iv) arg(zn) = n. arg(z) + 2fcjt, ke I, where proper value
of k must be chosen, so that RHS lies in (-71,7t]
Z] + z2 +2 Jzfa + Zj + z2 — 2 72i22 / \ /• \
(v) If arg | — | = 0, then arg — = 2rm - 6, where n e I.
2 2
J kZ2 /

- 2^1 + +| f) (vi) arg (z) = - arg (z)

= i.2{|^|!+|^|2} [ by property (xiv)] I Example 35.lf arg (z,) = — and arg (z2) =—, find
18 18
= |z1| + |z2|=lhs the principal argument of Z]Z2 and (/] /z2).
Sol. arg (ZjZ2) = arg (zj + arg (z2) + 2kn
I Example 34. zy andz2 are two complex numbers, 17n lit
i i ^Z 2 . . I, i .|
=---- + — + 2kn
18 18
such that------ — is ummodular, while z2 is not
2-ZrZ2 = — + 2hc
3
unimodular. Find |z, |. 4n n 2n
Zi - 2z2 - ------ 271 = [for k = -1]
Sol. Here, =1 3--------- 3
2 - ZiZ2
and arg = arg (Zj) - arg(z2) + 2ht
Zj — 2z2
=> =1 [by property (vi)]
2 - zxz2
18 18 18
| Zj - 2z2 2 - zYz2 571 . 57C
= —+ 0 = — [for k = 0]
9 9
14 Textbook of Algebra

! Example 36. If zy and z2 are conjugate to each [b] Trigonometric or Polar or

other, find the principal argument of (— Z]Z2 )- Modulus Argument Form of a
Sol. zx and z2 are conjugate to each other i.e., z2 = zlt there­
fore, ZjZ2 = zxzx = | Zj |2
Complex Number
Let z = x + iy, where x, y 6 R and i = 7-1, z is represented
/.arg (- Z] z2) = arg (-1 Zi |2) = arg [negative real number]
by P (x, y) in the argand plane.
= 71
yi
[ Example 37. Let z be any non-zero complex_
number, then find the value of arg (z)+ arg (z). P(x,y)
Sol. arg (z) + arg (z) = arg (zz) f
= arg (| z |2) = arg [positive real number] f y
e
=0
o X 7?--- *"x
M Real axis

(a) Mixed Properties of Modulus By geometrical representation,

OP = 7(x2+/)=|z|
and Arguments
(i) zx + z2 = + z2 «arg(z1) = arg(z2) Z.P0M = 0 = arg (z)
*1

(ii) zx+z2 = “ *2 «arg(z1)-arg(z2) = 7t In AOPM, x = OP cos (Z POM) = z cos(arg z)

*1
and y = OP sin (Z POM) = z sin (arg z)
Proof (i) Let arg (z,) = 0 and arg (z 2) = <|)
z = x + iy
| *1 + z2 |=|zi | + |z2 I
z = | z | (cos (arg z) + i sin (arg z))
On squaring both sides, we get
or z = r (cos 0 + i sin0)
|*1 + *2^ =|*1|2 +|*2|2 +2|Z1||Z2|
z = r(cos0 -isin0)
=> |*1|2+|*2|2+2|z1||z2| cos (0-0) where, r = | z | and 0 = principal value of arg (z).

= |zi|2 +|z2|2 + 2|z1||z2| Remark

1. cos© + / sin0 is also written as CiS 0.
=> cos (0 - 0) = 1 2. Remember
0 - 0 = 0 or 0 = 0 1 = cosO + / sin 0 -1 =cos7t + i sinn
arg(zj) = arg(z2) 7T . 71 . . 71
/ =cos - + / sin- => -1 =cos - - i sin —
2 2 2 2
(ii) v |zj + z2| = |z1|-|z2|
On squaring both sides, we get 1
i Example 38. Write the polar form of - - -
| *1 + *2 |2 =1*1 |2 +|*z|2 -2|zi||z2| 2 2
(where, i =
=> 1*1 |2 +1*2 |2 +2|zi ||z2 |cos(0-0) e I t . 1 / 73
Sol. Let z =------------ . Since,
J -----
73^
f3
lies in III quadrant.
= 1*1 |2 +1*2 |2 -2|zi ||z2 | 2 2 k 2
-’ 2J -73/2
.'. Principal value of arg (z) = - 7T + tan"1
=> cos (0 - 0) = -1 -1/2
0-0 = 71 or arg(zj-arg(z2) = 7t 71 271
= - it + tan-1 73 = -7C + — =
3 3
Remark 2
1
1. | z, -z21 = 1*, 1+ | *21 <=> ar9 (Zi)=arg(z2) and | z | = + 3 I = 1=1
-1 + -
2. j z, - z21 = | Zi | -1 z21 « arg (z,) - arg (z2) = n 2 4 4J

3. |z1-z2| = |zl + z2| « arg(z1)-arg(z2) = ±^,z1z2 .'. Polar form of z = | z | [cos (argz) + i sin(argz)]

and — are purely imaginary. • - —1_ 271 2lt

i.e. cos + i sin
*2 2 2 3 3
 Chap 01 Complex Numbers 15

Sol. Given, |z| = !

(c) Euler's Form
z„ _= e„ i' 0 -(i)
If9E R and i = J-l, then e '0 =cos9 + i sin9 is known as
arg (z) = 0 •••(ii)
Euler’s identity. "arg(zH
e~iQ = cos 9 - isin9
1 + i tan
Now, < 2 J _ 1 + t’tan (9/2) [from Eq. (ii)]
RHS =---------
Let z-eiQ rarg(z)^ 1 - i tan (0 / 2)
1 - i tan
k 2 J
| z | = 1 and arg (z) =9 ei0/2
cos0 12 + i sin0 /2
Also, e"
+e-ll0 and e 0-e-'0 21 sin 9
° =2cos9 ande' e-,e/2
cos0 12 - i sin0 /2
and if 9,4>6 R and i = then = ei0 = z = LHS [from Eq. (i)]
0+$
i /
10 2 9-^ 'a-ibY 2ab
(i) e + e* = e • 2 cos I Example 41. Prove that tan i In
2 ^O+ib )
eiQ + 6* = 2 cos (where a,b e R+ and i = J^T).
< 2 ,
a - ib a - ib
and arg(e'e +e,<!>) = Sol. v =1 [•.■|z| = |z|]
< 2 ) a + ib a + ib
0+0 a - ib i0
i
i0 2 9-<|)' Let ------ = e ...(i)
(ii) e -e* = e • 2i sin a + ib
2
By componendo and dividendo , we get
'9-4?
e10 -e1'* = 2 sin (a-a)-(a+,t)=£e2-i^. i0-l =.tan(0/2)
< 2 y
(a - ib) + (a + ib) e‘e + l a
and arg (e10 -e''*) = 9 + <|> ft [\’i = ein/2 0 _b
2 2 or tan (ii)
2 a
Remark /
1. e'0 + 1 = e'0/2-2cos (0/2) (Remember) 'a - ib ^
LHS = tan i In
2. e /e -1 = e'e/2-2/sin(0/2) (Remember) <a + ibL
= tan (i In (e10)) [from Eq. (i)]
3. = /tan (0/2) (Remember)
e'0 + 1 = tan (i-i0) = - tan 0
io
4. If z = r e'°; | z | = r, then arg (z) = 0, z = r e~ 2tan0/2
= — --------------
5. If|z-zo| = 1. thenz-z0 = e'9 l-tan20/2
2(-b/a)
I Example 39. Given that | z -11 = 1, where z is a point [from Eq. (ii)]
l-(-b/a)2
on the argand plane, show that -—- = i tan (arg z), 2ab
= RHS
Z a -b
where / =
Sol. Given, | z - 11 = 1 Applications of Euler's Form
z - 1 = e10 =s
= z = ei0 + l = e,0/2 • 2 cos (0 12) If x, y, 9 E R and i = -J-l, then
arg (z) = 0 / 2 —(i)
let z = x + iy [cartesian form]
z -2 1 + e,e - 2 e10
LHS = ----- —- = i tan (012) z (cos 9 + isin9) [polar form]
z 1 + e'0 e10 + 1
= i tan (arg z) = RHS [from Eq. (i)] z e10 [Euler’s form]

i Example 40. Let z be a non-real complex number (i) Product of Two Complex Numbers
"arg (zH . Let two complex numbers be
1 + i tan
< 2 ) = |Zi |e'0) and z2 = |z21 e^2,
lying on | z | = 1, prove that z =--------- *1
farg(zH where9!,92 E R and i = J-l
1 - i tan
(where, i = J^T). 2 J
16 Textbook of Algebra

Z] ’*2 = |Z1|e |22|e«= =

lzdhle i(9I+02) w (cos (arg w) - i sin (arg w))
w (cos (- arg w) + i sin (~ arg w))
Z1 z2 (cos(0; +02) + isin(01 +02))
= — w (cos (arg w) + i sin (arg w)) = - w
Thus, |zi z2| = *1 Z2
and arg(zj z2) =6i+0 2 = arg (zj) + arg (z2) I Example 44. Express (1+1)"', (where, i = V-1) in the
form A+ iB.
(ii) Division of Two Complex Numbers
Let two complex numbers be Sol. Let A + iB = (1 + i) 1

zi =|zi | On taking logarithm both sides, we get

and z2 =| z2 le*02,
loge (A + iB) = - i loge (1 + 0
where 0j, 0 2 G 7? and i =
= - i logf V2 -= + -7=
_lZ1L(o1-.e2) -J2
*2 [z^e^ 71 . . 71
= - i loge V2 cos—r i sin —
4
4 7
zd (cos(0j -02) + isin(0j -02))
= iI— = - i loge (^2 e'n/4) = - i (loge J2 + loge ein/4)
|Z2|
.(1.. „. in}
.fl in i . „ n
= - l - loge 2 + =---- loge 2 + —
Thus, £1 I Z1 I 2^o)
= L-J,(z I2
<2 4 J 2 4
*2 Iz2| A + iB = e 2
_ n
"■ 2 + —
--log, 4 =eK/4.ellog,2-'2
/ \
and arg -h = 0j-02 = arg (zj) - arg (z2) = en/4 •(cos( loge2” 1/2) + isin( loge2-1/2))
kZ2 7
= en/4 • COS loge "7T
n/ 4 • Ti
sin loge ~r f11 Y
(iii) Logarithm of a Complex Number I U2.
2 k kv2.
loge(z) = loge (| Z | 6 ) = loge | Z | + loge (e *)
I Example 45. If sin (loge /') = o+ ib, where I =
= loge I Z I +10 = loge I z I + i arg (z) find a and b, hence and find cos (loge /').
So, the general value of loge(z) Sol. a + ib = sin (loge i') = sin (i loge i)
= loge(z) + 2nni(-71 <argz <7i). = sin (i (loge | i | + i arg i))
= sin(i(loge l + (i7t/2)))
I Example 42. If m and x are two real numbers and = sin (i (0 + (i tt/2)))= sin ( - 7t/2) = -1
xm
( xi +1 a = -1, b = 0
i- , prove that e2m/cot ’x = 1.
k xi -1 y Now, cos( loge i') = -Jl - sin2 (loge i‘)
Sol. Let cot 1 x = 0, then cot 0 = x
\m
=- (-1)2 = /a -1)=o
/.LHS = e2 m'cot 1 xi + 1 2m iQ i cot0 +1 Aliter
=e
xi - 1, i cot0 - ly ^(e^2)' _ -n/2
m
= e2mi0
' i (cot0 - i) I „2 m 10 | cos 0 - i sin0
=e /. sin (loge i') = sin (l°ge e -It/2 ) = sin f - logee
J (cot0 + i) k cos 0 + i sm0?
- sin (- tc/2) = - 1 = a + ib [given]
fe"'eT
=e
2 miQ

e 2 miO
'br^—2 m i 9
_ e 2 mi0

= e° =1=RHS
•(e - 2 I0yn
and
a = - 1, b = 0
COS ( loge ) = COS ( loge e - rt/2)
Tt
= cos -ylog.e = cos =0
2
I Example 43. If z and w are two non-zero complex
numbers such that | z =| w | and arg(z) + arg(w) = 7t, I Example 46. Find the general value of log2 (5/),
prove that z = -w.'
where i =7“T
Sol. Let arg (w) = 0, then arg (z) = n - 0 _ log. 5i _ 1
Sol. log2 5i {loge 15i | + i arg (5z) + 2nni}
.'. z = z (cos (arg z) + i sin (arg z)) log. 2 10ge 2
z (cos (7t - 0) + i sin (7t - 0))
1 in
z (- cos0 + isin0)=-|z |(cos0 -isin0) {loge 5 + — + 2nni} ,ne I
10ge 2 2
I

