Mat 1120 Notes on Sets 2021-22 PDF

1.1 Sets Operations A set is a collection of well defined objects called elements.The elements of a set must be distinct and distinguishable in the sense that, given any object, one should be able to determine whether the object belongs to the set or does not belong to the set under consideration...

1.1 Sets Operations A set is a collection of well defined objects called elements.The elements of a set must be distinct and distinguishable in the sense that, given any object, one should be able to determine whether the object belongs to the set or does not belong to the set under consideration. Notation We shall use capital letters for names of sets and small letters for elements of a set. The most common notation for representing a set is the use of braces and listing the elements of the set between the braces. For example A = {1, 2, 3, 4} or B = {book, pen, pencil, ruler}. However, this notation is limited in its application. If for example one wants to write the set of all natural numbers using set notation, one may not be able to list all the natural numbers and put them in one set. In this case, we can give the set by describing the property or attribute that characterizes the elements of the set. We can thus write A = {x : x is a natural number}. This is read as ‘all elements x such that x is a natural number. The advantage of this notation is that it is more general and it can be used even in the case where we can list the elements. Example1.1.1 Let A = {2, 3, 4, 5, 6} then A = {n : n is a natural number and 1 < n < 7} If A is a set and x is an element, we shall write x ∈ A to mean ‘x belongs to A’ and x ∉ A to mean ‘x does not belong to A’. Example 1.1.2 Let A = {2, 4, 6, 8, 10} then 4 ∈ A but 5 ∉ A. Example1.1.3 Let S = {1, 2, 3, 4, ….., 10} and B = {k ∈ S : k = 3n + 1, n = 0, 1, 2, 3}. We can list the elements of B from the description given. B = {1, 4, 7, 10}. Consider now the two sets: A = {x : x is an odd number} B = {x : x is a natural number} Then all the elements of A are also contained in B. When this happens we say that A is a subset of B written A ⊂ B. If A and B are two sets, we shall say that ‘A is a subset of B’ written A ⊂ B or B ⊃ A if whenever x ∈ A then x ∈ B. That is, every element belonging to A also belongs to B. 5 Two sets A and B are equal (A = B) if they contain same elements. That is, A = B if both A ⊂ B and B ⊂ A hold. The empty set or null set, denoted by,∅ is the set having no elements in it. The null set is a subset of every other set. The universal set is a set which contains all the elements under discussion. The universal set will be denoted by the letter U or in some cases E and it contains every other set. If A and B are sets, then A – B is the set of all elements which belong to A but do not belong to B. That is; A – B = {x ⏐ x ∈ A and x ∉ B} Example1.1.4 Let A = {2, 3, 4, 5, 6, 7, 8}and B = {5, 6, 7, 8, 9, 10, d, f, g} then B − A = {9, 10, d , f , g} If U is the universal set and A is a set, then the complement of A denoted by Aʹ = U – A is a set which contains all elements in the universal set that are not in the set A. Example 1.1.5 Let U = {x1 , x2 , x3 , x4 , x5 , x6 , x7 } be the universal set and let A = {x2 , x3 , x5 }. Then Aʹ = {x1 , x4 , x6 , x7 } A venn diagram is a pictorial representation of a set. If for example U is the universal set containing the set A as a subset, then this relation can be represented in a venn diagram as below: Set Operations Let A and B be two sets, then the intersection of two sets A and B, denoted by A∩B is the set which contains elements which are common to both set A and set B. That is A∩B = {x ⏐ x ∈ A and x ∈ B}. 