Number Systems PDF

version: 1.1 CHAPTER 1 Number Systems Animation 1.1: Complex Plane Source & Credit: elearn.punjab 1. Quadratic Equations eLearn.Punjab 1. Quadratic Eq...

version: 1.1 CHAPTER 1 Number Systems Animation 1.1: Complex Plane Source & Credit: elearn.punjab 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab 1.1 Introduction 1.2 Rational Numbers and Irrational Numbers In the very beginning, human life was simple. An early ancient herdsman compared p sheep (or cattle) of his herd with a pile of stones when the herd left for grazing and again on We know that a rational number is a number which can be put in the form where p, q its return for missing animals. In the earliest systems probably the vertical strokes or bars such as I, II, III, llll etc.. were used for the numbers 1, 2, 3, 4 etc. The symbol “lllll” was used by qUZ / q ≠ 0. The numbers 16 , 3.7, 4 etc., are rational numbers. 16 can be reduced to the many people including the ancient Egyptians for the number of ingers of one hand. p 4 Around 5000 B.C, the Egyptians had a number system based on 10. The symbol form where p, qUZ, and q ≠ 0 because 16 = 4 =. q 1 for 10 and for 100 were used by them. A symbol was repeated as many times as it was needed. For example, the numbers 13 and 324 were symbolized as and p respectively. The symbol was interpreted as 100 + 100 +100+10+10+1+1+1 Irrational numbers are those numbers which cannot be put into the form where q +1. Diferent people invented their own symbols for numbers. But these systems of notations 7 5 proved to be inadequate with advancement of societies and were discarded. Ultimately the p, qUZ and q ≠ 0. The numbers 2, 3, , are irrational numbers. 5 16 set {1, 2, 3, 4,...} with base 10 was adopted as the counting set (also called the set of natural numbers). The solution of the equation x + 2 = 2 was not possible in the set of natural numbers, So the natural number system was extended to the set of whole numbers. No 1.2.1 Decimal Representation of Rational and Irrational Numbers number in the set of whole numbers W could satisfy the equation x + 4 = 2 or x + a = b , if a > b, and a, b, UW. The negative integers -1, -2, -3,... were introduced to form the set of 1) Terminating decimals: A decimal which has only a inite number of digits in its decimal integers Z = {0, ±1, ±2 ,...). part, is called a terminating decimal. Thus 202.04, 0.0000415, 100000.41237895 are examples Again the equation of the type 2x = 3 or bx = a where a,b,UZ of terminating decimals. and b ≠ 0 had no solution in the set Z, so the numbers of the form Since a terminating decimal can be converted into a common fraction, so every terminating decimal represents a rational number. a where a,b,UZ and b ≠ 0, were invented to remove such diiculties. The set b 2) Recurring Decimals: This is another type of rational numbers. In general, a recurring or a Q = { I a,b,UZ / b ≠ 0} was named as the set of rational numbers. Still the solution of equations periodic decimal is a decimal in which one or more digits repeat indeinitely. b It will be shown (in the chapter on sequences and series) that a recurring decimal can such as x2 = 2 or x2 = a (where a is not a perfect square) was not possible in the set Q. So the be converted into a common fraction. So every recurring decimal represents a rational irrational numbers of the type ± 2 or ± a where a is not a perfect square were introduced. number: This process of enlargement of the number system ultimately led to the set of real numbers A non-terminating, non-recurring decimal is a decimal which neither terminates nor _ = Q~Q’ (Q’ is the set of irrational numbers) which is used most frequently in everyday life. it is recurring. It is not possible to convert such a decimal into a common fraction. Thus a non-terminating, non-recurring decimal represents an irrational number. version: 1.1 version: 1.1 2 3 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab Example 1: Squaring both sides we get;.25 ( = 25 p2 i) ) is a rational number. 