Chap 01 Complex Numbers 17

§ Exercise for Session 2

1 lfk“ = a - ib and a2 + b2 = 1, where a, b e R and / = then x is equal to
1 + /X
29 /(b) 2b
(a) ----------------- y
m ---------2b------- z- (C) 23
(c) ----------------- — (d)
(1+a)2 + b2 (1+ a)2 + b2
(1+a) (1+b)2 + a2 (1+b)2 + a2

r 1+/
(1 +i 2 . 1 . \
2. The least positive integer n for which ----- = — sec-1 — + sin'1 x (where, x *0,-1<x <1and/' = 7~1). is
U-/
k1 - i j n x
(a) 2 (b)4 (c)6 (d)8
3 If z =(3 + 4/)'>6 + (3 - 4/ )6, where i = then Im(z) equals to
(a)-6 (b)0 (c)6 (d) None of these

4 lf(x+/y))1/1 3 = a + ib, where i = ^-1, then is equal to

a b
(a) 4a2b2 (b)4(a2-b2) (c)4a2-b2 (d)a2 + b2

5 If --------- - --------- = a + ib, where / = and a2 + b2 = Xa - 3, the value of X is

2 + cos 0 + / sin 0
(a) 3 (b)4 (c)5 (d)6

6 If------is purely imaginary, then | z | is equal to

(b)1 (c)V2 (d)2

7 The complex numbers sin x + / cos2x and cos x -i sin2x, where / = J-\ are conjugate to each other, for

(a) x = nn,n el (b) x = 0 (c)x = |n + - Ln el (d)no value ofx

8 If a and 0 are two different complex numbers with 101 = t then

0- a is equal to
1-a0
(a)0 (c)1 (d)2
(b>5
9 If x = 3 + 4/ (where, / = 7~1). the value of x4 -12x 3 + 70x 2 - 204 x + 225, is
(a)-45 (b)0 (c) 35 (d)15

10 If | z1 -11 < 1,| z2 -21 <2,| z3 - 31 <3, the greatest value of |z, +z2 +z3 |is
(a) 6 (b)12 (c) 17 (d)23
8n I + / sin ^-(
11 The principal value of arg (z), where z = f 1 + cos — 8n where, i =1/=d)is given by
5 5
(a)-I (b)-^ (c)5 (d)^
5 5 5 0
12 If | z1|=2, | z21 = 3, | z 3| =4 and | z-i +z2 +z 3| =5, then 14z2z3 + 9z3z^ + 16z1z21 is
(a) 24 (b)60 (c) 120 (d) 240

13 If | z - i | < 5 and z} = 5 + 3/ (where, i = the greatest and least values of | iz + z are

(a) 7 and 3 (b)9and1 (c)10and0 (d) None of these

14 IfZi.Za andZ3-z4 are two pairs of conjugate complex numbers, then arg | — | + arg — equals to
z4 \Z3J
(a)0 <b)| (C)7t
Session 3
- umM-n —ir - -*-.fwr-n im < -ii ni'i ■ ■>n—rr«rm>rr- r»w «inanim >nw j.jwu—■mhhiui—■« !■■■■ iim nniiwrn—rm -[Trwii~ini ~ i ~ —— — ——— - - **"■*■ — ■. i » ——

amp(z) - amp (-z) = ± n; According as amp (z) is Positive or

Negative, Square Root of a Complex Number, Solution of
Complex Equations, De-Moivre's Theorem, Cube Roots of Unity

amp(z)-amp (-z) = ±7t, From Eq. (ii), we get

Zj = | z2 | (cos (71 + arg (z2)) + i sin (7t + arg (z2)))
According as amp (z) is Positive [from Eq. (i) and | Zj | = | z21 ]
or Negative = I Z2 | (- cos (arg z2 )-i sin (arg z2)) = - z2
[from Eq. (iii)]
Case I amp (z) is positive.
Z] + z2 = 0
If amp (z) = 0, we have
Y I Example 48.Let z and w be two non-zero complex
numbers, such that | z | = | w I and
amp (z)+amp(w) = 7t, then find the relation between
P=z
z and w.
X Sol. Given, amp (z) + amp(w) = 7t
r 0/ => amp (z) - amp (w) = 7t
-(n-0)
Here, |z|=w|=|w| .[given |z| = |w|]
r=-z
and amp (z) > 0
Then, z + w =0
amp(-z) = -( ZP' OX) = — (7t — 0)
amp (z) - amp (-z) = it
Case II amp (z) is negative.
[here, OP = OPQ
Square Root of a Complex Number
Let z = x + iy,
If amp (z) =-0
where x, y G R and i = yf-l.
We have, amp (—z) = ZP'OX = 7t-0
amp (z) - amp (~z) = - it [here, OP = OP'] Suppose y(x + iy) = a + ib
Y On squaring both sides, we get
(x + iy) =(a2 -b2) + 2iab
P = —z
7t-0 On comparing the real and imaginary parts, we get
2 <2
o -e a -b =x -(ii)
P sz and 2ab=y .■■(Hi)

az,2 -+l bf,2 = 7 (a2 -fc2)2 +4a2b2 =7 (x2 +y2)

a + b = |z| -(iv)
I Example 47.If Jz, | = |z21 and arg (z1/z2) = ti, then
From Eqs. (ii) and (iv), we get
find the value of z, + z2.
( z + x^ Z -x
Sol. v arg — = 71 a = ±^ , b=±
Uz \
2
7
2
7
arg(zj) - arg(z2) = ti -(■) 7 z - Re (z/
Zj = | zj (cos ( arg zj + i sin (arg Zj)) ■■■(ii)
or a=± l2l+Re(z)) , hb _= +±
'(---------------
and z2 = | z2 | (cos (arg z2) + i sin (arg z2)) ...(in) 2 2
\ 7
 Chap 01 Complex Numbers 19

Now, from Eq. (i), the required square roots, 2 X2

3 "l
+ + 2.A.
;f l|z|
i -, + Re(z) J z| - Re (z) £7 V2 /2
±+ x I LJ
■ ---------- + i LJ------- , if Im (z) > 0
X
N 2 V 2
7 2
z =• [ 3+i 3+z
/I| z| + Re (z) Il z I - Re (z) =±
+ J'—i-------— - i J—------ — , if Im (z) < 0 72
V 2 V 2
k 7 (ii) Let z = — 5 + 12i
Aliter |z | = 13, Re (z) = -5, Im (z) = 12>0
If + zy), where x, y e R and z = -1, then
| z | + Re (z) z -Re(z)
(i) If y is not even, then multiply and divide in y by 2, z =+ +i
2 2
then 7(x + zy) convert in k
7 /
13-5
yjx+y^-i = x +2
4
V" 2
+i
13+5^
2
x 7

v2
(ii) Factorise: - — say a, p (a < P). Aliter
4
Take that possible factor which satisfy
7(-5 + 12Q =7(~5 + 12 7~1)
x =(ocz)2+P2,ifx >0 or x =a2+(zP)2,if x <0 = 7(-5+277Z36)
(iii) Finally, write x + zy = (ocz)2 + p2 + 2zaP = 7(-5+27(“9x4))
or a2 +(zp)2 +2z‘ap = 7(-9 + 4 + 2/(-9x4))
and take their square root.
= 7(3z)2 +22 +2 • 3z - 2
± (ai + P) + (P - ia)
(iv)7(x + zy) = ■ and 7(x - zy)
or±(a + zp) or ± (a-zP) = 7(2 + 3z)2 =±(2+30
(iii) Let z=-8-15i
Remark
| z | = 17, Re (z) = -8, Im (z) =-15 <0
1. The square root of i is ± i where i = x
f f/17-8 (17+8^
V(-8-15i)=± -i
2. The square root of (- /) is | 1. 2 2
IV2; k

i Example 49. Find the square roots of the following = ± p-5f

I ^2 >
(i) 4 + 31 (ii) - 5 + 12/
(iii) —8 — 15/ (iv) 7 - 24/(where, / = 7~1) Aliter 7(-8-15i) = -8 -15 7^)
Sol. (i) Let z = 4 + 3z
-8-2
2255 -8-2
( 25
- — x-
9
/. |z| = 5,Re(z) = 4, Im(z) =3>0
l 2 2zy
>|z| + Re(z)
z -Re(z)
■x
'9 25 9 ]
2 2 ”-2 ----- x-

7(4+3z) = ±
5+4
+i =+
f3 + z I 12 2 2 2jJ

Aliter V
2 k 2 k 3j
+
_sf
2
Z2.4 5i
7(4 +3z) =^4 + 3 -1 = 4 + 2 9 J2, 72>
4
3-5r 3-5/
’-l+2 9 =±
72 > x/2
2 2 4
20 Textbook of Algebra

(iv) Letz=7-24z xz+x+r fx2 -x+1

| z | = 25, Re (z) = 7, Im (z) = - 24 < 0 2 7 k 2

z + Re (z) Jz7! - Re (z) 7 2

X 2i + X + 1
T
fx2 - x + P"1
Vz = ± -i +2 x-
2 2 2 2
x 7
25+7a 25-7
/.7(7-24i) =± -i 2 2
2 J 2 \2+x + P + i
x2-x + P
7
2 7
2 7,

IfxVx + lY . Ifx2 - X + 1^1

Aliter

7(7 - 24 0 = 7(7 ~ 24x/~T) = ^7 - 2 7(- 144) +2{—J-JHH

=77~27(16x-9)
= 716-9-27 (16 x-9)
= 7(4)2 + (3z)2 - 2 -4-3i

= 7(4 -3i)2 =±(4-30

I Example 50. Find the square root of

4
Solution of Complex Equations
x+ -X -X Putting z = x + iy, where x, y G R and z = 7~^ hi the given
4
So/. Let - x2 - 0 equation and equating the real and imaginary parts, we
x4 + x2 +1) get x and y, then required solution is z = x + iy.

••• X4 +x2 +1) I Example 51. Solve the equation z2 +1 z | = 0.

Sol. Let z = x + iy, where x, y e R and i = 7~4
= 7( x4 + 2x 2 + o=7u2+i)2
.-. | z | = (x2 + 1) z2 = (x + iy)2 = x2 - y2 + 2ixy

Re(z)= x and |z | = 7(x2 + y2)

Im(z) = 7tx4 + x2 + 1) >0 Then, given equation reduces to
( |z|-Re(z/ x2 - y2 + 2ixy + 7(x2 + y2) = 0
| z| + Re(z) )
2 2 On comparing the real and imaginary parts, we get
7
X + 7(- X4 - x2z - 1)
x2 -y2 + 7(x2 +y2) = 0 ■•(ii)
and 2xy = 0 .(iii)
f Ifx2 + 1 + x) . ffx2 + l-x^
From Eq. (iii), let x = 0 and from Eq. (ii),
-y2 +7/=0
Aliter • -jy|2+|y|=o
+ -J(-x'4 - X 2-0 = x+2
-x4 - x2 - r |y| = 0.1
4 => y = 0, ± 1
From Eq. (iii), let y = 0 and from Eq. (ii),
- (x2 + x + 1) (x2 - X + 0
x+2 x2 + [^=0
1 N 4
=> x2L +|x|=0
2
X +X+1 ■x2 -x + l^l | x |2 +1 x | = 0 => x = 0
x+2 x - ■ =>
1 2 2 7 x + iy are 0 + 0-z, 0 + i, 0 - i
i.e. z = 0, i, - i are the solutions of the given equation.
 Chap 01 Complex Numbers 21

I Example 52. Find the number of solutions of the

equation z2 + |z|2 =0. De-Moivre’s Theorem
z2 + | z ||22 = 0 or z2 + zz = 0 Statements
Sol. v
(i) If01,02,03,...,0n G R and i = ^/-l.then
z (z + z) = 0
(cos0j + isin0x) (cos02 +isin02)
z =0 ...(i) (cos03 + isin03)...(cos0„ + isin0„)
and z + z = 0 => 2Re(z) = 0
= cos(0j +02 + 03 +...+ 0n)
Re(z)=0
+ isin (0i +02 +03 + ... + 0n)
If z = x + iy [v x = Re (z)]
(ii) If 0 G R, n G I (set of integers) and i = 7~T, then
= 0 + iy, y G R
and i = ^1 -(H) (cos0 + isin0)n = cosn0 + isinn0

On combining Eqs. (i) and (ii), then we can say that the (iii) If 0 G R, n&Q (set of rational numbers)
given equation has infinite solutions. and i = 7~1, then cos n 0 + i sin n 0 is one of the values