6 Two sets are disjoint if their intersection is a null set. That is, A and B are disjoint implies that A∩B = ∅. The union of two sets A and B, denoted by A∪B, is the set of all the elements which either belong to the set A or belong to the set B or belong to both sets A and B. That is A∪B = {x⏐ x ∈ A or x ∈ B} A∪B Example 1.1.6 Let A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {3, 6, 9, 12, 15}, C = {2, 5, 7}. Then (a) A∩B = {3, 6} (b) B∪C = {2, 3, 5, 6, 7, 9, 12, 15} (c) B∩C = ∅. Another relation that can be represented by venn diagram is that of ‘is a subset of’. If A and B are two set such that B is a subset of A, then this is shown in the diagram below: B⊂A Note that if B ⊂ A then A ʹ ⊂ Bʹ Some properties of intersection and union of sets are given below: (1) Commutativity For any sets A and B we have; (a) A∪ B = B∪ A (b) A∩ B = B∩ A (2) Associativity 7  For any sets A, B and C, (a) A ∪ (B ∪ C ) = (A ∪ B) ∪ C (b) A ∩ (B ∩ C ) = (A ∩ B) ∩ C (3) Distributive properties For any sets A, B and C, (a) A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ) (b) A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ) (4) De Morgan’s Laws For any sets A and B we have; (a) ( A ∪ B)ʹ = Aʹ ∩ B ʹ (b) ( A ∩ B)ʹ = Aʹ ∪ B ʹ Example1.1.7 Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 3, 4, 5}, B = {2, 3, 5, 7}and C = {3, 4, 5, 6, 7, 8}. Verify the following: (i) ( A ∪ B)ʹ = Aʹ ∩ B ʹ (ii) ( A ∩ B)ʹ = Aʹ ∪ B ʹ (iii) A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ) (iv) A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ C ) Solution: (i) A ∪ B = {1, 2, 3, 4, 5, 7}so that ( A ∪ B)ʹ = {6, 8, 9, 10} I On the other hand Aʹ = {6, 7, 8, 9, 10} and B ʹ = {1, 4, 6, 8, 9, 10}so that Aʹ ∩ B ʹ = {6, 8, 9, 10} II Comparing I and II we have ( A ∪ B)ʹ = Aʹ ∩ B ʹ as required. (ii) A ∩ B = {2, 3, 5} so that ( A ∩ B)ʹ = {1, 4, 6, 7, 8, 9, 10} III Also from (i) Aʹ ∪ B ʹ = {1, 4, 6, 7, 8, 9, 10} IV Comparing III and IV gives ( A ∩ B)ʹ = Aʹ ∪ B ʹ (iii) B ∩ C = {3, 5, 7}so that A ∪ ( B ∩ C ) = {1, 2, 3, 4, 5, 7} V Now, A ∪ B = {1, 2, 3, 4, 5, 7} and A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}. Thus ( A ∪ B) ∩ ( A ∪ C ) = {1, 2, 3, 4, 5, 7} VI V and VI shows that A ∪ ( B ∩ C ) = ( A ∪ B) ∩ ( A ∪ C ) (iv) B ∪ C = {2, 3, 4, 5, 6, 7, 8}so that A ∩ ( B ∪ C ) = {2, 3, 4, 5} VII Also A ∩ B = {2, 3, 5} while A ∩ C = {3, 4, 5} so that ( A ∩ B) ∪ ( A ∩ C ) = {2, 3, 4, 5} VIII VII and VIII gives A ∩ ( B ∪ C ) = ( A ∩ B) ∪ ( A ∩ B) as required. We can prove these properties analytically if we so wish. 8 Example1.1.8 Prove that if A, B, C are any sets, then A∩(B∪C) = (A∩B)∪(A∩C) Proof: We must show that A∩(B∪C) ⊂ (A∩B)∪(A∩C) and also that(A∩B)∪(A∩C) ⊂ A∩(B∪C). To show the first inclusion, let x∈A∩(B∪C). Then x∈A and x∈(B∪C). Thus x∈A and x∈B or x∈C. If x∈B then x∈A∩B or if x∈C then x∈A∩C and in either of the two cases we have x∈(A∩B)∪(A∩C). Thus A ∩(B∪C) ⊂ (A∩B)∪(A∩C). For the reverse inclusion we note that since B ⊂ B∪C then A∩B ⊂ A∩(B∪C). Similarly A∩C ⊂ A∩(B∪C). Therefore (A∩B)∪(A∩C) ⊂ A∩(B∪C)∪A(B∪C) = A∩(B∪C). Combining the two inclusions we conclude that A ∩(B∪C) = (A∩B)∪(A∩C) as required. Exercise 