2 = 2 or p2 = 2q2 (1) 100 q.333...( = ) is a recurring decimal, it is a rational number. 1 ii) The R.H.S. of this equation has a factor 2. Its L.H.S. must have the same factor. 3 Now a prime number can be a factor of a square only if it occurs at least twice in the iii) 2.3(= 2.333...) is a rational number. square. Therefore, p2 should be of the form 4p‘2 0.142857142857... ( = so that equation (1) takes the form: 1 iv) ) is a rational number. 4p‘2 = 2q2....(2) 7 i.e., 2p’2 = q2....(3) v) 0.01001000100001... is a non-terminating, non-periodic decimal, so it is an In the last equation, 2 is a factor of the L.H.S. Therefore, q2 should be of the form 4q’2 so irrational number. that equation 3 takes the form vi) 214.121122111222 1111 2222... is also an irrational number. 2p‘2 = 4q’2 i.e., p’2 = 2q’2....(4) vii) 1.4142135... is an irrational number. viii) 7.3205080... is an irrational number. From equations (1) and (2), 3.141592654... is an important irrational number called it p(Pi) which ix) 1.709975947... is an irrational number. p = 2p‘ x) and from equations (3) and (4) denotes the constant ratio of the circumference of any circle to the length q = 2q‘ p 2 p′ of its diameter i.e., ∴ = p= circumference of any circle q 2 q′ length of its diameter. An approximate value of p is ,a better approximation is 22 355 p and a still better This contradicts the hypothesis that is in its lowest form. Hence 2 is irrational. 7 113 q approximation is 3.14159. The value of p correct to 5 lac decimal places has been Example 3: Prove 3 is an irrational number. determined with the help of computer. Solution: Suppose, if possible 3 is rational so that it can be written in the form p/q when Example 2: Prove 2 is an irrational number. p,q U Z and q ≠ 0. Suppose further that p/q is in its lowest form, then 3 = p/q , (q ≠ 0) Solution: Suppose, if possible, 2 is rational so that it can be written in the form p/q where p,q U Z and q ≠ 0. Suppose further that p/q is in its lowest form. Squaring this equation we get; p2 3= or p2 = 3q2........(1) Then 2 = p/q, (q ≠ 0) q2 version: 1.1 version: 1.1 4 5 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab The R.H.S. of this equation has a factor 3. Its L.H.S. must have the same factor. same Universal set. Now a prime number can be a factor of a square only if it occurs at least twice in _ usually denotes the set of real numbers. We assume that two binary operations the square. Therefore, p2 should be of the form 9p‘2 so that equation (1) takes the form: addition (+) and multiplication (. or x) are deined in _. Following are the properties or laws 9p‘2 = 3q2 (2) for real numbers. ’2 2 i.e., 3p = q (3) 1. Addition Laws: - In the last equation, 3 is a factor of the L.H.S. Therefore, q2 i) Closure Law of Addition should be of the form 9q’2 so that equation (3) takes the form [ a, bU_, a + bU_ ([stands for “for all” ) 3p‘2 = 9q2 i.e., p’2 = 3q’2 (4) ii) Associative Law of Addition From equations (1) and (2), [ a, b,cU_, a + (b + c) = (a + b ) + c P = 3P’ iii) Additive Identity and from equations (3) and (4) [ aU_, \ 0U_ such that a+0=0+a=a q = 3q’ (\ stands for “there exists”). p 3 p′ 0(read as zero) is called the identity element of addition. ∴ = q 3q′ iv) Additive Inverse [ aU_, \ (- a)U_ such that a + ( - a ) = 0 = ( - a) + a p This contradicts the hypothesis that is in its lowest form. q Hence 3 is irrational. v) Commutative Law for Addition [ a, bU_, a + b = b + a Note: Using the same method we can prove the irrationality of 2. Multiplication Laws 5, 7,...., n where n is any prime number. vi) Closure I.aw of Multiplication [ a, bU_, a. bU_ (a,b is usually written as ab). vii) Associative Law for Multiplication 1.3 Properties of Real Numbers [ a, b, cU_, a(bc) = (ab)c viii) Multiplicative Identity We are already familiar with the set of real numbers and most of their properties. We [ aU_, \ 1U_ such that a.1 = 1.a = a now state them in a uniied and systematic manner. Before stating them we give a prelimi- 1 is called the multiplicative identity of real numbers. nary deinition. ix) Multiplicative Inverse Binary Operation: A binary operation may be deined as a function from A % A into A, but for the present discussion, the following deinition would serve the purpose. A binary oper- [ a(≠ 0)U_, \ a-1U_ such that a.a-1 = a-1.a = 1 (a-1 is also written as 1 ). ation in a set A is a rule usually denoted by * that assigns to any pair of elements of A, taken a in a deinite order, another element of A. x) Commutative Law of multiplication Two important binary operations are addition and multiplication in the set of real num- [ a, bU_, ab = ba bers. Similarly, union and intersection are binary operations on sets which are subsets of the version: 1.1 version: 1.1 6 7 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab 3. Multiplication - Addition Law c) [ a,b,c,dU_ and a,b,c,d are all positive. xi) [ a, b, c U _, i) a > b / c > d ⇒ ac > bd. ii) a < b / c < d ⇒ ac < bd a (b + c) = ab + ac (Distrihutivity of multiplication over addition). (a + b)c = ac + bc Note That: In addition to the above properties _ possesses the following properties. i) Order Properties (described below). 1. Any set possessing all the above 11 properties is called a ield. ii) Completeness axiom which will be explained in higher classes. 2. From the multiplicative properties of inequality we conclude that: - If both the sides The above properties characterizes _ i.e., only _ possesses all these properties. of an inequality are multiplied by a +ve number, its direction does not change, but Before stating the order axioms we state the properties of equality of numbers. multiplication of the two sides by -ve number reverses the direction of the inequality. 4. Properties of Equality 3. a and (-a) are additive inverses of each other. Since by deinition inverse of -a is a, Equality of numbers denoted by “=“ possesses the following properties:- ∴ - (-a) = a i) Relexive property [ aU_, a = a 4. The left hand member of the above equation should be ii) Symmetric Property [ a,bU_, a = b ⇒ b = a. read as negative of ‘negative a’ and not ‘minus minus a’. iii) Transitive Property [ a,b,cU_, a = b/b = c ⇒ a = c [ a,b,cU_, a = b ⇒ a + c = b + c 1 iv) Additive Property 5. a and are the multiplicative inverses of each other. Since by v) Multiplicative Property [ a,b,cU_, a = b ⇒ ac = bc / ca = cb. a a-1 1 vi) Cancellation Property w.r.t. addition deinition inverse of is a (i.e., inverse of is a), a≠0 [ a,b,cU_, a + c = b + c ⇒ a = b a ∴ vii) Cancellation Property w.r.t. Multiplication: (a-1)-1 = a or [ a,b,cU_, ac = bc ⇒ a = b, c ≠ 0 1 =a 1 5. Properties of Ineualities (Order properties ) a 1) Trichotomy Property [ a,bU_ Example 4: Prove that for any real numbers a, b i) a.0 = 0 ii) ab = 0 ⇒ a = 0 0 b = 0 [ 0 stands for “or” ] either a = b or a > b or a< b 2) Transitive Property [ a,b,cU_ a > b / b > c ⇒ a > c ii) a b / c > d ⇒a+c > b+d = a.1-a.1 (Distributive Law) =a-a b) i) ii) a < b ⇒ a + c < b + c a < b / c < d ⇒ a+c < b+d (Property of multiplicative identity) ii) = a + (-a) (Def. of subtraction) 4) Multiplicative Properties: =0 (Property of additive inverse) a) [ a,b,cU_ and c >0 a > b ⇒ ac > bc a < b ⇒ ac < bc. Thus a.0=0. i) ii) ii) Given that ab = 0 (1) b) [ a,b,cU_ and c < 0. a > b ⇒ ac < bc a < b ⇒ ac > bc Suppose a ≠ 0, then exists i) ii) version: 1.1 version: 1.1 8 9 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab = ,(k ≠ 0) 1 1 iv) a ka (Golden rule of fractions) (1) gives: (ab) =.0 (Multiplicative property of equality) a a b kb ⇒ (.a)b =.0 1 1 a (Assoc. law of %) a a b ad v) = (Rule for quotient of fractions). The symbol ⇔ stands for if i.e.. if and only if. ⇒ 1.b = 0 c bc (Property of multiplicative inverse). d ⇒b=0 (Property of multiplicative identity). Thus if ab = 0 and a ≠ 0, then b = 0 Similarly it may be shown that Solution: = (= bd ) ( bd ) if ab = 0 and b ≠ 0, then a = 0. Hence ab = 0 ⇒ a = 0 or b = 0. ⇒ a c a c i) ( bd ) = ( bd ) b d b d ⇒ a.1 c.1 b d ⇒ a.