I Example 53. Find all complex numbers satisfying of(cos0 + isin0)n.

the equation 2 |z|2 + z2 - 5+/ 73 = 0, where / = ^-1 • Proof
Sol. Let z = x + iy, where x, y 6 R and i = 7“T 19 =cos0 + isin0
(i) By Euler’s formula, e®

=> z 2 =(x + iy)2 = x2 -y2 +2ixy LHS =(cos0! + isin0i) (cos02 + isin02)

and | z | = J( x2+y2) (cos03 + isin03)...(cos0n + isin0n)

_ . p'Oj +9j+0j+--+Q1)
C C C •• • C c
Then, given equation reduces to
2 (x2 + y2) + x2 - y2 + 2ixy - 5 + i Ji = 0 -COsfOj + 02 +03 +”. + 0n)
=> (3x2 +y2 - 5) + i (2xy + a/3) = 0 = 0 + i'0 + i sin (0j + 02 +83 +... + 0n) = RHS
On comparing the real and imaginary parts, we get (ii) If = 02 = 03 = ... = 0„ = 0, then from the above
3x2 +y2 -5 = 0 ...(i) result (i), (cos0 + i sin0) (cos0 + isin0)
and 2xy + Ji = 0 •••(ii) (cos0 + i sin0)... upton factors

On substituting the value of x from Eq. (ii) in Eq. (i), we get = cos (0 + 0 + 0 + ... upton times)

3
' 2^V + y2 - 5 = 0
+ isin(0 + 0 + 0 + ... upton times)

2y' 9 i.e., (cos0 + isin0)n =cos n0 + isin n0

9 + /=S
A-
=>
4yz (iii) Let n = —, where p,qE I and q 0, from above result (ii),
<7
or 4y4 - 20y2 +9=0
(2y2-9)(2y2-l) = 0
/
we have cos — P 0 | + i sin — 0 ( wq
y 2 = 9 y 2 = 1 or y = ±,3 y = ±,11
I V?
// X / \ \
Ci Ct y Lt2
Lt y
= cos — 0 q + isin — 0 q = cos p 0 + isin p 0
or = __3_ 2___ L _L A? ' > I? J J
y Ji’Ji’ Ji’Ji r 00^
From Eq. (ii), we get => cos + i sin — is one of the values of
1 l/<7
X = ~r (cosp 0 + isin p 0)
P05)
z = x + iy => cos + i sin — is one of the values of
__________
1
V6
3i 1 T_ 3i |3_±_ |3
+ -=
Ji’ >/6T^’v2 Ji’ V2
+4 UJ
2 [(cos 0 + isin 0)p]1/9
are the solutions of the given equation.
22 Textbook of Algebra

To Find the Roots of (a + ib)p,q, where a,beR;

=> cos + i sin is one of the values of
9 p,qe I,q 0 and i =
(cos0 + isinO) P Let a + ib = r (cos0 + isin0) [polar form]
(a + ib)p /g = {r (cos (2 rm + 0)
Other Forms of De-Moivre's Theorem + i sin (2rm + 0))}p /?, n e I
1. (cos0 - / sin0)n = cosn0-/sinn0,V net
Proof (cos0 - / sin0)n = (cos (- 0) + / sin (- 0)),n
= rp,q (cos (2hti + 0) + i sin (2mt + 0))p/?
= cos (- r?0) + / sin (- n0) = cos n0 - i sin nO
/ \ / A
_ rP cos 2
— (2n7r+0) +i sin — (2rm+0)
2. (sin0 + i cos0)n = (/)” (cos n0 - i sin n 0), V n e I
7
Proof (sin0+ i cos0)n = (/ (cos0 - / sin0))n
where, n = 0,1,2,3,.... g-l
= in (cos0 - z sin0)n = (z)n (cos/?0 - / sin n 0)
[from remark (1)] L Example 56. Find all roots of x5 -1 = 0.
3. (sin0 - z cos0)n = (-/)n (cos n0 + /sin n0), V r? e /
Proof (sin0-/cos0)° =(-/(cos0+ / sin0))n Sol. x5-l = 0=>x5=l
= (-/)n (cos0 + / sin0)n ’ = (cosO + i sin0)1/5,
x = (l)|1/5
= (- i)n (cos o0 + / sin r?0)
4. (cos0 + i sin0)n * cos n0 + /' sin/70. V ne/ where i = -7-1
[here, 0 * 0.-. De-Moivre's theorem is not applicable] = [cos (2n7t + 0) + i sin (2n7t + 0)]1/5
5. -------- --------- = (cos0 + i sin 0)-1
cos0+/sin0 2n7t 2nn
= cos + i sin
= cos (-0) + /' sin(- 0) = cos0 - i sin0 5 5
where, n = 0,1,2,3, 4
7t A . . < 71 A Roots are
I Example 54. If zr =cos — +1 sin — , where
3rJ k33rr J 1, cos
2ti f 271 ] f 4tt ]
+ i sin — , cos -— + i sm
I 471
i= prove that z1 z2 z3... upto infinity = i. 5 I 5 5 I5
71 6tc (671 ) 8tc 8tt >
Sol. We have, zr = cos + i sin I 71 cos + i sin — , cos + i sin
7 <3r 5 I5J 5 5J
'n n 71 67t 67t
Zj z2 Z3...oo = COS —+—+—+ + °° Now, cos + i sin
.3 32 33 5 5
. . [ 71 7t 71 f 471 I . • L 471
+ i sin — + — + — +... + °° = cos 27T------ + i sin 271---------
<3 32 33 5 5
f n \ 71
4n 47T
3 3 71 71 = cos - i sin
= cos + i sin = cos + i sin 5 5
1-1 1--
2 2
8n 8tc
k 37 \ 3> and cos + i sin
5 5
= 0 + i-l = i
27t
(cos0 + i sin 6)4 = cos L271------
27C |1 + i . sin• | L 271
271-------
I Example 55. Express in o+ ib I 5 5
(sin6 + i cos0)5 27C 271
= cos - i sin
form, where i = -1. 5 5
Sol. v (sin0 + icos0)5 = (i)5 (cos0 - i sin0)5 (271) 2tt
Hence, roots are 1, cos — ± i sin
= i (cos0 + i sin0)“5 I 5J 5
4tt 471
(cos0 + i sin0)4 (cos0 + i sin 0)4 and cos ± i sin
5 5
(sin0 + i cos0)5 i (cos0 + i sin 0)“ 5
(cos0 + i sih0 )9 Remark
Five roots are 1, z,, z2, z]t z2 (one real, two complex and two
cos90 + i sin90 conjugate of complex roots).
= - i cos90 + sin90
= sin90 - i cos90
 Chap 01 Complex Numbers 23

[i Example .57. Find all roots of the equation

x5 - x5 + x4 - x3 + x2 - x + 1 = 0.
Properties of Cube Roots of Unity
2 J , „4___ 5 , „6
(i) 1 + (O + (02 =0 and CD3 =1
Sol. v 1- X+X - x" + x ‘ - X + X° =0
1 [1 —(—x)7] (ii) To find the value of (i)”(n >3).
= 0,1 + x*0
l-(-x) First divide n by 3. Let q be the quotient and r be the
7
or .1 + x' = 0, x - 1 or x7' = - 1 remainder. 3)n(q
x = (- 1)1/7 = (cosn + i sin 7t)1/7, i = 7-1 -3q

= [cos (2n + 1) 7t + i sin (2n + 1)ti]1/7 r

(2n + 1)ti i.e. n = 3q + r, where 0 < r < 2

(2n + 1)71
= cos + i sin
7 7 (0n=0)3<?+r=(w3)<?.(0r=(0r

for n = 0,1,2, 4,5,6. In general, (O3n = 1, (03n+1 = (0, (o3n + 2 = (o2

Remark 3, when n is a multiple of 3

(iii) 1 + (Or + (o2r
For n = 3 x = -1 but here x + - 1 0, when n is not a multiple of 3
n+ 3
(iv) Cube roots of -1 are -1, - co and - w2.
(v) a + b (0 + c to2 = 0 => a = b = c, if a, b, c G R.
Cube Roots of Unity (vi) If a, b, c are non-zero numbers such that
Let z =(1)1/3 => z3 = 1 => z3 — 1=0 a + b + c = 0 =a2 +b2 + c2, then a: b: c = 1: (0: (02.
=»(z-l)(z2+z+l)=0 => z -1 = 0 or z22+z+l = 0 (vii) A complex number a + ib (where i = 7-1), for which
-1±>^4) -l±iV3 | a: b | = 1: 73 or 73 :1 can always be expressed in
z = 1 or z =------- ----------- =-----------
2 2 terms of (0 or (D2.
l + iV3 -l-i73 -
Therefore, z = 1, -----------,-------------- , where 1 = v-1. For example,
2 2
(a) l + iV3 =-(-1-173)
If second root is represented by (O (omega), third root will
be co2. <-l-i73>
=-2 = -2(O2
Cube roots of unity are 1, (0,(02 and (0, (O2 are called
non-real complex cube roots of unity.

Remark 2^
I o
1. 0) = CD2, (CD)2 = CD 2. Vco = ± co,22, CD — ± CD
2 7
3. | CD | = | CD2 | = 1
2(0
= — = -2ico
Aliter
Let z =(1)1/3 =(cos0 + isin0)1/3,i = 7"-1 (viii) The cube roots of unity when represented on complex
plane lie on vertices of an equilateral triangle
il/3
= [cos (2 mt + 0) + i sin (2 mt + 0)] inscribed in a unit circle, having centre at origin. One
(2 mt . . I 2 nit ] , vertex being on positive real axis.
= cos +1 sm ----- L where, n = 0,1,2
3 I 3 J Y-
Therefore, roots are
CO,
2n/3
( 27t'l 271^ 471 471^ 2n/3
1, cos — + i sin — ,cos + i sin
I3J 3J / X
u 1
1 »2ni73 „ 4ni/3
or 1, e ,e
CD2 2n/3
If second root is represented by (0, then third root will be (02
or if third root is represented by (O, then second root will be co2.
24 Textbook of Algebra

Important Relations in Terms i Example 59. If a, p and y are the roots of

of Cube Root of Unity x3 - 3x2 + 3x + 7 = 0, find the value of
(i) a2 + ab + b2 = (a - few) (a - bat2) a— P -1 +-----
-1 +----- y
(ii) a2 - ab + b2 = (a + few) (a + few2) P-1 y-1 ex — 1
(iii) a3 + fe3 =(a + fe)(a + few) (a + few2) Sol. We have, x3 - 3x2 + 3x + 7 = 0
(iv) a3 -fe3 = (a-fe) (a-few) (a-few2) => (x -l)3 + 8 = 0
(v) a2 + fe2 + c2 - ab - be - ca
(x-l)3+23 =0
= (a + few + cw2) (a + few2 + cw)
(vi) a3 +fe3 +c3 -3abc => (x - 1+ 2)(x - 1+ 2co)(x - 1+ 2co2) = 0
= (a + b + c) (a + bw + cw2) (a + few2 + exo) => (x + l)(x - 1+2co)(x - 1 + 2(O2) = 0

[ Example 58. If co is a non-real complex cube root of x = - 1,1 - 2CQ 1 - 2W2

unity, find the values of the following. a = -1,P = 1 — 2(0, y -1 ~2(O2
(i) w1999 B-l y-1 “
-22 —
- 2C0
2(0 - 2(O2
(ii) w’998 Then, ----- +
-- +------- F------ =------ + ------ - +------
P-1 y-1 a-1 -— 2(0 2co - 2co 2C02 -2
3n + 2
^-1+/V3
(iii) , n e N and i = = - + - + co2 = co2 + co2 + co2 = 3wz
< 2 j co co

(iv) (1 + w) (1 + w2) (1 + wA) (1 + co8)...upto 2n factors V3 + /

I Example 60. If z = - , where / = -1, find the
a + Pco+yco2+ 8co2> 2
(v) where cc,P,y,8gR value of (z101 + /103)105.
k p + aco2 +yco+ 8co J
Sol.
73 + i 1 I' i 73
z =------- = - — [vi2=-i]
(vi) 1-(2-w)(2-w2)+2-(3-w)(3-w2)+3- 2 2
(4-co)(4-co2)++ (/i — 1) (n — co)(n — co2) — 1 + i 73
= -i = - ico
Sol. (i) co199’ = (o3 x 666 + 1 = (o < 2
1 co 101 • coJ01
z101 = (- ico)101 = - i■ 101 101 = - i co2 and i103 = i3=-i
(ii) co-"8 =-----
999
= co
W998 CO Then, z101 + i103 = - ico2 - i = - i (co2 + 1)
( x 3n + 2