1.1.9 Let A, B and C be any sets, prove that (a) A∪(B∩C) = (A∪B)∩(A∪C) (b) A–B⊂A (c) (A∩B)ʹ = Aʹ∪Bʹ and (A∪B)ʹ = Aʹ∩Bʹ. Example1.1.10 Simplify (A∩B)∪(A – B). (A∩B)∪(A – B) = (A∩B)∪(A∩Bʹ) = [(A∩B)∪A]∩[(A∩B)∪Bʹ] by distributive property = A ∩[(A∪Bʹ)∩(B∪Bʹ)] since (A∩B)∪A = A = A∩[(A∪Bʹ)∩U] since B∪Bʹ = U. = A ∩(A∪Bʹ) since (A∪Bʹ)∩U = A∪Bʹ = (A∩A)∪(A∩Bʹ) by distributive property = A ∪(A∩Bʹ) = A. 1.2 Sets of Numbers We are familiar with the set of natural numbers denoted by N and the set of integers denoted by Z. Thus N = {1, 2, 3, 4, ……………..} Z = {………., -3, -2, -1, 0, 1, 2, 3, 4, …………..} The set of positive integers denoted by Z+ is just the set of natural numbers N. Note that the number zero is neither negative nor positive. 9 We now introduce the set R of real numbers which will be fundamental to our treatment of functions of calculus. Before we do that, there are two sets we need to introduce, the set of rational numbers and the set of irrational numbers. 1.2.1 Rational Numbers The set of rational numbers denoted by Q consists of all numbers of the a form where a and b are integers and b ≠ 0. These are numbers such b 1 3 2 11 356 78 as , ,− , ,− , e.t.c. 2 1 3 9 1 10096 The decimal representation of any rational number is either terminating 1 decimal or it is a repeating decimal. For example, = 0.125 , 8 1 = 0.1428571428571428571......... When a decimal is a repeating decimal we 7 shall put a bar on top of a block which repeats. For 1 1 example = 0.142857142857..... can be written as = 0.142857 , 7 7 19 35 = 1.727272727......... = 1.72 , = 5.833..... = 5.83. Because of these 11 6 properties, when a rational number is given in decimal form we can find a two integers a and b such that we can express it in the form. b Example1.2.1.1 Show that 4.83 is a rational number. a Solution: We need to express 4.83 in the form where a and b are b integers. Recall that 4.83 = 4.83333333…… Now let x = 4.83. Then 10 x = 48.3333.... 100x = 483.3333….. Subtracting 100x minus 10x will now eliminate the decimal part as follows: 435 87 100 x − 10 x = 435.00000..... ⇒ 90 x = 435. Thus x = =. 90 18 a Example1.2.1.2 Express 0.354 in the form where a and b are b integers. 10 Solution: Let x = 0.354 Then 1000x = 354.354. Subtracting now yields 354 118 1000x – x = 354 so that 999x = 354 ⇒ x = = 999 333 Example1.2.1.3 Show that 2.75 is a rational number. Solution: Let x = 2.75 Note here that since the decimal is terminating there is no need for subtraction. Multiplication by a suitable power of 10 will remove the decimal part. Now 100x = 275 11 275 ⇒ x = =. 100 4 1.2.2 Irrational Numbers An irrational number is a numbers which cannot be expressed in the a form for some integers a and b. The decimal representation of an b irrational number is non terminating and non repeating. The set of irrational numbers will be denoted by Irr. Before we show that there exists some numbers which are not rational numbers we first show that if p is an integer such that p2 is divisible by 2 then p itself is divisible by 2. Proof: Since every integer is either divisible by 2 or leaves a remainder of 1 when divided by 2, we observe that every integer can be written in one of the following ⎧ 2n p=⎨ where n is an integer. Therefore, ⎩2 n + 1 ⎧ 4n 2 p2 = ⎨ 2 ⎩4n + 4n + 1 Since p2 is divisible by 2, then p2 must be 4n2. Hence p = 2n which is divisible by 2. We shall prove that 2 is not a rational number. We shall prove