(.b).d = Example 5: For real numbers a,b show the following by stating the properties used. (-a) b = a (-b) = -ab 1 1 i) ii) (-a) (-b) = ab c.(.bd ) b d = c(bd. ) 1 ⇒ ad = Solution: i) (-a)(b) + ab = (-a + a)b (Distributive law) d ∴ ad = cb ∴ (-a)b + ab = 0 = 0.b = 0. (Property of additive inverse) bc ad =bc ⇒ ( ad ) ×. = 1 1 1 1 Again b.c.. ∴ (-a)b = -(ab) = -ab (Q -(ab) is written as -ab) i.e.. (-a)b and ab are additive inverse of each other. b d b d ⇒ a..d = 1 1 1 1 (-a) (-b) -ab = (-a)(-b) + (-ab) b.. c. b d b d ⇒ =. ii) a c = (-a)(-b) + (-a)(b) (By (i)) b d = (-a)(-b + b) (Distributive law) ( ab )..= (a. ).(b = = 1 = (-a).0 = 0. (Property of additive inverse) 1 1 1 1 (-a)(-b) = ab ii) ) 1.1 a b a b Example 6: Prove that 1 1 Thus ab and. are the multiplicative inverse of each other. But multiplicative inverse = ⇔ ad =bc a c a b i) (Principle for equality of fractions 1 b d of ab is ab =∴ 1 1 1 ii). = 1 1 1 a b ab.. ab a b. =. = (a. ).(c. ) a c ac iii) (Rule for product of fractions). a c 1 1 b d bd iii) b d b d version: 1.1 version: 1.1 10 11 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab = ( ac ) (. ) 2. Name the properties used in the following equations. 1 1 (Letters, where used, represent real numbers). (Using commutative and associative laws of multiplication) b d (a + 1) + = a + (1 + ) 3 3 i) 4+9=9+4 ii) = ac = 1 ac 4 4.. ( 3 + 5) + 7 = 3 + ( 5 + 7) iv) bd bd = = a c ac iii) 100 + 0 = 100. b d bd v) 1000 % 1 = 1000 vi) 4.1 + ( - 4.1 ) = 0 = a a = = a k ak vii) a-a=0 viii) a(b - c) = ab - ac x) (x - y)z = xz - yz iv).1. b b b k ak ix) ∴ =. xii) a(b + c - d) = ab + ac - ad. a ak xi) 4%(5 % 8) = (4 % 5) % 8 b bk 3. Name the properties used in the following inequalities: a a 1 v) = b= = (bd ) ad (.b) -3 < -2 ⇒ 0 < 1 -5 < - 4 ⇒ 20 > 16 b b ad. i) ii) 1 > -1 ⇒ -3 > -5 a < 0 ⇒ -a > 0 c c 1 bc (bd ) cb(.d ) d d d iii) iv) Example 7: Does the set {1, -1 } possess closure property with respect to v) a>b ⇒ < 1 1 vi) a > b ⇒ -a < -b i) addition ii) multiplication? a b i) 1 + 1 = 2, 1 + (-1) = 0 = -1 + 1 4. Prove the following rules of addition: - a b a+b a c ad + bc Solution: -1 + (-1) = -2 i) + = ii) + = c c c b d bd But 2, 0, -2 do not belong to the given set. That is, all the sums do not belong to the 7 5 -21 - 10 given set. So it does not possess closure property w.r.t. addition. 5. Prove that - - = ii) 1.1= 1, 1. (-1) = -1, (-1).1 = -1, (-1). (-1) = 1 12 18 36 Since all the products belong to the given set, it is closed w.r.t multiplication. 6. Simplify by justifying each step: - Exercise 1.1 + 4 + 16 x 1 1 4 5 - 1. Which of the following sets have closure property w.r.t. addition and multiplication? i) ii) 4 1 1 (0 , - 1 ) { 1, -1 } 4 5 i) {0} ii) {1} iii) iv) version: 1.1 version: 1.1 12 13 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab Thus 2i, -3i, 5i , - i are all imaginary numbers, i which may be 11 + - a c 1 1 2 written 1.i is also an imaginary number. b d a b - 1-. iii) iv) Powers of i : i2 = -1 (by deination) a c 1 1 i3 = i2.i = -1.i = -i b d a b 1.4 Complex Numbers i4 = i2 % i2 = (-1)(-1) = 1 Thus any power of i must be equal to 1, i,-1 or -i. For instance, i13 = (i2)6.2 = (-1)6.i = i i6 = (i2)3 = (-1)3 = -1 etc. The history of mathematics shows that man has been developing and enlarging his concept of number according to the saying that “Necessity is the mother of invention”. In the remote past they stared with the set of counting numbers and invented, by stages, the negative numbers, rational numbers, irrational numbers. Since square of a positive as well 1.4.1 Operations on Complex Numbers as negative number is a positive number, the square root of a negative number does not exist in the realm of real numbers. Therefore, square With a view to develop algebra of complex numbers, we state a few deinitions. The symbols a,b,c,d,k, where used, represent real numbers. 