(iii) = co3" - co2 = (co=3)COn 3n- co+ 22

2 = — i (— co) = ico
7
Hence, (z101 + i,03)105 =(ico)105 = i105 •coI05 = i-l = i
= (1)" ■ co2 = co2
50
(iv) (1 + co)(1 + co2)(l + co4)(l + co8)... upto2n factors 3 /V3A
! Example 61. If — +---- = 325(x-/y), where
= (1 + co) (1 + CO2)(1 + co)(1 + co2)... upto 2n factors 2 2 J
= (- C02)(- co)(- C02)(- co)... upto2n factors x,yeR and i = 7~T find the ordered pair of (x,y).
= (CO3)(co3)... upto n factors = 1-1- l>...upto n factors + i2^
Sol. v
= (l)n=l I 2 7 2 7

a + Pco+ yw2 + 8co2> co(a 4- pco + y o>2 + Seo2) = -iT3-

(V) = - i 73 co
P + aco2 + yco + Sco J (Pco + aco3 + yco2 + 8co2) 2

co(a + Pw + yco2 + 8co2) /3+£^ v° = (—i73co)50 = i50 •325 •co50

(Pco + a + yco2 + 8co2) "I2 2 }
(vi) Z(n - l)(n - co)(n - co2)= L(n 3 - 1) = Z n3 - X1
= -l-325-(O2 =-325
-i-i7T|
2
fn (n + 1) 2 7
-n
”I 2
 Chap 01 Complex Numbers 25

Then, f (- co) = 0 and f (- co2) = 0

= 325 [given]
12 2 J - 7co3 - aco + b = 0 and - 7CO6 - aco2 + b = 0
1 or -7 -aco+ 6 = 0
x = -,y = -
2 and - 7 - aco2 + b = 0
(1 On adding, we get
=> Ordered pair is •—
2 J - 14 -a(co + co2)+ 26 = 0
or -14 + a + 26 = 0 or a + 2b = 14 (i)
I Example 62. If the polynomial 7x3 + ax+b is
and on subtracting, we get
divisible by x2 -x+1, find the value of 2a+b. -a(co - co2) = 0
Sol. Let f(x) = 7x 3 + ax +6 => a=0 [•/ CO - CO2 *0]
and x2 - x +1 = (x + co)(x + co2) From Eq. (i), we get 6 = 7
V f (x) is divisible by x2 -x + 1 2a + 6 = 7

g Exercise for Session 3

1 The real part of (1-/) ', where/ = 7~1 is

(a)e-"/41 cos I - loge 21 (b) -e"*'4 sin loge 2

f1.
(c)en/4 cos I - loge 21 (d)e’n/4 sin [ — loge 21
12 )

2 The amplitude of e 6 °, where 0gR and/ = ^/-l is

(a) sinQ (b)- sinO
(cje0050 (d)esine

3 \iz'=i loge(2 -73), where/ = then the cos z is equal to

(a)/ (b)2 (C)1 (d)2
4 Ifz =/'\where/=7-~1, then | z | is equal to
(a)1 (b)e-’t/2 (c)e-" (d)e“

5 7(-8-6/) is equal to (where, / = V-1)

(a)1±3/ (b)±(1-3/) (c)±(1+3/) (d)±(3-/)

7(5+12/-)+ 7(5-12/).
6 , —7 is equal to (where, / = V- 1)
7(5 + 12/) - 7(5^-127)
(d)-|
(a)-1 (b)j<
4
I \
(c) ” 3;
-1

7 If 0 < amp (z) < rc, then amp (z) - amp (- z) is equal to
(a)0 (b)2amp(z) (c)n (d)-n

8 If |z1| = |z2|andamp(z1) + amp(z2) = 0,then

(a)z1=z2 (b)z1=z2 (c) + z2 = 0 (d)z1 = z2

9 The solution of the equation | z | - z = 1 + 2/, where / = ’-i, is

(a)2-| (b)| + 2 . 3 .
(c)i-2f (d)-2+|
2
26 Textbook of Algebra

10 The number of solutions of the equation z2 + z = 0, is

(a) 1 (b) 2
(c) 3 (d)4
ra
11 If zr= cos + i sin [ ~ , where r = 1,2,3 n and / = M, then lim zyz2z3 ...zn is equal to
In J n —°°

(a)e« (b)e"'“/2
(c)e”/2 (d)^“

n
1 + sin 0 + / cos 0
12 If 0eR and/ = 7-t then I is equal to
1 + sin 0 - / cos0
nrc - n0
(a) cos | — - nO j + / sin (— “9 (b) cos • |inn
+ n0j + /• sin ^ +. n0
12 J I. 2 J 12 J
(c) sin - noj + / cos m - nO
“oj (d) cos + 20 + / sinfnf- + 20
2 J I 12
13 If / z4 + 1 = 0, where / = then z can take the value
n + / sin f—
(a)l±i (b) cos |-|
18 J 18
V2
(C)44
(d)/

14 If co(*1)is a cube root of unity, then(1 - w + co2)(1- o>2 + w4)(1- co4 + to8)... upto2n factors, is
(a) 2" (b)22n
(c)0 (d)1
xa+yB + zy. ..
15 If a, p and y are the cube roots of p (p < 0), then for any x, y and z, -------- —------ - is equal to
x0 + yy + z a

(b)-(1+/V3),/=7^1
2
(c)J(1-i 73)./=^ (d) None of these
Session 4 v .r; v. »e«

nth Root of Unity, Vector Representation of Complex Numbers,

Geometrical Representation of Algebraic Operation on Complex Numbers,
Rotation Theorem (Coni Method), Shifting the Origin in Case of Complex
Numbers, Inverse Points, Dot and Cross Product, Use of Complex Numbers
in Coordinate Geometry

nth Root of Unity Remark

W-aLoC' = (-1) is the basic concept to be
Let x be the nth root of unity, then
understood.
x = (1)1 /n = (cos 0 + i sin 0)1/n
(c) If a is an imaginary nth root of unity, then other roots
= (cos (2 &7C + 0) + i sin (2 for + 0)1/n are given by a2 ,a3 ,a4,...,an.
[where k is an integer]
= (cos 2bt + i sin 2k7t)1/n Imaginary
axis
X = cos
f 2kn + i sin 2kn Y+
k n 7 n
i
where, k = Q, 1,2,3,...,n-l A=(a2)
r/^A,(a)
27t 2ti
Let a = cos — + i sin —, the n, nth roots of unity are V
n n ^9
ak (k = 0,1,2,3,... ,n -1) i.e, the n, nth roots of unity are X' <e ■X Real axis

l,a,a2,a3,...,an-1 which are in GP with common ratio /W

-e 2rti/n -i

(a) Sum of n, nth roots of unity Y'

_i-(i-qn) (d) v 1 + a +a2 +... + a" 1 -0

1 + a +a2 +a3 +...+a" 1
(l-a) n -1

1 - (cos 2ti + i sin 2tt) X ak =0

k=0
l-a
n -1 n -i .
2nk ] . y . [ 2nk 2ltk
= 1-<1+°)=o or S cos
k-0
-----| + i Zj sm |------ = 0
n fc = o n
l-a
n -1

Remark X cos 271 k = 0

1 + a + a2 + a,33 + + a.n "1 = 0 is the basic concept to be
k=0 n
understood. n -1
y . [ 2itk
(b) Product of n, nth roots of unity and Zj sm|-----=0
k=o n
lxaxa 2 xa x...xa"-1 =a1+2+3+-+<n-1)
xa3 x...xa
These roots are located at the vertices of a regular
(n-l)n
(n-l)n
271 . . 271 2 plane polygon of n sides inscribed in a unit circle
=a 2 cos — +1 sin — having centre at origin, one vertex being on positive
n n
real axis.
= cos (n -1) 7t + i sin (n -1) 71 (e) xn -l=(x-l)(x-a)(x-a2)...(x-an-1).
= (cos 7t + isin it)n 1 = (-l)n-1
28 Textbook of Algebra

niQ
Important Benefits e* • e 2 -2isin
1. If 1. (XpOo, a3 an_: are the noth root of unity, then
= Im ■
e10
Im
27
(1)p + (aj)p + (tto)^ + ... + (an_!)p
TT-T f0Y
k
[0. ifp is not an integral multiple of n e'e/2 • 2i sin
[n, if pis an integral multiple of n
0. ifnis even
2. (1 + a1)(1+ &>)... (1 + an_1) = fnG nG

3. (1-O|)(1-a2)...(1-an_1) = n
1, ifnisodd
= Im
sin
k •e
2 I
sin
2 ,
• sin
n+P
e
4. zn-1 = (z-1)(z+1)
(n - 2) / 2
n z2 2z cos — + 1
sin (e'
I “ sin T
<2,
2 7

r =1 n
if ‘ii is even.
(n-2) Z2 Remark
5. zn + i= n z2 -2z cos \2r +1) 7t + 11 if n is even.
r=0 n ForQ = —, we get
n
(n-3)/2
\2r + 1) 71 « 1 i 2ti 4 71 6n (2n-2)7i A _
6. Z" + 1 = (Z+1) n z2 - 2z cos + 1 1.1 +cos — + cos + cos + ... + cos 0
r =0 n kn < n n n J
if 'ri is odd. n . (27lA . i*+;i
471 . (Guy (2n-2)n
2. sin — | + sin | — + sin — + ... + sin =0
kn n I nJ n

The Sum of the Following 2 Example 63. If 1,co,(o2,...,con 1 are n,nth roots of
Series Should be Remembered unity, find the value of (9-co)(9-(o2)...(9-G)n'1)-
(i) cosG + cos20 +cos30 + ... + cosnG Sol. Let x = (l),/n => xn - 1 = 0
fnG has n roots 1, cn, cd2, ..., of “1
sin
k
fey
.cos
n+P
2
0
xn -1 = (x - l)(x - cd)(x - cd2)...(x - of “’)

sin 7 On putting x = 9 in both sides, we get

9n -1
= (9 - CD) (9 - CD2) (9 - CD3)... (9 - Of " *)
(ii) sin0 +sin20 + sin30 + ... +sinnG 9 -1

sin
nff or (9-CD)(9 - CD2)...(9-af-1) =
9n -1
2 , n+P 8
• sin e
2 Remark
sin
<2, xn -1 = (X - co) (x - CO2) ... (x - w'

Proof x -1
n -1-= lim (x - co) (x - co2)... (x - ci/’~1)
lim -x------
(i) cosG + cos20 +cos30 + ... + cosnG
x 1 x —1 * 1
= Re {e*0 + e2® +
+ee 3,e + ... +en10}, where i = 7~1 n = (1-(D)(1-CD2)...(1-(Dn-1)
=>

n0Y
P0 e n'0/2 • 2i sin ( 2tt
6 Example 64.If a = cos I — 1+ i sin
( 2 7t A
, where
e'9{(e'0)" -1} V2)
= Re I = Re( 7
e*-l e,0/2
• 2i sin (0 /2) i = V-1, find the quadratic equation whose roots are
a = a+o2 + O4 and 0 = a3 + a5 + a6.
fnGA (nQ' Sol. •.•
271 27t

= Re
sin
I2J
(0A
■ e
2 )
sin
k
f gA
• cos
I7 n + P
~2~>
0
a = cos
7
+ i sin

a7 = cos2n + isin 2it =1+0 = 1

7

sinl — sini —
V2J or a = (l)1/7

(ii) sinG +sin20 + sin30 + ... + sinnG /. 1, a, a2, a3, a4, a5, a6 are 7,7 th roots of unity.
= Im{e'e + e2'0 +
+ee 3/0 + ... + en'0}, where z = V-l 1 + a + a.2£ + a„3J + a44 + a5 + a6 = 0 ...(i)
=> (a + a:2 + a4) + (a3 + a5 + a6) = - 1 or a + 0 = - 1
 Chap 01 Complex Numbers 29

andaP = (a + a2 + a4)(a3 + a5 + a6) I Example 67.If n> 3 and 1,al,a2,a3,...)an_1 are