this by contradiction. To prove a mathematical statement by contradiction you assume something (usually what it is not) about the statement. Work through the statement according to the assumption. If you come up with the conclusion which is not consistent with the assumption, then this is the 11 contradiction you are looking for. This contradiction then implies that your earlier assumption about the statement is false. For the proof that 2 is not a rational number, we make an assumption about 2 and if our assumption leads to contradictory statements, we conclude that our assumption is false. Proof: Assume that 2 is a rational number. Then there are integers m m and n with no common factors and m > 0, n > 0 such that 2 =. n Squaring both sides yields 2 ⎛m⎞ 2 2 ⎜ ⎟ = 2 implying that m = 2n. Since m2 has a factor 2, it is even and so n ⎝ ⎠ is divisible by 2. By the above proof m must itself be divisible by 2. Thus we can write m in the form m = 2k for some integer k. This then gives 2n2 = m2 = 4k2 so that n2 = 2k2. This shows that n is also divisible by 2. But then this means both m and n are divisible by 2 contradicting our assumption that m and n have no common factor. This contradiction proves that 2 cannot be a expressed in the form for some integers a and b. Hence 2 is not a b rational number. Numbers such as 2 , are called irrational numbers. Some examples of irrational are 2 , 3 , 5 , π , e e.t.c Example 1.2.2.1 Show that 2 + 3 is not a rational number. Solution: Suppose 2 + 3 is rational, then there exists integers a and b a with b ≠ 0 such that 2 + 3 =. Then b a a − 2b 3 = −2⇒ 3 =. But a – 2b is an integer so that the b b right side is rational while the left side is irrational. This contradiction proves that 2 + 3 is not a rational number. 1.2.3 Real Numbers We have seen that a rational number is a number which can be expressed as a quotient of two integers while an irrational number 12 cannot be expressed in such a form. Therefore, since no number can be both rational and irrational at the same time, the two sets are disjoint. The union of the set of rational numbers with the set of irrational numbers is the set of real numbers R. Real numbers form a continuous line and so they cannot be listed using braces. However, set of real numbers also contains the set of natural numbers and the set of integers. Here, we shall deal mainly with special subsets of R called intervals. We shall use brackets to represent an interval. We use open bracket ( or ) when the number at the boundary is not included and a closed bracket [ or ] when the number at the boundary is included. These intervals can be represented using a number line. For example, the set [– 5, 4] can be represented on a number line as follows This number line represents all real numbers between – 5 and 4 inclusive. The dots at the ends of the line representing the set indicate that – 5 and 4 are also included. We can also use set notation to write the set of real numbers. The set above can for example be given as { x∈R ⏐ – 5 ≤ x ≤ 4 } Note that whether we use set notation or interval notation to describe the set of real numbers we do not attempt to list the elements or members of the set. This is because there is an uncountable number of real numbers between any two given real numbers. Here are some examples where a and b are real numbers: Interval notation Set notation Number line ( a, b ) {x∈R ⏐ a < x < b} [a, b ) {x∈R ⏐ a ≤ x < b} ( a, b ] {x∈R ⏐ a < x ≤ b} 13 ( a, ∞ ) {x∈R ⏐ a < x} (– ∞, b ] {x∈R ⏐ x ≤ b} Example1.2.3.1 Let A = { x∈R ⏐ – 7 ≤ x < 3 } and B = { x∈R ⏐ x ≥ – 1 }.Find (a) A∩B (b) Aʹ Solution: (a) We have From the diagram, the shaded part gives the intersection of the sets. Therefore, A∩B = [– 1, 3). Note that since 3∉A so 3∉A∩B (b) Again using the number line we have the following: Aʹ = (– ∞ , – 7) ∪ [3 , ∞) Note that when the symbol ∞ or – ∞ appear we shall always use an open bracket because ∞ is not a real number and so it cannot be in the set of real numbers. Example1.2.3.2 Let U = (−7 , 10] be the universal set, A = [−3 , 7] , B = (−2 , 5) and C = (−6 , 10]. Find the following sets: (i) A∩ B (ii) U −C (iii) Bʹ ∪ A (iv) (A ∪ C )ʹ Solution: (i) A ∩ B = (−2 , 5) It can be seen here that B is a subset of A. (ii) U − C = (−7 , − 6] Note that this set is C ʹ , the complement of C. (iii) Since B ʹ = (−7 , − 2] ∪ [5 , 10] we have B ʹ ∪ A = U (iv) (A ∪ C )ʹ = (−7 , − 6] 14 Properties of real numbers Real numbers have certain properties which allow us to compare any two real numbers or to perform some algebraic operations. These are grouped into algebraic properties and order properties. A. Algebraic properties of R If a , b , c are any real numbers, then (i) a + b = b + a, ab = ba commutative law of addition and multiplication (ii) a + (b + c) = (a + b) + c a(bc) = (ab)c associative law of addition and multiplication (iii) a+0=0+a=a The real number 0 is additive identity (vi) a.1 = 1.a = a The real number 1 is the multiplicative identity (v) For each a ∈ R, there corresponds – a ∈ R such that a + (– a) = (– a) + a = 0. Thus every real number has an additive inverse. (vi) Each non zero real number has multiplicative inverse. (vii) a( b + c) = ab + ac distributive law. B. Order relations Let x and y be any two real numbers, then we say that (i) x is greater than y written x > y if x – y > 0. For example, 5 > 2 since 5 – 2 = 3 > 0 Similarly 4 > – 17 since 4 – (– 17) = 4 + 17 = 21 > 0. (ii) x is less than y written x < y if x – y < 0. (iii) x is equal to y written x = y if x – y = 0 The law of trichotomy holds for R. That is, if x , y ∈ R, then either x > y or x = y or x < y For all x , y , z ∈ R, x > y ⇒ x + z > y + z. If z > 0 then xz > yz. Examples 1.2.3.3 (A) Linear equations in one variable A linear equation is one of the form ax + b = 0 or ax + by + c = 0 where a, b and c are constants and x and y are unknown variables. Note that in a linear equation there is no cross – product of variables. The equations of the form ax 2 + bx + c = 0 or axy + bx + cy + d = 0 , are not linear. Here we shall 15 concentrate on the equation of the form ax + b = c where a , b and c are constants and x is the unknown. Using the algebraic properties (v) and (vi) above we can solve this equation for x. For example, applying (v) and adding – b both sides we have: ax + b + (−b) = c + (−b) ⇒ ax = c − b. Since a is not zero, applying (vi) we have: 1 1 c−b (ax ) = (c − b) ⇒ x = a a a (a) Solve each of the equations below: 2 (i) 7 x − 23 = 5 (ii) 5 x − 2 = 2 x − 17 (iii) ( x + 7) = x 3 Solutions: (i) Adding 23on both sides we have 7 x − 23 + 23 = 5 + 23 ⇒ 7 x = 28. Dividing by 7 both sides we have: 7 x 28 = ⇒x=4 7 7 (ii) Using the same operations in (v) above we collect the like terms as follows: 5 x − 2 x = −17 + 2 ⇒ 3 x = −15 It follows after dividing by 3 both sides that x = −5 (iii) To deal with this problem we can first use the distributive law to remove the brackets. Thus, 2 2 x 14 ( x + 7) = + , it follows then that 3 3 3 2 x 14 2x 14 x 14 + =x⇒ −x=− or − = − ⇒ x = 14 after 3 3 3 3 3 3 multiplying by – 3 on both sides. There are fractional equations which can be reduced to linear equations. 