1) a + bi = c + di ⇒ a = c b = d. roots of negative numbers were given no attention for centuries together. However, recently, properties of numbers involving square roots of negative numbers have also been discussed in detail and such numbers have been found useful and have been applied in many branches 2) Addition: (a + bi) + (c + di) = (a + c) + (b + d)i 3) k(a + bi) = ka + kbi 4) (a + bi) - (c + di) = (a + bi) + [-(c + di)] o f pure and applied mathematics. The numbers of the = a + bi + (-c - di) form x + iy, where x, y U_ , and i = ,are called complex numbers, here x is called real = (a - c) + (b - d)i part and y is called imaginary part of the complex number. For example, 3 + 4i, 2 - 5) (a + bi).(c + di) = ac + adi + bci + bdi = (ac - bd) + (ad + bc)i. 6) Conjugate Complex Numbers: Complex numbers of the form (a + bi) and (a - bi) which i etc. are complex numbers. have the same real parts and whose imaginary parts difer in sign only, are called conjugates of each other. Thus 5 + 4i and 5 - 4i, -2 + 3i and - 2 - 3i,- Note: Every real number is a complex number with 0 as its imaginary part. 5 i and 5 i are three pairs of Let us start with considering the equation. conjugate numbers. x2 + 1 = 0 (1) ⇒ x2 = -1 Note: A real number is self-conjugate. ⇒ x=± -1 -1 does not belong to the set of real numbers. We, therefore, for convenience call it 1.4.2 Complex Numbers as Ordered Pairs of Real Numbers imaginary number and denote it by i (read as iota). The product of a real number and i is also an imaginary number We can deine complex numbers also by using ordered pairs. Let C be the set of ordered pairs belonging to _ % _ which are subject to the following properties: - version: 1.1 version: 1.1 14 15 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab i) (a, b) = (c, d ) ⇔ a = c ∧ b = d.  a -b  ( a, b)  2 2  a +b a +b  ii) (a, b) + (c, d) = (a + c, b + d) 2 , 2 = (1, 0), the identity element  a -b  iii) If k is any real number, then k(a, b) = (ka, kb) iv) (a, b) (c, d) = (ac - bd, ad + bc) = 2 2  Then C is called the set of complex numbers. It is easy to sec that (a, b) - (c, d) a +b a +b  2 , 2 ( a, b) = (a - c, b - d) v) (a, b)[(c, d ) ± (e, = f )] (a, b)(c, d ) ± (a, b)(e, f ) Properties (1), (2) and (4) respectively deine equality, sum and product of two complex Note: The set C of complex numbers does not satisfy the order axioms. In fact there is no numbers. Property (3) deines the product of a real number and a complex number. sense in saying that one complex number is greater or less than another. Example 1: Find the sum, diference and product of the complex numbers (8, 9) and (5, -6) 1.4.4 A Special Subset of C Solution: Sum = (8 + 5, 9 - 6) = (13, 3) = (8 - 5, 9 - (-6)) = (3, 15) We consider a subset of C whose elements are of the form (a, 0) i.e., second component Diference = (8.5 - (9)(-6), 9.5 + (-6) 8) of each element is zero. Product = (40 + 54, 45 - 48) Let (a, 0), (c, 0) be two elements of this subset. Then = (94, -3) i) (a, 0) + (c, 0) = (a + c, 0) ii) k(a, 0) = (ka, 0) iii) (a, 0) % (c, 0) = (ac ,0) 1  Multiplicative inverse of (a, 0) is  , 0  , a ≠ 0. 1.4.3 Properties of the Fundamental Operations on Complex a  iv) Numbers Notice that the results are the same as we should have obtained if we had operated on It can be easily veriied that the set C satisies all the ield axioms i.e., it possesses the the real numbers a and c ignoring the second component of each ordered pair i.e., 0 which properties 1(i to v), 2(vi to x) and 3(xi) of Art. 1.3. has played no part in the above calculations. By way of explanation of some points we observe as follows:- On account of this special feature wc identify the complex number (a, 0) with the real i) The additive identity in C is (0, 0). number a i.e., we postulate: ii) Every complex number (a, b) has the additive inverse (-a, -b) i.e., (a, b) + (-a, -b) = (0, 0). (a, 0)= a (1) Now consider (0, 1) iii) The multiplicative identity is (1, 0) i.e., (a, b).