= a4 + a6 + a7 + a5 + a7 + as + a7 + a9 + a10 the n,nth roots of unity, then find the value of
= a4 + a6 + 1 + a5 + 1 + a + 1 + a2 + a3 [•••a7 = l] ZZ a, a,.
= (1 + a + a2 + a3 + cP + cP + a6) + 2 1<i < j <n-l
=0+2 I [from Eq. (i)] Sol. Let x = (l),/n
=2
xn=l or xn - 1 = 0
Therefore, the required equation is
1 + CLi + tt2 + CL3 + ... + 0tn _ j = 0
x2-(a+P)x + ap=0 or x2 + x + 2 = 0
or + a2+a3+... + an_! =-1
On squaring both sides, we get
I Example 65. Find the value of
10 r ' (2^ cl2 + a22 +a32 +...+ a2n-i + 2(a,a2 +a,a3
(2nk\
Z sin - / COS ------ , where i = -1. + ... + a1an_1 + a2a3+... + a2an-i
k=l < 11, I 11 J +... + a n-2an_1) = l
10 r
_ . V . I 2nk
ZTC/C 2nk orf + faj)2 + (a2)2 +(a3)2 + ... + (a„_1)2
Sol. 2j sm |----- - i cos
k=i k 11 11
io
+2 ZZ c^a j = 1 +12
1SK jSn-l
2nk 2nk
= -iZ cos + i sin
k=l 11 11 0+2 ZZ a,a; =2
1ii<jSn -1
io
= -i z
k =0
2itk , . . I2nk
cos __ + ism|-----
I 11
-1
zz
[here, p is not a multiple of n]
CLjCLj = 1
1 5 i < jn - 1
= -1(0-1) [sum of 11,11th roots of unity] Aliter
v x" -l = (x-l)(x-a1)(x-a2)...(x-a„_1)
I Example 66. If ao,oci,a2,...,an_1 are the n.nth On comparing the coefficient of x" ~2 both sides, we get
roots of the unity, then find the value of Z — 0= ZZ a, a. j +«! + a2 + + an-i
7 i=o2-aj
0Si<j£n -1

Sol. Letx = (l)I/n =>xn=l .*. xn-1=0 0= ZZ a^j -1

1< i < j < n - 1
or x” -l = (x-a0)(x-a1)(x-a2)...(x-an_1) + a„_i = 0]
[vl + aj + a2 +
n -1
ZZ a, a, = 1
= H(x-a,) Hi < jin-I
i=0

On taking logarithm both sides, we get

n -i Vector Representation of
loge(xn -1)= Z loge(x-a,)
i =0
Complex Numbers
On differentiating both sides w.r.t. x, we get If P is the point (x, y) on the argand plane corresponding
n -1 to the complex number z = x + iy, where x, y e R and
nxn-1
xn-l
=z
i =0
1
x - a, j
On putting x = 2, we get P(x,y)
1 n -1
__ 1
•(i)
2n - 1 •' =o (2-a,) iy
n -1 0 i

a'
"-1 ( 2 n >x
Now, Z = Z -1 + 0 x M
' i = o (2-OCj) i=0 2-aG
n -1 n-1
2-n-2"~1 OP = x i +y j => OP = ^x2 +y2) = | z |
-__ 1 Then,
=- Z 1+2 Z
^o(2-a,)
= -(n) + [from Eq. (i)]
i=0 i=C 2n -1 -i1 (y /x) =0
and arg (z) = direction of the vector OP = tan-
n-2n n
= - n +------- Therefore, complex number z can also be represented by OP.
2n -1 2n -1
 30 Textbook of Algebra

Geometrical Representation (c) Product

Zj = rj (cos 0! + z’sin©!) = rj e'6‘
of Algebraic Operation on Let
| Zj | = n and arg (zj =0j
Complex Numbers and z2 = r2 (cos 02 + z sin02) = r2 e
te2

[a] Sum | z2 | = r2 and arg (z2) = 02

Let the complex numbers Zj = X] + iy} =(Xj,yj)and Then, zyz2 =r1r2(cos01 + zsin©!) (cos 02 +zsin02)
z2 = x2 + iy2 = (x2> y>2) be represented by the points P and = r!r2{cos(0! +02) + zsin(0i +02)}
Q on the argand plane. | ZjZ2 1 = ^ andarg(zjz2) =0t +02
YA
YA fl(ZiZ2)
fl (Z1 + z2)
O(z2)

fl(zi) O(z2)
--------- X ^/fl(zl)
<2
0 ®1 <1
'2 01
Complete the parallelogram OPRQ. Then, the mid-points o +X
A
of PQ and OR are the same. The mid-point of
Let P and Q represent the complex numbers zx and z2,
+ *2 ?i + y2 1
PQ = respectively.
2 ’ 2 )
OP = rltOQ = r2
Hence, R = (xj +x2,yi +y2)
^POX = Gl and XQOX = Q2
Therefore, complex number z can also be represented by
----> Take a point A on the real axis OX, such that OA = 1 unit.
OR = (x, + x2) + i(yi +y2)=(xj + zyj) +(x2 + zy2) Complete the XOPA
= Zj +z2 =(x1,y1)+(x2,y2) Now, taking OQ as the base, construct a AOQR similar to
In vector notation, we have DJ? DP
AOPA, so that —= —
OQ OA
Zj + z2 =OP+OQ=OP+PR=OR
i.e. OR = OP ■ OQ = r2 [since, OA = 1 unit]
and XROX = XROQ + XQOX + 02
(b) Difference
Hence, R is the point representing product of complex
We first represent - z2 by Q', so that QQ' is bisected at O. numbers Zj and z2.
Complete the parallelogram OPRQThen, the point R
represents the difference zr -z2. Remark
1. Multiplication by/
Ya
Since, z = r (cos0 + i sin 0) and / = cos - + i sin -
Q(z2) I 2 2
n + 01 + / sin f- + 0^
iz = r cos | -
Pfr) 2 2 J.
0
X+ ------ >~X Hence, multiplication of z with /, then vector for z rotates a
right angle in the positive sense.
R^-Z2) 2. Thus, to multiply a vector by (-1) is to turn it through two
Q (-z2) right angles.
ty' 3. Thus, to multiply a vector by (cos0 + i sin0) is to turn it
through the angle 0 in the positive sense.
We see that ORPQ is a parallelogram, so that OR - QP
We have in vectorial notation, (d) Division
Z1 -z2 =OP-OQ=OP+QO Let zx =T] (cos0j +isin0!) = ri
|zi | = H andarg(z1)=01
=OP+PR=OR=QP and z2 ~r2 (cos02 +isin02) -r2 e <02
 Chap 01 Complex Numbers 31

lz2 | = r2 and arg(z2)=02

Then, we have AC = z3-Z] and AB = z2 -z.
Then = — • (cos91 +fsin9i)
[z2 £0, r2 £0]
z2 r2 (cos02 + isin02) and let arg AC = arg(z3 -z1)=0

— = —[cos(0j —02) + isin(0! -02)]

and arg AB = arg(z2 -Zj) = <|)
Z2 r2
zi ri f zi Let ZCAB=a
= — .arg — -0j ~02
z2 r2 Z.CAB = a = 0 - 0 = arg AC - arg AB
Let P and Q represent the complex numbers zx and z2, = arg(z3 -Z!)-arg(z2 -zj
respectively.
fz
2 3 -zZ1
OP = ri>OQ = r2i ZPOX =0] and ZQOX =02 = arg
kZ2 “Z1 ,
Let OS be new position of OP, take a point A on the real
axis OX, such that OA = 1 unit and through A draw a line or angle between AC and AB
making with OA an angle equal to the XOQP and meeting r affix of C - affix of A
OS in R. = arg k affix of B - affix of A J
Then, R represented by (z i /z2).
For any complex number z, we have
yt O(z2)
I («g -)
Z =| Z | e
f2. ?3 -Z1
i arg
z3 ~Z1 = z3 ~zi 22 -Z]
Similarly, e L
02^ JP(Z1) Z2 -Zj
<z2 "zl;
A ■>X
"701-02
z3 ~Z1 _IZ3 ~21 I gi(ZCAB) -ACc ia
or
R z2 -Z1 | z2 -Z1 | AB

Now, in similar AOPQ and AOAR. Remark

1. Here, only principal values of the arguments are considered.
(<Z_1 ~z_ 2 \
OA OQ r2 2. arg = 0, if AB coincides with CD, then
73 “ z4 ,
since OA = 1 and Z.AOR = Z.POR - Z.POX = 02 - 0i
arg | Zl ~ 22 | == 00 or
orn. so that
n. so that ——— is real. It follows that
Hence, the vectorial angle of R is - (02 - 0 j) i.e., 0 j - 02. lZ3-Z4J Z3"Z4
if ^—£1 is real, then the points A B.CDare collinear.
Remark z3"z4

If 0j and 02 are the principal values of zt and z2, then 0, + 02 and D

0, - 02 are not necessarily the principal value of arg (z,z2) and
arg(z,/z2). /Pfa)
S(z4) 9
A 8
Rotation Theorem [Coni Method] H(z3)

Let zt ,z2 and z3 be the affixes of three points A,B and C O(z2)
respectively taken on argand plane. C

YA 3. If AB is perpendicular to CD. then

C(z3)
/ /0(Z2) n iSo£lzA is purely imaginary.
= ±!!
arg
a Z3 " 2i > 2 z3 - z4
4. It follows that, if z, -z2 = ± k (z3 - z4), where k is purely
imaginary number, then AB and CD are perpendicular to
al
each other.
0 RQ
->-X
0
32 Textbook of Algebra

f Example 68.Complex numbers zbz2 and z3are the From Coni method, Z1 ~ z2 ft/3
G)
z3 - z2 a
vertices A, B, C respectively of an isosceles right angled
triangle with right angle at C. Show that and XBAC = -
(z1-z2)2=2(z1-z3)(z3-z2). 3
. i 2 3 “ Z1
Sol. Since, Z.ACB = 90° and AC - BC, then by Coni method From Coni method, -------- ■(ii)
z2 ~ zi a
21 ~ z3 _ ft/2
=i
z2 - z3 BC A(z0
fi(z2)
ji/4

/^Tt/4£j-
4(Zj) C(Z3) S(z2) a C(z3)
=> z,-z3= i(z2-z3) From Eqs. (i) and (ii), we get ——— = — ---- -
On squaring both sides, we get z3 — z2 z2 - Zj
(zi -z3)2 = “(z2 (Z! - z2)(z2 - Z]) = (z3 - Z])(z3 - Z2)

Zj2 +z32 -2Z]Z3=-(z22 + z32 -2z2z3) zf + z2 + z2 = zxz2 + z2 z3 + z 3z,

zi +z22 -2zjZ2 = 2(zjZ3 -ZjZ2 -z3 +z2z3)
Remark
Therefore, (zj - z2 )z = 2(z, - z3) (z3 - z2) Triangle with vertices zx, z2, z3, then
(i) (z, - z2)2 + (z2 - z3)2 + (z3 - z, )2 = 0
Aliter CA = CB = BA
■42 (ii) (z1-z2)2 = (z2-z3)(z3-zl)
B(z2)
(iii) £(Zt -z2)(z2-z3) =0 WJ_±_=o
(z,-z2)
n/4

Complex Number as a Rotating Arrow

£ in the Argand Plane
C(z3) Let z = r(cos0 + zsin0) = reidiQ ...(i)
XBAC=(n/4) be a complex number representing a point P in the argand
z2 ~ Z1 _ PA e (in/ 4) plane.
- ’ . y
z3 — zi CA Q(zeif)

or Z1~Z2 ...(i) r/
Z1 - z3
P(z)
and XCBA =(rr/4) X' X
0
1
. z3 ~ z2
or z3 ~z2 = ~i= e
_C£e(in/4) or (i n/ 4)
•••(ii)
Zi -z2 AB zi~z2 ■42
Y' ■
On dividing Eq. (i) by Eq. (ii), we get
(zj -z2)2 =2(zj -z3)(z3 -z2) Then, OP = | z | = r and XPOX = 0
Now, consider complex number zx = ze
I Example 69. Complex numbers zb z2, z3 are the or zx = reiQ • e"^ =re‘^e+^ [from Eq. (i)]
vertices of A, B, C respectively of an equilateral Clearly, the complex number zx represents a point Q in the
triangle. Show that argand plane, when OQ = r and Z.QOX = 0 + 0
Z12+Z^+Z^=Z1Z2+Z2Z3+Z3Zb
Clearly, multiplication of z with e rotates the vector OP
Sol. Let AB = BC = CA = a through angle (J) in anti-clockwise sense. Similarly,
XABC = - multiplication of z with e-1* will rotate the vector OP in
3
clockwise sense.
 I