5 1 Consider for example the equation = 3 , if x ≠ then the 2x − 1 2 denominator is not zero, therefore, multiplying both sides by 2 x − 1 we get (2 x − 1) 5 = 3(2 x − 1) ⇒ 5 = 6 x − 3 which is a linear equation and now can 2x − 1 be solved as above. (b) Solve each of the following equations: 3 x 1 2x 2 (i) = (ii) = −7 (iii) = 2x − 5 2 2x + 3 x x +1 Solution: (i) Multiplying both sides by 2(2 x + 3) which is the lowest common multiple of the two denominators yields 3(2x + 3) = 2x.Multiplying using distributive law 16  gives 6 x + 9 = 2 x. Now solving this equation we 9 get x = −. 4 (ii) Multiplying both sides by x and solving the resulting 1 linear equation we get x = −. 7 (iii) Here, even though the equation contains a power of x, after multiplying both sides by x + 1 and subtracting 2x 2 from both sides of the resulting equation, the result is a linear equation. We have 2 (x + 1) 2 x = (x + 1)(2 x − 5) x +1 ⇒ 2 x = 2 x 2 − 3x − 5 2 ⇒ 0 = −3 x − 5 ⇒ 3 x = −5 5 ⇒x=− 3 (B) Absolute value of a real number Definition: An absolute value of a real number is its distance from the origin. It is therefore always a positive quantity. Absolute value is ⎧ x if x ≥ 0 denoted by two short lines⏐ ⏐. In general x = ⎨ ⎩− x if x < 0 For example ⏐7⏐ = 7 and ⏐– 7⏐ = – (– 7) = 7 (a) Find the value of a − b when (i) a = 7, b = 4 (ii) a = −6, b = 8 (iii) a = 2, b = 11 Solution: (i) 7−4 = 3 =3 (ii) − 6 − 8 = − 14 = −(− 14) = 14 (iii) 2 − 11 = − 9 = 9 p−a 3 (b) Given that a < b , express p in terms of a and b if = and b−a 4 p 0 by order relation property (i). p−a p−a 3 Hence, b − a = b − a. We have = =. Since b − a ≠ 0 , b−a b−a 4 17  3 we have p − a = (b − a ). But p − a < 0 so that 4 3b 3a 3b 3a − ( p − a) = − or − p + a = −. Solving for p we 4 4 4 4 7a 3b get p = −. 4 4 Using the properties of absolute value can sometimes simplify calculations of equations that may otherwise result in quadratic equations because such calculations are reduced to dealing with linear equations. (c) Solve the equation ⏐2x – 3 ⏐ = 5. Solution: There are two possibilities since we are considering the absolute value of 2x – 3. Either 2x – 3 is positive or it is negative. If 2x – 3 > 0, then we have 2x – 3 = 5 ⇒ 2x = 8 ⇒ x = 4 If 2x – 3 < 0, then we have – ( 2x – 3 ) = 5 ⇒ 2x – 3 = – 5 ⇒ 2x = – 2 ⇒ x = – 1 Another point worth noting with absolute value problems is when an inequality is involved. Observe that if ⏐x⏐ < k for some positive real number k, then, if x > 0 we have x < k and if x < 0 then – x < k which implies that x > – k. Thus from the two solutions we see that ⏐x⏐ < k implies that – k < x < k. (d) Solve the inequality ⏐2x – 3 ⏐ ≤ 5. Solution: From the above discussion we have – 5 ≤ 2x – 3 ≤ 5. Solving each of the inequalities separately we get – 1 ≤ x ≤ 4 as our solution set. 18 1.2.4 Complex Numbers Define i = − 1 , then i 2 = −1. Thus, − 25 = − 1× 25 = − 1 × 25 = i × 5 = 5i. Numbers of the form bi where b is a real number are called imaginary numbers. A complex number is a number of the form a + bi where a and b are real numbers, a is called the real part and b is the imaginary