(1, 0) = (a.1 - b.0, b.1 + a.0) = (a, b). (0, 1). (0, 1) = (-1, 0) = -1 (by (1) above). = (1, 0) (a, b) If we set (0, 1) = i (2) then (0, 1)2 = (0,1)(0,1) = i.i = i2 = -1 iv) Every non-zero complex number {i.e., number not equal to (0, 0)} has a multiplicative inverse. We are now in a position to write every complex number given as an ordered pair, in  a -b  The multiplicative inverse of (a, b) is  2 2  terms of i. For example a +b a +b  2 , 2 (a, b) = (a, 0) + (0, b) (def. of addition) version: 1.1 version: 1.1 16 17 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab = a(1, 0)+ b(0, 1) (by ( 1) and (2) above) 16. Separate into real and imaginary parts (write as a simple complex number): - 2 - 7i (-2 + 3i ) 2 = a.1 + bi i 4 + 5i (1 + i ) 1+ i = a + ib i) ii) iii) Thus (a, b) = a + ib where i2 = -1 This result enables us to convert any Complex number given in one notation into the other. 1.5 The Real Line Exercise 1.2 In Fig.(1), let X ′ X be a line. We represent the number 0 by a point O (called the origin) 1. Verify the addition properties of complex numbers.  2. Verify the multiplication properties of the complex numbers. 3. Verify the distributive law of complex numbers. of the line. Let |OA| represents a unit length. According to this unit, positive numbers are represented on this line by points to the right of O and negative numbers by points to the left of O. It is easy to visualize that all +ve and -ve rational numbers are represented on this (a, b)[(c, d) + (e, f)] = (a, b)(c, d) + (a, b)(e, f) (Hint: Simplify each side separately) 4. Simplify’ the following: line. What about the irrational numbers? The fact is that all the irrational numbers are also represented by points of the line. (- 21 Therefore, we postulate: - Postulate: A (1 - 1) correspondence can be established between the points of a line l and i) i9 ii) i14 iii) (-i)19 iv) ) 2 5. Write in terms of i -16 the real numbers in such a way that:- -1b -5 1 -4 i) ii) iii) iv) i) The number 0 corresponds to a point O of the line. 25 ii) The number 1 corresponds to a point A of the line. iii) If x1, x2 are the numbers corresponding to two points P1, P2, then the distance between P1 and P2 will be |x1 - x2|. Simplify the following: 6. (7, 9 ) + (3, -5) 7. (8, -5 ) - (-7, 4) 8. (2, 6)(3, 7) 9. (5, -4) (-3, -2) 10. (0, 3) (0, 5) 11. (2, 6)'(3, 7). It is evident that the above correspondence will be such that corresponding to any real number there will be one and only one point on the line and vice versa.  (2,6) 2 + 6i 3 - 7i  When a (1 - 1) correspondence between the points of a line x′x and the real numbers (5, -4) '(-3, -8)  Hint for 11: = × etc.   (3,7) 3 + 7i 3 - 7i  12. has been established in the manner described above, the line is called the real line and the real number, say x, corresponding to any point P of the line is called the coordinate of the 13. Prove that the sum as well as the product of any two conjugate point. complex numbers is a real number. ( ) 14. Find the multiplicative inverse of each of the following numbers: 1.5.1 The Real Plane or The Coordinate Plane i) (-4, 7) ii) 2, - 5 iii) (1, 0) 15. Factorize the following: We know that the cartesian product of two non-empty sets A and B, denoted by A % B, i) a2 + 4b2 ii) 9a2 + 16b2 iii) 3x2 +3y2 is the set: A % B = {(x, y) I xUA / yUB} version: 1.1 version: 1.1 18 19 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab The members of a cartesian product are ordered the coordinate plane and every point of the plane will represent one and only one complex pairs. number. The components of the complex number will be the coordinates of the point The cartesian product _%_ where _ is the set of real representing it. In this representation the x-axis is called the real axis and the y-axis is called numbers is called the