Chap 01 Complex Numbers 33

Remark
1. If z,, z2 and z3 are the affixes of the C(z3)
Shifting the Origin in Case
three points A B and C, such that
AC = AB and /CAB - 0. Therefore,
of Complex Numbers
AB = z2-z,, AC = z3-zv Z B(Z2) Let 0 be the origin and P be a point with affix z0. Let a
--- > point Q has affix z with respect to the coordinate system
Then, AC will be obtained by rotating A(zi) passing through 0. When origin is shifted to the point P
— y
AB through an angle 0 in anti­ (z0), then the new affix Z of the point Q with respect to
clockwise sense and therefore. new origin P is given by Z = z - z0.
AC = ABe'e i.e., to shift the origin at z
z0o., we should replace z by Z + z0.
/O z3 -z}
or (z3-z1) = (z2-z1)e or —---- - = e'° yk
z2-z, Y1
2. If A B and Care three points in argand plane, such that •0
AC = AB and /CAB = 0, then use the rotation about /Ito find
e'e, but if AC * AB, then use Coni method. □_ +-X
P(zo)
I Example 70. Let zy and z2 be roots of the equation
~l
z2 + pz + q = 0, where the coefficients p and q may be 0 IM
■>x

complex numbers. Let A and B represent zy and z2 In I

the complex plane. If ZAOB = a / 0 and OA = OB,

where O is the origin, prove that p2 = 4 qcos2 (a/2). § Example 71. If z1 ,z2 and z3 are the vertices of an
equilateral triangle with z0 as its circumcentre, then
Sol. Clearly, OB is obtained by rotating OA through angle a.
changing origin toz0, show that Z}2+ Z22 + Z32 = 0, where
OB=OAefa Z, ,Z2 ,Z3 are new complex numbers of the vertices.
=> z2=z1e,a Sol. In an equilateral triangle, the circumcentre and the
z2 _ „ ia
centroid are the same point.
—=e ...(i) Zj + z2 + z3
Z1 So, ——
fl(z2)
Zj + z2 + z3 = 3z0 ...(i)
To shift the origin at z0, we have to replace zu z2, z3 and z0
A(zi) byZt + z0, Z2 + z0,Z3 + z0 andO + z0.
a Then, Eq. (i) becomes
O (Zj + zo) + (Z2 + zo)+ (^3 + zo)= 3(0 + Zq)
or —+ l = (e ia + l) => Z] + Z2 + Z3 = 0
Z1 On squaring, we get
(Z1 +z2) = e'a/2 Zj + Z2 + Z3 + 2 (ZjZ2 + Z2Z3 + Z3Zj) = 0 ...(ii)
=> ■ 2 cos (a / 2)
Z1
But triangle with vertices Zj, Z2 and Z3 is equilateral, then
On squaring both sides, we get Zj2 + Z22 + Z32 = Z,Z2 + Z2 Z3 + Z3Zj ...(iii)
(z1+z2)2=eia.(4cos2a/2)
From Eqs. (ii) and (iii), we get
Z1 3(Zj2 +Z22 +Z32) = 0
(zi +z2)2
= —-(4cos2 a/2) [from Eq. (i)] Therefore, Zj2 +z22 +Z32 =0
z? Z1

(z{ + z2)2 = 4 ztz2 cos2 (a 12)

(- p)2 = 4 q cos2 (a/2)
Inverse Points
v zt and z2 are the roots of z2 + pz + q = 0 (a) Inverse points with respect to a line Two points P
and Q are said to be the inverse points with respect to
:. Zj + z2 = - p and z1 z2 = q
the line RS. If Q is the image of P in RS, i.e., if the line
or p2 = 4 q cos2 (a/2) RS is the right bisector of PQ.
34 Textbook of Algebra

I Example 72. Show that z},z2 are the inverse points fc Example 73. Show that inverse of a point a with
with respect to the lineza + az = b, ifz}a + az2=b. respect to the circle | z - c | = R (a and c are complex
............................. R2
Sol. Let RS be the line represented by the equation, inumbers, centre c and radius R) is the point c + zz~.
za +az=b ,..(i) a— c
Let P and Q are the inverse points with respect to the line RS. Sol. Let a' be the inverse point of a with respect to the circle
The point Q is the reflection (inverse) of the point P in the | z - c | = R, then by definition,
line RS, if the line RS is the right bisector of PQ. Take any
point z in the line RS, then lines joining z to P and z to Q are
equal.
P c» a
a \z - c | = Fl
*

The points c, a, a are collinear.

R _ii S We have, arg (a' - c) = arg(a - c)
= -arg(a-c) [••• arg z = - argz]
I
I
=> arg (a' - c) + arg (a - c) = 0
Q arg {(a'-c)(a-c)} = 0
.♦. (a' - c) (a - c) is purely real and positive.
i.e., |z-z,| = | z - z21 or | =1 z - z212 By definition,| a' - c 11 a - c | = R>2‘ [vCP-Ce = r!)
i.e„ (z - zj(z - zj = (z - z2)(z - z2)
=>z(z2 -zj) + z(z2 -zJ + CziZj -z2z2) = 0 ...(ii) => | a' - c 11 a - c | = R2 [•■•1*1 = 1*11
Hence, Eqs. (i) and (ii) are identical, therefore, comparing |(a'-c)(a -c)| = R2
coefficients, we get
=> (a' - c) (a -c ) = Rz
[•.■(a' - c)(a - c) is purely real and positive]
, R2 , R2
I a - c = ——- => aa' = c + —---- -
a-c a-c
L
I

R S
A(z)
I
I
I
Dot and Cross Product
O = z2 Letzj =Xj + iyt ^(x^y^) andz2 = x2 + iy2 = (x2,y2),
where xl,yl,x2,y2 G R and i = ^/-l, be two complex
a a -b
Z2 - Z] Z2 - Zj Z]Z] — z2 z2 numbers.
If Z.POQ =0, then from Coni method,
_ az2
So that, Z2 -o _| z2 I
Zi(z2-Zj) Z2(Z2-Z])
e'9 Qfo)
-b _ Zyd f az2 - b Z1-0 |zj|

Z]Zj — z2 z2 0
22^1 _IZ2 Lie
[by ratio and proportion rule] => --- - - ----- - e P(Zl)
ZiZj |zj
z1a + az2-b = 0 or z}a+az2 = b
Z2Z) _|Z2 |
(b) Inverse points with respect to a circle If C is the e®
centre of the circle and P, Q are the inverse points
with respect to the circle, then three points C, P, Q are
Pi I2 hl
collinear and also CP ■ CQ = r2, where r is the Z2Z1 =|Z1 ||z2[e IQ

radius of the circle. Z2Zj = z i z2 (cos0 + isin0)