part of the complex number a + bi. Let z = x + yi be the complex number, then Re z = x (for real part of z) Im z = y (for imaginary part of z). A complex number can be represented geometrically on the xy-plane by taking the x – axis as the real axis and the y – axis as the imaginary axis. A diagram such as this one which represents a complex number is called an Argand diagram. We define the equality and addition of complex numbers in the following way: Let z1 = x1 + iy1 and z 2 = x2 + iy2 be two complex numbers. Then, (a) z1 = z 2 if and only if x1 = x2 and y1 = y2 (b) z1 ± z 2 = ( x1 ± x2 ) + i( y1 ± y2 ) e.g (4 + 3i) + (7 – 5i) = (4 + 7) + i(3 – 5) Example1.2.4.1 Find x and y such that (a) x + iy = 7 − 4i (b) (3x − iy) + (2 + 13i) = −7 + 3i Solution: (a) Equating the real parts and the imaginary parts we have x = 7 and y = – 4 19  (b) First we add the real parts and the imaginary parts on the left side to get (3x + 2) + i(13 − y ) = −7 + 3i. Now equating the real and imaginary parts we have 3x + 2 = – 7 ⇒ x = – 3, 13 – y = 3 ⇒ y = 10. To define the product of two complex numbers, let z1 = x1 + iy1 and z 2 = x 2 +iy2 be two complex numbers, then z1 z 2 = ( x1 + iy1 )(x2 + iy2 ) = x1 x2 + ix1 y 2 + ix2 y1 + i 2 y1 y 2 = x1 x2 + i( x1 y2 + x2 y1 ) − y1 y2 = x1 x2 − y1 y2 + i( x1 y2 + x2 y1 ) Definition: If z = x + iy is a complex number, then the modulus of z 2 denoted by ⏐z⏐ is the real number z = x 2 + y 2 so that z = x 2 + y 2 Example1.2.4.2 Find the modulus of each complex number below: (i) 5 + 12i (ii) − 8 + 6i (iii) 15i Solution: (i) 5 + 12i = 5 2 + 12 2 = 25 + 144 = 169 = 13 2 (ii) − 8 + 6i = (− 8) + 6 2 = 64 + 36 = 100 = 10 (iii) Note that 15i = 0 + 15i since the real part is zero. Therefore, 15i = 0 2 + 15 2 = 15 2 = 15 Definition 1.2.4.1 The conjugate of a complex number z = x + iy denoted by z is the complex number z = x − iy. Thus, the conjugate of a complex number is obtained by changing the sign of the imaginary part of the complex number. As an example, the conjugate of 7 + i is 7 − i , the conjugate of − 13 − 37i is − 13 + 37i. Take note that if z a real number, i.e z = k for some real number k, then z is its own conjugate, z = z. If z is purely imaginary, i.e z = ik for some real number k, then the conjugate of z is the negative of z, z = − z Note also that if z = a + ib , then z z = zz = (a + ib)( a − ib) = (a − ib)( a + ib) = a 2 + b 2 is a real number since it has no imaginary part. We can now define the division of two complex numbers as follows: Let z1 = x1 + iy1 and z 2 = x2 + iy2. Then z1 z1 z 2 ( x1 + iy1 )( x2 − iy2 ) ( x1 x2 + y1 y 2 ) + i( x2 y1 − x1 y 2 ) = = = z 2 z 2 z 2 ( x2 + iy2 )( x2 − iy2 ) x22 + y 22 x x + y1 y 2 x y −x y = 1 22 2 + i 2 21 12 2 x2 + y 2 x2 + y 2 20 Thus, to divide two complex numbers we multiply both the numerator and the denominator by the conjugate of the denominator. Example1.2.4.3 Express each of the following complex numbers in the form a + ib where a and b are real numbers. 2 − 5i (i) (3 + 5i ) + (2 − 3i ) (ii) (4 + i )2 (iii) 3 + 2i Solution: (i) Adding the real parts and the imaginary parts we get (3 + 5i ) + (2 − 3i ) = (3 + 2) + (5 − 3)i = 5 + 2i. (ii) (4 + i )2 = (4 + i )(4 + i ) = 16 + 4i + 4i + i 2 = 16 + 8i − 1 since 2 i 2 = −1 This gives (4 + i ) = 16 − 1 + 8i = 15 + 8i (iii) Multiplying both the numerator and the denominator by the conjugate of the denominator we have: 2 − 5i (2 − 5i )(3 − 2i ) 6 − 10 + (− 4 − 15)i − 4 − 19i 4 19 = = = =− − i 3 + 2i (3 + 2i )(3 − 2i ) 9+4 13 13 13 3 + 2i Example1.2.4.4 Let z =. 