cartesian plane. the imaginary axis. The coordinate plane itself is called the complex plane or z - plane. By taking two perpendicular lines x′ox and y′oy as By way of illustration a number of complex numbers have been shown in igure 3. coordinate axes on a geometrical plane and choosing a convenient unit of distance, elements of _%_ can be represented on the plane in such a way that there is a (1-1) correspondence between the elements of _%_ and points of the plane. The igure representing one or more complex numbers on the complex plane is called an The geometrical plane on which coordinate system has been speciied is called the Argand diagram. Points on the x-axis represent real plane or the coordinate plane. real numbers whereas the points on the y-axis Ordinarily we do not distinguish between the Cartesian plane _%_ and the coordinate represent imaginary numbers. plane whose points correspond to or represent the elements of _%_. If a point A of the coordinate plane corresponds to the ordered pair (a, b) then a, b are called the coordinates of A. a is called the x - coordinate or abscissa and b is called the y - coordinate or ordinate. In the igure shown above, the coordinates of the pointsB, C, D and E are (3, 2), (-4, 3), In ig (4), x, y are the coordinates of a point. (-3, -4) and (5, -4) respectively. It represents the complex number x + iy. Corresponding to every ordered pair (a, b) U_%_ there is one and only one point in The real number x 2 + y 2 is called the modulus the plane and corresponding to every point in the plane there is one and only one ordered of the complex number a + ib. In the igure MA ⊥ o x pair (a, b) in _%_.  There is thus a (1 - 1) correspondence between _%_ and the plane. ∴ OM= x, MA= y In the right-angled triangle OMA, we have, 1.6 Geometrical Representation of Complex Numbers The by Pythagoras theorem, = OM + MA Complex Plane 2 2 2 OA ∴ OA = x2 + y 2 We have seen that there is a (1-1) correspondence between the elements (ordered pairs) of the Cartesian plane _%_ and the complex numbers. Therefore, there is a (1- 1) correspondence between the points of the coordinate plane and the complex numbers. We can, therefore, represent complex numbers by points of the coordinate plane. In this Thus OA represents the modulus of x + iy. In other words: The modulus of a complex representation every complex number will be represented by one and only one point of number is the distance from the origin of the point representing the number. version: 1.1 version: 1.1 20 21 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab - z =-a - ib, z =a - ib and - z =-a + ib The modulus of a complex number is generally denoted as: |x + iy| or |(x, y)|. For convenience, So, a complex number is denoted by z. ∴ - z= (-a ) 2 + (-b) 2= a 2 + b2 If z = x + iy = (x, y), then = x +y (1) = a 2 + b2 2 2 z Example 1: Find moduli of the following complex numbers : z (2) (i) 1 - i 3 (ii) 3 (iii) -5i (iv) 3 + 4i z = (a ) 2 + (b) 2 = a 2 + b 2 (3) -z = (-a ) 2 + (b) 2 = a 2 + b2 Solution: (4) Let z = 1 - i 3 or z =1 + i ( - 3 ) i) ii) Let z = 3 By equations (1), (2), (3) and (4) we conclude that - z =z =z =- z or z = 3 + 0.i (ii) Let z = a + ib ∴= z (1) + (- 3) 2 2 ∴= z (3) + (0)= 3 2 2 So that z= a - ib = 1+ 3 = 2 Let z = -5i Taking conjugate again of both sides, we have iii) iv) Let z = 3 + 4i or z = 0 + (-5)i ∴= (3) 2 + (4) 2 z =a + ib =z z = a + ib so that z= a - ib z (iii) Let ∴= 02 + (-5)= ∴ z.z = (a + ib)(a - ib) =a 2 - iab + iab - i 2b 2 2 z 5 = a 2 - (-1)b 2 Theorems: [z, z1, z2 U C, = a 2 + b2 = z 2 (iv) Let z1 = a + ib and z2 = c + id, then i) -z = z = z = -z ii) z=z z1 + z2 = (a + ib) + (c + id) = (a + c) + i(b + d) iii) z z = z z1 + z2 = z1 + z2 z1 + z2 = (a + c) + i (b + d ) (Taking conjugate on both sides) 2 iv) so, = (a + c) - i(b + d) = (a - ib) + (c - id) = z1 + z 2 z  v) =  , z2 ≠ 0 z1.z2 = z1. z2 Let z1 = a + ib and z2 = c + id, where z2 ≠ 0, then z1 (v)  z2  1 vi) z1 a + ib z2 = Proof :(i): Let z= a + ib, z2 c + id version: 1.1 version: 1.1 22 23 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab a + ib c - id (vii) Algebraic proof of this part is tedius. Therefore, we prove it geometrically. = × c + id c - id (Note this step) (ac + bd ) + i (bc - ad ) ac + bd bc - ad = = + c2 + d 2 c2 + d 2 c2 + d 2 i  z  ac + bd bc - ad ∴ 1 = +  z2  c + d c2 + d 2 2 2 i ac + bd bc - ad = 2 -i 2 c +d c + d2 2 (1) z1 a + ib a - ib Now = = z 2 c + id c - id a - ib c + id = × c - id c + id (ac + bd ) - i (bc - ad ) = In the igure point A represents z = a + ib and point C represents z = c + id. We complete c2 + d 2 1 2 the parallelogram OABC. From the igure, it is evident that coordinates of B are (a + c, b + d), ac + bd bc - ad = 2 -i 2 c +d c + d2 (2) therefore, B represents z + z = (a + c) + (b + d)i and OB= z1 + z2. 2 From (1) and (2), we have 1 2  z1  z1 Also OA = z1 , = OC = z2.  = AB  z2  z 2 In the 3OAB; OA + AB > OB (OA = mOA etc.) (vi) Let z = a + ib and z = c + id, then z1.z2 =(a + ib)(c + id ) 1 2 ∴ |z |+ |z | > |z + z | = (ac - bd ) + (ad + bc)i (1) 1 2 1 2 = (ac - bd ) 2 + (ad + bc) 2 Also in the same triangle, OA - AB < OB = a 2c 2 + b 2 d 2 + a 2 d 2 + b 2c 2 ∴ |z |- |z | < |z + z | = (a + b )(c + d ) (2) 2 2 2 2 1 2 1 2 = z1. z2 Combining (1) and (2), we have This result may be stated thus: - |z |- |z | < |z + z | < |z |+ |z | (3) The modulus of the product of two complex numbers is equal to the product of their 1 2 1 2 1 2 moduli. version: 1.1 version: 1.1 24 25 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab + + - - = = z2 3 - 2i 3 - 2i (2 - 3) + (-6 - 1)i -1 - 7i = = 3 - 2i 3 - 2i (-1 - 7i )(3 + 2i ) = which gives the required results with inequality signs. (3 - 2i ) (3 + 2i ) (-3 + 14) + (-2 - 21)i = = - 11 23 32 + 22 i Results with equality signs will hold when the 13 13 points A and C representing z and z Example 3: Show that, " z1 , z2 ∈ C , z1 z2 = 1 2 become collinear with B. This will be so when = a c z1 z2 b d Let z+ 1 = += (see ig (6)). Solution: a bi , z2 c di z1zz1 2z2== (a(a++bibi)()(cc++didi) )==(ac--bd ad++bc In such a case z1 + z2 = OB + OA (ac bd)()(ad bc)i)i ==(ac (ac--bd bd) )--(ad (ad++bc bc)i)i (1) = OB + BC z1.zz1.2z= (a(a+= 2 = ) += +bibi )= += +didi )+) = OC (c(c == (a(a-= - bi ) --)(bi +-)bc - cdi (cad di))i = z1 + z2 (abi )( = = (ac(-acbd-)bd + )(-+ad - bc-)bc (-ad i )i Thus z1 + z2 = z1 + z2 z1.z2 = +(a bi )+(c di ) = -(a bi )-(c di ) The second part of result (vii) namely = (ac - bd ) + (-ad - bc)i (2) z1 + z2 ≤ z1 + z2 Thus from (1) and (2) we have, z1 z2 = z1 z2 is analogue of the triangular inequality*. In words, it may be stated thus: - Polar form of a Complex number: Consider adjoining diagram The modulus of the sum of two complex numbers is less than or equal to the sum of the representing the complex number z = x + iy. From the diagram, we moduli of the numbers. see that x = rcosq and y = rsinq where r = |z| and q is called argument of z. Example 2: If z = 2+ i, z = 3 - 2i, z = 1 + 3i then express 1 3 in the z z Hence x + iy = rcosq + rsinq....(i) 1 2 3 z2 where= and q = tan x x 2 + y 2 and form a + ib - -1 y r (Conjugate of a complex number z is denoted as z ) Equation (i) is called the polar form of the complex number z. z1 z3 (2 + i ) (1 + 3i ) (2 - i ) (1 - 3i ) Solution: = = *In any triangle the sum of the lengths of any two sides is greater than the length of the third z2 3 - 2i 3 - 2i side and diference of the lengths of any two sides is less than the length of the third side. (2 - 3) + (-6 - 1)i -1 - 7i = = - - version: 1.1 version: 1.1 - - + 26 = - + 27 1. Quadratic Equations eLearn.Punjab 1. Quadratic Equations eLearn.Punjab 1. Number Systems eLearn.Punjab 1. Number Systems eLearn.Punjab Example 4: Express the complex number 1 + i 3 in polar form. ii) Let x1 + iy1 = r1 cosq1 + r1 sinnq1 and x2 + iy2 = r2 cosq2 + r2 sinnq2 then,  x1 + iy1   r1 cos q1 + r1 i sinq1  r1n (cosq1 + i sinq1 ) n