Re(z2zi) = zi z2 COS0
<
and Im(z2z1)=|z1 || z2 sin 9 •••(ii)
0 The dot product zx and z2 is defined by,
C P
zi-z2 =|zj 11 z2 |cos0
= Re(zj z2) = x1x2 +yjy2 [from Eq. (i)]
Exploring the Variety of Random
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on which those citizens have granted power to that legislature to
command them.
That we may intelligently so insist, and that our insistence may be
made in the proper place and at the proper time, let us briefly
consider on what subjects, in the making of our Constitution, our
predecessors, as American citizens, granted their enumerated
national powers to our only government of all Americans. Like those
predecessors, assembled in their conventions, we find all those
enumerated powers in the First Article of the Constitution proposed
from Philadelphia.
In substance they are the war power; the power of making treaties;
the power of regulating commerce between ourselves and all people
outside of America and between the citizens of the different states;
the power of taxation; and all other incidental and supplementary
powers necessary to make laws in the execution of these
enumerated and granted powers.
Noticeably absent from these enumerated powers granted to the
only general government of the citizens of America is that power,
then existing and still in the national government of each nation or
state, known (rather inaccurately) as the police power or the power
to pass any law, in restraint of individual human freedom, reasonably
designed, in the judgment of that particular legislature, to promote
the general welfare of its own citizens. It seems hardly necessary, at
this moment, to refer to the innumerable decisions of the Supreme
Court that such power was not among those enumerated and
granted to the American government by its citizens. It was solely
because such power had definitely not been granted by them to it
that the government of the American citizens made its famous
proposal that a portion of such power, in relation to one subject, be
granted to it in the supposed Eighteenth Amendment of our
Constitution.
As a matter of fact, the police power of any government is really all
its power to pass any laws which interfere with the exercise of
individual freedom. In that respect, the American people made a
marked distinction between the quantum of that kind of power which
they granted to their one general national government and the
quantum they left in the national government of the citizens of each
state. The quantum they granted to their own government was
definitely enumerated in the First Article. On the other hand, except
for the limitations which they themselves imposed upon the
respective governments of each state, they left the citizens of each
state to determine what quantum the government of that state should
have.
In other words, the police power of the American Congress is
strictly limited to the enumerated powers of that kind granted by the
citizens of America. And, although the fact does not seem to be
generally known, it is because the First Article vests in the sole
Legislature of the whole American people nothing but enumerated
powers to interfere with the freedom of the individual American that
our American government has received its universal tribute as a
government of nothing but enumerated powers over a free people,
who are its citizens.
In the Constitution are provisions in separate Articles for the
three great departments of government,—legislative,
executive, and judicial. But there is this significant difference
in the grants of powers to these departments: the First Article,
treating of legislative powers, does not make a general grant
of legislative power. It reads: “Article one, section one. All
legislative powers herein granted shall be vested in a
Congress,” etc.; and then, in Article 8, mentions and defines
the legislative powers that are granted. By reason of the fact
that there is no general grant of legislative power it has
become an accepted constitutional rule that this is a
government of enumerated powers. (Justice Brewer, in the
Supreme Court, Kansas v. Colorado, 206 U. S. 46.)
Among the national powers, which are enumerated in the First
Article, there is one which (whenever operative) approximates the
extensive police power of a state government to interfere with the
freedom of its citizens. That is the war power of the Government of
America. As the purpose of the Constitution of the American
Government is to protect the freedom of the American and as such
freedom needs effective protection from foreign attack, the
Americans of that earlier generation made the war power of their
government almost as unlimited as that of a despotic government. All
history and their own human experience had taught them that the
war power, if it was to be effective for their protection, must be
practically unlimited. If we grasp this extent of the American war
power, we realize why our sole American government, without the
grant of a new national power to it, could validly enact what we know
as the War Time Prohibition Statute, although without such a new
grant, it was powerless to enact what we know as the Volstead Act
or National Prohibition for time of peace. It is because the citizens of
each state, in their Constitution of their national government, had
given to it a general (although specifically limited) ability to interfere
with their own human freedom in most matters, that each state
government could validly make prohibition laws for its own citizens. It
is because the American citizens had not given to their government
any such general ability to interfere with their freedom, that the
American Government, for any time except that of war, could not
validly enact National Prohibition for the American people without a
new grant of a new national power directly from its own citizens. In
the days of those earlier Americans, the legal necessity of deriving
such power directly from the American citizens themselves was “felt
and acknowledged by all.” In our day, among our leaders and our
“constitutional” lawyers, there was none so humble as to know or
honor this basic legal necessity.
The other enumerated national powers, which American citizens
ever gave their national government, are few in number, although
they vested a vast and necessary ability in that government to
protect the freedom of its citizens and promote their happiness and
welfare by laws in certain matters. For our present purpose, they
need only be mentioned. They require no present explanation. They
are the power to make all treaties with foreign nations or
governments; the power to regulate commerce, except the
commerce within any one particular state; and the power of taxation.
Having now some accurate conception of the limited and specific
quantum of national power which American citizens consented to
grant in those earlier days, it is pertinent to our inquiry, as to whether
we (their posterity) have again become subjects, to dwell briefly
upon the reluctance with which they made even those grants. In
considering that attitude, it is essential always to keep in mind the
status of the citizens of each state, at that time, and their relation to
their own national government and the relation of each state to the
federal government of all the states. Under the existing system of
governments, the citizens of each state were subject to no valid
interference whatever with their own individual freedom except by
laws of a legislature, every member of which they themselves
elected and to which they themselves granted every power of such
interference which that legislature could validly exercise. To those
free men in those free states, men educated in the knowledge of
what is real republican self-government, these two facts meant the
utmost security of their human rights. No government or
governments in the world, except their own one state government
could interfere at all directly with those rights, and they had given to,
and they could take from, that government any power of that kind. As
for the respective states and the relation of each to the federal
government of all, each state had an equal voice in the giving to or
taking from that government any federal power and each had an
equal voice, in the federal legislature, in exercising each valid federal
power. These existing facts, respectively of vast importance to the
citizens of each state and to its government, influenced, more than
any other facts, the framing of the new Articles, particularly the First
Article, at Philadelphia and the opposition to those Articles in the
conventions in which the people of America assembled.
The First Article, as we know it, starts with the explicit statement
that all national powers, which are granted by Americans in that
Constitution, are granted to the only American legislature, Congress.
It then provides how the members of each of the two bodies in that
legislature shall be elected. It then enumerates the granted powers,
confining them to specific subjects of interference with the human
freedom of the American citizen. It then, for the particular security of
that human freedom, imposes specific restraints upon that legislature
even in the exercise of its granted national powers. Finally, it
prohibits the further exercise of specific powers by any state
government.
No American, who reads the debates of the Philadelphia
Convention of 1787, can fail to realize that the grant of any national
power,—power to interfere with human freedom—is the constitution
of government. The First Article was the subject of almost all the
discussion of those four months at Philadelphia. Seemingly invincible
differences of desire and opinion, as to who should elect and the
proportion (for citizens of the new nation and for states of the
continuing federation) in which there should be elected the members
of the legislature which was to exercise the granted national powers,
almost ended the effort of that Convention. This was in the early part
of July. For exhausting days patriotic men had struggled to reconcile
the conflict of desire and opinion in that respect. One element,
mainly from the larger states, insisted that the members (from each
state) of both branches of the new legislature should be proportioned
to the number of Americans in that state. The other element, mainly
from the smaller states, insisted that the Americans in each state
should have an equal representation in each branch of the new
legislature. Each element was further divided as to who should
choose the members of that legislature. Some held that the people
should choose every member. Others held that the state legislatures
should choose every member. Still others held that each state
should, by its legislature, choose the members of one branch, so that
those members might speak for that state, and that the American
people themselves, divided into districts, should choose the
members of the other branch, so that those members might speak
for the general citizens of America.
Mason of Virginia, later one of the great opponents of the adoption
of all the Articles, insisted that election by the people was “the only
security for the rights of the people.” (5 Ell. Deb. 223.)
Madison “considered an election of one branch, at least of the
legislature by the people immediately, as a clear principle of free
government.” (5 Ell. Deb. 161.)
 Wilson of Pennsylvania “wished for vigor in the government, but
he wished that vigorous authority to flow immediately from the
legitimate source of all authority.” (5 Ell. Deb. 160.) Later he said, “If
we are to establish a national government, that government ought to
flow from the people at large. If one branch of it should be chosen by
the legislatures, and the other by the people, the two branches will
rest on different foundations, and dissensions will naturally arise
between them.” (5 Ell. Deb. 167.)
Dickenson of Delaware “considered it essential that one branch of
the legislature should be drawn immediately from the people, and
expedient that the other should be chosen by the legislatures of the
states.” (5 Ell. Deb. 163.)
Gerry of Massachusetts, consistent Tory in his mental attitude
toward the relation of government to people, insisted that “the
commercial and moneyed interest would be more secure in the
hands of the state legislatures than of the people at large. The
former have more sense of character, and will be restrained by that
from injustice.” (5 Ell. Deb. 169.)
On June 25, Wilson, at some length, opposed the election of
senators by the state legislatures. He stated that: “He was opposed
to an election by state legislatures. In explaining his reasons, it was
necessary to observe the two-fold relation in which the people would
stand—first, as citizens of the general government; and, secondly, as
citizens of their particular state. The general government was meant
for them in the first capacity; the state governments in the second.
Both governments were derived from the people; both meant for the
people; both therefore ought to be regulated on the same
principles.... The general government is not an assemblage of
states, but of individuals, for certain political purposes. It is not meant
for the states, but for the individuals composing them; the
individuals, therefore, not the states, ought to be represented in it.”
(5 Ell. Deb. 239.)
There came a day, early in that memorable July, when all hope of
continuing the Convention was almost abandoned, by reason of the
difference of desire and opinion on this one subject. Let us average
Americans of this generation remember that this one subject was
merely the decision whether the people were to choose all the
members of the legislature which was to exercise granted national
powers to interfere with the human freedom of the citizens of
America. Happily for all of us, there were many patriotic as well as
able leaders at Philadelphia. From their patriotism and ability they
evolved the compromise, on that question, which is expressed in
their First Article. When it came from Philadelphia, it provided that
each state should have equal representation in the Senate, senators
to be chosen by the state legislatures, and that the House of
Representatives should consist of members chosen directly by the
citizens of America, in districts proportioned to the number of those
citizens in it.
No one has read the recorded debates of the Convention which
proposed and the conventions which adopted our Constitution
without learning that the Americans in those conventions knew that
the grant of enumerated national powers in the First Article was the
constitution of the American government of men. In and out of the
Philadelphia Convention, the greatest and most persistent attack
upon its proposal was the insistent claim that it had acted wholly
without authority in proposing an Article which purported to grant any
such national power to interfere with the human freedom of all
Americans. Since July 4, 1776, no legislature or legislatures in the
world had possessed any national powers over all Americans. The
Americans in each existing nation elected every member of the one
legislature which had any such power over them. It was felt and
stated at Philadelphia, it was felt and urged and insisted upon,
sometimes with decency and reason, sometimes with bitterness and
rancor and hatred, between the closing day at Philadelphia and the
assembling of various Americans in each state, that the Americans
in each state would be unwilling to give any such national power
over themselves to any legislature whose members were not all
elected by the people in that state. In all the conventions which
adopted the Constitution, the one great object of attack was the
grant even of enumerated powers of a national kind to a legislature
whose members would not all be chosen by the Americans in the
state in which the convention was held. The record of the Virginia
convention fills one entire volume of Elliot’s Debates. Almost one-
half of the pages of that volume are claimed by the eloquent attacks
of Patrick Henry upon those grants of enumerated powers in that
First Article. The basis of all his argument was the fact that this grant
of national power in the First Article would make him and all his
fellow Virginians, for the first time since the Declaration of
Independence, citizens of a nation—not Virginia—who must obey
the laws of a legislature only some of whose members Virginians
would elect.
“Suppose,” he says, “the people of Virginia should wish to alter”
this new government which governs them. “Can a majority of them
do it? No; because they are connected with other men, or, in other
words, consolidated with other states. When the people of Virginia,
at a future day, shall wish to alter their government, though they
should be unanimous in this desire, yet they may be prevented
therefrom by a despicable minority at the extremity of the United
States. The founders of your own Constitution made your
government changeable: but the power of changing it is gone from
you. Whither is it gone? It is placed in the same hands that hold the
rights of twelve other states; and those who hold those rights have
right and power to keep them. It is not the particular government of
Virginia: one of the leading features of that government is, that a
majority can alter it, when necessary for the public good. This
government is not a Virginian, but an American government.” (3 Ell.
Deb. 55.)
How forceful and effective was this objection, we average
Americans of this generation may well realize when we know that the
Constitution was ratified in Virginia by the scant majority of ten votes.
In New York and Massachusetts and other states, the adoption was
secured by similar small majorities. In North Carolina, the first
convention refused to adopt at all.
Furthermore, it is recorded history that, in Massachusetts, in
Virginia, in New York, and elsewhere, the vote of the people would
have been against the adoption of the Constitution, if a promise had
not been made to them by the advocates of the Constitution. It was
the historic promise that Congress, under the mode of procedure
prescribed in Article V, would propose new declaratory Articles,
suggested by the various conventions and specifically securing
certain reserved rights and powers of all Americans from all ability of
government to interfere therewith. This historic promise was fulfilled,
when the first Congress of the new nation proposed the suggested
declaratory Articles and ten of them were adopted. These are the
Articles now known as the first ten Amendments. It has been settled
beyond dispute, in the Supreme Court, that every one of the
declarations in these ten Articles was already in the Constitution
when it was originally adopted by the citizens of America.
The most important declaration in those amazingly important ten
declarations, which secured the adoption of our Constitution, is the
plain statement that every national power to interfere with the human
freedom of Americans, not granted in Article I, was reserved to the
American people themselves in their capacity as the citizens of
America. That is the explicit statement of what we know as the Tenth
Amendment. In itself, that statement was but the plain and accurate
echo of what was stated by the American people (who made the
enumerated grants of such powers in Article I) in the conventions
where they made those grants. Their statement was nowhere more
accurately expressed, in that respect, than in the resolution of the
Virginia Convention, which ratified the Constitution. That resolution
began, “Whereas the powers granted under the proposed
constitution are the gift of the people, and every power not
granted thereby remains with them, and at their will, etc.” (3 Ell.
Deb. 653.)
After the same statement had been expressly made (with
authoritative effect as part of the original Constitution) in that Article
which we know as the Tenth Amendment, it was again and again
echoed, in the plainest language, from the Bench of the Supreme
Court.
As far back as 1795, in the case of Vanhorne’s Lessee vs.
Dorrance, 2 Dall. 304, Justice Patterson stated that the Constitution
of England is at the mercy of Parliament, but “in America, the case is
widely different.”... A Constitution “is the form of government,
delineated by the mighty hand of the people, in which certain first
principles of fundamental laws are established. The Constitution is
certain and fixed; it contains the permanent will of the people, and is
the supreme law of the land; it is paramount to the power of the
legislature, and can be revoked or altered only by the authority that
made it. The life-giving principle and the death-dealing stroke must
proceed from the same hand.... The Constitution fixes limits to the
exercise of legislative authority, and prescribes the orbit within which
it must move.... Whatever may be the case in other countries, yet in
this there can be no doubt, that every act of the legislature,
repugnant to the Constitution, is absolutely void.”
To us average Americans, who have lived with those earlier
Americans through the days in which they constituted their nation
and distributed all granted national powers between governments in
America and reserved all other general American national powers
exclusively to themselves, the Virginia Resolution, the Tenth
Amendment, and the quoted language of the Circuit Court are in
strict conformity with the education we have received.
What, however, are we to think of the Tory education of so many of
our leaders and “constitutional” lawyers, who have calmly accepted
and acted upon the amazing assumption that state governments in
America can exercise and can grant to other governments any or all
general national powers to interfere with the human freedom of
American citizens, including even the national powers expressly
reserved by those citizens to themselves in the Tenth Amendment?
If they adopt their familiar mental attitude that all these statements
were made more than a hundred years ago and have no meaning or
weight now, we refer them to the Supreme Court, in 1907, when it
stated:
The powers the people have given to the General
Government are named in the Constitution, and all not there
named, either expressly or by implication, are reserved to the
people and can be exercised only by them, or upon
further grant from them. (Justice Brewer in Turner v.
Williams, 194, U. S. 279.)
 For ourselves, we average Americans turn now to examine in
detail how clearly the Americans at Philadelphia in 1787 did know
and obey the basic law of America that all national powers to
interfere with individual freedom are the powers of the people
themselves and can be exercised only by them or upon direct grant
from them. We find their knowledge, in that respect, evidenced by an
examination of the reasoning by which they reached the correct legal
conclusion that their proposed grants of general national powers, in
their First Article, could only be made by the citizens of America
themselves, assembled in their “conventions”—that grants of such
powers could not be made even by all the legislatures of the then
independent states.
 CHAPTER VII
PEOPLE OR GOVERNMENT?—CONVENTIONS
OR LEGISLATURES?