4 − 3i (a) Express z in the form a + ib where a and b are rational numbers. (b) Find ⏐z⏐ (3 + 2i)(4 + 3i) (12 − 6) + i(9 + 8) 6 17 Solution: (a) z= = = + i (4 − 3i)(4 + 3i) 16 + 9 25 25 2 2 2 ⎛ 6 17 ⎞⎛ 6 17 ⎞ ⎛ 6 ⎞ ⎛ 17 ⎞ 36 289 (b) z = z z = ⎜ + i ⎟⎜ − i ⎟ = ⎜ ⎟ + ⎜ ⎟ = + ⎝ 25 25 ⎠⎝ 25 25 ⎠ ⎝ 25 ⎠ ⎝ 25 ⎠ 625 625 2 325 13 Thus z = =. Then 625 25 13 13 z = zz = =. 25 5 Example1.2.4.5 Solve each of the following equations x + iy 1 3 − 2i (i) = 3 − 5i (ii) = 2+i x + iy 1 + i Solution: (i) Cross-multiplying gives: x + iy = (3 − 5i )(2 + i ) = (6 + 5) + (3 − 10)i = 11 − 7i. Equating the real parts and the imaginary parts we get: x = 11 and y = −7 (ii) If x + iy ≠ 0 , cross-multiplying gives: 1 + i = (3 − 2i )(x + iy). Dividing both sides by 3 − 2i we get 21  x + iy = 1+ i = (1 + i )(3 + 2i ) = 1 + 5 i. Thus x = 1 and 3 − 2i (3 − 2i )(3 + 2i ) 13 13 13 5 y=. 13 1.2.5 Surds Numbers of the form 2 , 3 , 5 ,......... are called surds. We have already seen that these numbers are irrational numbers, as such, they cannot be a expressed in the form where a and b are integers. Here, we shall take b 7 for example to mean the positive squareroot of 7 and so on. Surds occur quite frequently in solutions of equations, as such, it is important that we know how to simplify them. 1 Take note that 2 can be written as 2 2 so that ( ) ( ) 2 1 2 1 2 × 2 = 2 = 2 2 = 2 2 = 21 = 2. ×2 It is convenient to leave such numbers in surd than any other form. Example1.1.5.1 Express each of the following as the simplest possible surd: (i) 12 (ii) 63 (iii) 80 Solution: (i) 12 = 4 × 3 = 4 × 3 = 2 3 (ii) 63 = 9 × 7 = 9 × 7 = 3 7 (iii) Similarly, 80 = 4 5 Example1.2.5.2 In each of the following, expand the given expression and simplify: (i) (1 + 2 5 )(2 + 5 ) (ii) (7 − 3 7 )(3 + 2 7 ) Solution: (i) (1 + 2 5 )(2 + 5 ) = 2 + 5 + 4 5 + 2 5 × 5 = 2 + 5 5 + 10 = 12 + 5 5 (ii) ( )( ) 7 − 3 7 3 + 2 7 = 21 + 14 7 − 9 7 − 6(7 ) = 21 − 42 + 5 7 = − 21 + 5 7 When a fraction contains a surd in the denominator we can simplify it by removing the surd from the denominator so that the denominator contains only an integer. This process of removing the surd from the denominator is called rationalizing the denominator. For example, to 22  3 rationalize the denominator of we multiply both the numerator and 7 3 3× 7 3 7 the denominator by 7 to get = =. Similarly, to 7 7× 7 7 6 rationalize the denominator of we have 45 6 6 6 6 2 5 = = = =. 45 9×5 9× 5 3 5 5 When the denominator is a sum or difference of two terms, we achieve this by multiplying both the numerator and the denominator by a term which makes the denominator a difference of squares. Example1.2.5.3 Rationalize the denominator 1− 5 2 2 (i) (ii) 3− 2 2 3+5 1 y (iii) (iv) ( 2+3 2 5− 3 )( 2+ y ) Solution: (i) Multiplying both the numerator and the denominator by 3 + 2 we get 1− 5 2 = (1 − 5 2 )(3 + 2 ) = − 7 − 14 2 = − 7 (1 + 2 2 ) = −1 − 2 2 3− 2 (3 − 2 )(3 + 2 ) 9 − 2 7 2 2(2 3 − 5) 2(2 3 − 5) 10 4 3 (ii) = = = − 2 3 + 5 (2 3 + 5)(2 3 − 5) 12 − 25 13 13 (iii) To deal with this problem we do it step by step. 1 5+ 3 5+ 3 = =. Dealing ( 2+3 2 5− 3 )( ) ( 2 + 3 2 (25 − 3) 22 2 + 3 2 ) ( ) with the other term we get: 5+ 3 = 5+ 3 2−3 2 (=− )( 5+ 3 2−3 2. ) ( )( ) ( 22 2 + 3 2 ) 22(4 − 18) 308 (iv) y = ( y 2− y ) = ( y 2− y ) 2+ y ( 2+ y )( 2− y ) 2− y 23