It is no longer open to question that by the Constitution a

nation was brought into being, and that that instrument was
not merely operative to establish a closer union or league of
states. (Justice Brewer, in Supreme Court, Kansas v.
Colorado, 206 U. S. 46 at page 80.)
Instructed by living through the education of the earlier Americans
to their making of that Constitution, we accurately know that they
themselves, by their own direct action, brought that new nation into
being. Through our course in their education, we have their
knowledge that only the men, who are to be its first members, can
create a new political society of men, which is exactly what any
American nation is. “Individuals entering into society must give up a
share of liberty to preserve the rest.” So said the letter which went
from Philadelphia with the proposed Articles whose later adoption
created the new nation and vested the delegated and enumerated
national powers of its government to interfere with the liberty of its
citizens, (1 Ell. Deb. 17.)
Furthermore, through our own personal experience, we
understand how all societies of men are brought into being. There
are few of us who have not participated in the creation of at least one
society of men. Most of us have personally participated in the
creation of many such societies. For which reason, we are quite well
acquainted with the manner in which all societies of men are brought
into being. We know that ourselves, the prospective members of the
proposed society, assemble and organize it and become its first
members and constitute the powers of its government to command
us, its members, for the achievement of the purpose for which we
create it.
For one simple reason, the Americans, through whose education
we have just lived, were “better acquainted with the science of
government than any other people in the world.” That reason was
their accurate knowledge that a free nation, like any other society of
individuals, can be created only in the same manner and by its
prospective members and that the gift of any national powers to its
government can only be by direct grant from its human members.
This is the surrender “of a share of their liberty, to preserve the rest.”
The knowledge of those Americans is now our knowledge. For
which reason, we know that they themselves created that new nation
and immediately became its citizens and, as such, gave to its
government all the valid and enumerated national powers of that
government to interfere with their and our human freedom. We know
that they did all these things, by their own direct action, “in the only
manner, in which they can act safely, effectively or wisely, on such a
subject, by assembling in conventions.”
Thus, whatever may have been the lack of knowledge on the part
of our leaders and “constitutional” lawyers for the last five years, we
ourselves know, with knowledge that is a certainty, that the ratifying
conventions of 1787 and 1788 were the American people
themselves or the citizens of the new nation, America, assembled in
their respective states.
Our Supreme Court has always had the same knowledge and
acted upon it.
The Constitution of the United States was ordained and
established, not by the states in their sovereign capacities
[the respective peoples or citizens of each State] but
emphatically, as the preamble of the Constitution declares, by
“the people of the United States” [namely the one people of
America].... It was competent to the people to invest the
general government with all the powers which they might
deem proper and necessary; to extend or restrain these
powers according to their own good pleasure, and to give
them a paramount and supreme authority.... The people had a
right to prohibit to the states the exercise of any powers which
were, in their judgment, incompatible with the objects of the
general compact [between the citizens or members of the
new nation], to make the powers of the state governments, in
given cases, subordinate to those of the nation, or to reserve
to themselves those sovereign authorities which they might
not choose to delegate to either. (Supreme Court, Martin v.
Hunter’s Lessee, 1 Wheat. 304, at p. 324.)
Instructed by experience, the American people, in the
conventions of their respective states, adopted the present
Constitution.... The people made the Constitution and the
people can unmake it. It is the creature of their will, and lives
only by their will. But this supreme and irresistible power to
make or to unmake resides only in the whole body of the
people, not in any subdivisions of them. (Marshall, in
Supreme Court, Cohens v. Virginia, 6 Wheat. 264.)
The Constitution was ordained and established by the
people of the United States for themselves, for their own
government, and not for the government of the individual
states. Each state established a constitution for itself, and in
that constitution provided such limitations and restrictions on
the powers of its particular government as its judgment
dictated. The people of the United States framed such a
government for the United States as they supposed best
adapted to their situation, and best calculated to promote their
interests. The powers they conferred on this government were
to be exercised by itself; and the limitations on power, if
expressed in general terms, are naturally, and, we think,
necessarily, applicable to the government created by the
instrument. They are limitations of power granted in the
instrument itself; not of distinct governments, framed by
different persons and for different purposes. (Marshall, in
Supreme Court, Barron v. Mayor of Baltimore, 7 Peters, 243.)
When the American people created a national legislature,
with certain enumerated powers, it was neither necessary nor
 proper to define the powers retained by the states. These
powers proceed, not from the people of America, but from the
people of the several states; and remain, after the adoption of
the Constitution, what they were before, except so far as they
may be abridged by that instrument. (Marshall, in the
Supreme Court, Sturges v. Crowninshield, 4 Wheat. 122.)
We average Americans know and will remember the clear
distinction, the substantial distinction, recognized by the great jurist,
between “the people of America” and “the people of the several
states,” although they happen to be the same human beings acting
in different capacities, as members of different political societies of
men. It is a matter of constant mention in the Supreme Court that we
ourselves, in addition to our capacity as human beings, have two
other distinct capacities, that of citizen of America and that of citizen
of our respective state; that, as citizens of America, we alone validly
give to its government any power to command us, and, as citizens of
our particular state, we alone validly give to its government all its
national power to command us. The decisions of the Supreme Court,
in that respect, are mentioned elsewhere herein. Meanwhile, we
average Americans understand these matters perfectly and will not
forget them. We are quite accustomed, while retaining our status as
free human beings, to be members of many different societies of
men and, as the members of some particular society, to give to its
government certain powers to interfere with our freedom.
We have in our political system a government of the United
States and a government of each of the several states. Each
one of these governments is distinct from the others, and
each has citizens of its own who owe it allegiance, and whose
rights, within its jurisdiction, it must protect. The same person
may be at the same time a citizen of the United States and a
citizen of a state, but his rights of citizenship under one of
these governments will be different from those he has under
the other.... Experience made the fact known to the people of
the United States that they required a national government for
national purposes.... For this reason, the people of the United
States ... ordained and established the government of the
 United States, and defined its powers by a Constitution, which
they adopted as its fundamental law, and made its rules of
action. The government thus established and defined is to
some extent a government of the states in their political
capacity. It is also, for certain purposes, a government of the
people. Its powers are limited in number, but not in degree.
Within the scope of its powers, as enumerated and defined, it
is supreme and above the states; but beyond, it has no
existence. It was erected for special purposes and endowed
with all the powers necessary for its own preservation and the
accomplishment of the ends its people had in view.... The
people of the United States resident within any state are
subject to two governments, one state, and the other national;
but there need be no conflict between the two. Powers which
one possesses, the other does not. They are established for
different purposes, and have separate jurisdictions. Together
they make one whole, and furnish the people of the United
States with a complete government, ample for the protection
of all their rights at home and abroad. (Justice Waite, in
Supreme Court, United States v. Cruikshank, 92 U. S. 542.)
It must seem remarkable to us average Americans, with the
education we have acquired at this point, to realize that our leaders
and “constitutional” lawyers have not known why only we ourselves,
in our capacity as citizens of America, can give any new national
power to interfere with our freedom and that we, for such new giving,
must act, in the only way in which the citizens of America “can act
safely, effectively, or wisely, on such a subject, by assembling in
convention,” in our respective states, the very “conventions”
mentioned for valid grant of such national power in the Fifth Article of
the Constitution made by the citizens of America, so assembled in
such “conventions.” Before dwelling briefly upon the accurate
appreciation of that legal fact displayed by those first citizens in
everything connected with the making of that Constitution and that
Fifth Article, let us realize how well the leaders and great
constitutional lawyers of other American generations between that
day and our own did know this settled legal fact.
 After the Americans in nine states had created the new nation and
had become its citizens and had (in that capacity) granted the
national powers of its First Article, the Americans in Virginia
assembled to determine whether they also would become citizens of
the new nation. As the president of the convention, in which they
assembled, they chose Edmund Pendleton, then Chancellor of
Virginia.
Very early in the debates, Henry and Mason, great opponents of
the Constitution, attacked it on the ground that its Preamble showed
that it was to be made by the people of America and not by the
states, each of which was then an independent people. Henry and
Mason wanted those peoples to remain independent. They wanted
no new nation but a continuance of a mere union of independent
nations. They knew that a constitution of government ordained and
established by the one people of America, assembled in their
respective “conventions,” as the Preamble of this Constitution
showed it to be, created an American nation and made the ratifying
Americans, in each state, the citizens of that new nation. For this
reason, the opening thunder of Henry’s eloquence was on that
Preamble. “My political curiosity, exclusive of my anxious solicitude
for the public welfare, leads me to ask, Who authorized them to
speak the language of We, the people, instead of, We, the states?
States are the characteristics and the soul of a confederation. If the
states be not the agents of this compact, it must be one great,
consolidated, national government, of the people of all the states.”
(Henry, 3 Ell. Deb. 22.)
The learned Pendleton, sound in his knowledge of basic American
law and quick to grasp the plain meaning of the Fifth Article of the
new Constitution, quickly answered Henry. “Where is the cause of
alarm? We, the people, possessing all power, form a government,
such as we think will secure happiness; and suppose, in adopting
this plan, we should be mistaken in the end; where is the cause of
alarm on that quarter? In the same plan we point out an easy and
quiet method of reforming what may be found amiss. No, but say
gentlemen, we have put the introduction of that method in the hands
of our servants, who will interrupt it for motives of self-interest. What
then?... Who shall dare to resist the people? No, we will assemble in
convention; wholly recall our delegated powers or reform them so as
to prevent such abuse; and punish those servants who have
perverted powers, designed for our happiness, to their own
emolument.... But an objection is made to the form; the expression,
We, the people, is thought improper. Permit me to ask the gentlemen
who made this objection, who but the people can delegate
powers? Who but the people have the right to form government?...
What have the state governments to do with it?” (3 Ell. Deb. 37.)
We average Americans know and will remember that this learned
American lawyer, only twelve years earlier a subject of an
omnipotent legislature, already knew the basic American principle to
be that the delegation of national power was the constitution of
government of a free people and that only the people, assembled in
convention, could delegate such power and that the state
governments, under basic American law, never can have the ability
to delegate that kind of power. We regret that our “constitutional”
lawyers, all born free citizens of a free republic, have not the same
accurate knowledge of basic American law.
But the knowledge of Henry and of Pendleton, that the document
under consideration was the Constitution of a nation whose citizens
alone could give to its government any valid power to interfere with
their human freedom, was the knowledge of all in that and the other
“conventions,” in which the one people of America assembled and
adopted that Constitution. Let us note another distinct type in that
Virginia convention, the famous Light-horse Harry Lee of the
Revolution. “Descended from one of the oldest and most honorable
families in the colony, a graduate of Princeton College, one of the
most daring, picturesque, and attractive officers of the Revolution, in
which by sheer gallantry and military genius he had become
commander of a famous cavalry command, the gallant Lee was a
perfect contrast to the venerable Pendleton.” (Beveridge, Life of
Marshall, Vol. I, page 387.) Lee also replied to Henry’s attack on the
expression “We, the people” and not “We, the states.” In his reply,
there was shown the same accurate knowledge of basic American
law. “This expression was introduced into that paper with great
propriety. This system is submitted to the people for their
consideration, because on them it is to operate, if adopted. It is not
binding on the people until it becomes their act.” (3 Ell. Deb. 42.)
In the Massachusetts convention, General William Heath, another
soldier of the Revolution, showed his accurate conception of the
legal fact of which we average Americans have just been reading in
the decisions of our Supreme Court. “Mr. President, I consider
myself not as an inhabitant of Massachusetts, but as a citizen of the
United States.” (2 Ell. Deb. 12.)
In the North Carolina convention, William Goudy seems to have
had some prophetic vision of our own immediate day. Speaking of
the document under discussion and clearly having in mind its First
Article, this is the warning he gave us: “Its intent is a concession of
power, on the part of the people, to their rulers. We know that private
interest governs mankind generally. Power belongs originally to the
people; but if rulers [all governments] be not well guarded, that
power may be usurped from them. People ought to be cautious in
giving away power.... Power is generally taken from the people by
imposing on their understanding, or by fetters.” (4 Ell. Deb. 10.)
In that same North Carolina convention, James Iredell, later a
distinguished judge of our Supreme Court, in replying to the common
attack that the Constitution contained no Bill of Rights, displayed
clearly the general accurate knowledge that, in America, any grant of
national power to interfere with human freedom is the constitution of
government and that the citizens of any nation in America are not
citizens but subjects, if even a single power of that kind is exercised
by government without its grant directly from the citizens themselves,
assembled in their conventions. “Of what use, therefore, can a Bill of
Rights be in this Constitution, where the people expressly declare
how much power they do give, and consequently retain all that they
do not? It is a declaration of particular powers by the people to their
representatives, for particular purposes. It may be considered as a
great power of attorney, under which no power can be exercised but
what is expressly given.” (4 Ell. Deb. 148.)
 When we average Americans read the debates of those human
beings, the first citizens of America, one thing steadily amazes us, as
we contrast it with all that we have heard during the past five years.
Some of those first citizens were distinguished lawyers or statesmen,
quite well known to history. Some of them bore names, then
distinguished but now forgotten. Most of them, even at that time,
were quite unknown outside of the immediate districts whence they
came. All of them, twelve years earlier, had been “subjects” in an
empire whose fundamental law was and is that its legislative
government can exercise any power whatever to interfere with
human freedom and can delegate any such power to other
governments in that empire. The object of the American Revolution
was to change that fundamental law, embodying the Tory concept of
the proper relation of government to human being, into the basic law
of America, embodying the American concept of that relation
declared in the great Statute of ’76, that no government can have
any power of that kind except by direct grant from its own citizens.
During that Revolution, human beings in America, in conformity with
their respective beliefs in the Tory or the American concept of the
relation of human being to government, had been divided into what
history knows as the Tories and the Americans. Many of the human
beings, assembled in those conventions of ten or twelve years later,
had been sincere Tories in the days of the Revolution.
Yet, if we average Americans pick up any volume of their recorded
debates in those “conventions,” we cannot scan a few pages
anywhere without finding the clearest recognition, in the minds of all,
that the American concept had become the basic American law, that
the Tory concept had disappeared forever from America. All of them
knew that, so long as the Statute of ’76 is not repealed and the result
of the Revolution not reversed, no legislatures in America can
exercise any power to interfere with human freedom, except powers
obtained by direct grant from the human beings over whom they are
to be exercised, and that no legislatures can give to themselves or to
another legislature any such power. It was common in those
“conventions” of long ago to illustrate some argument by reference to
this admitted legal fact and the difference between the fundamental
law of Great Britain and of America, in these respects. In that North
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