Math 0 PDF - Number Systems and Inequalities

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RaptJadeite8896

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Faculty of Computers and Information, Arish University

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set theory mathematics number systems inequalities

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This document is chapter 1 of a mathematics textbook, focusing on number systems and inequalities, including sets, subsets, and set operations. It introduces concepts such as finite and infinite sets, and De Morgan's rules, often used in secondary school mathematics.

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Chapter 1 Number Systems and Inequalities Chapter 1 Number Systems and Inequalities 1.1. Sets (A) Introduction to Sets Mathematics deals with objects of very different kinds; from your previous experience, you are familiar with many of them: Numbers, p...

Chapter 1 Number Systems and Inequalities Chapter 1 Number Systems and Inequalities 1.1. Sets (A) Introduction to Sets Mathematics deals with objects of very different kinds; from your previous experience, you are familiar with many of them: Numbers, points, lines, planes, triangles, circles, angles, equations, functions and many more. Often, objects of a similar nature or with a common property are collected into sets; these may be finite or infinite (For the moment, it is enough if you have an intuitive understanding of finite, resp. infinite; a more rigorous definition will be given at a later stage). The objects which are collected in a set are called the elements of that set. If an object a is an element of set M, we write 𝑎 ∈ 𝑀, which is read as a (is an) element of M. If a is not an element of M, then we write 𝑎 ∉ 𝑀 which is read as a is not an element of M. Example 1.1.1. 1. If G is the set of all even numbers, then 6 ∈ 𝐺 and 3 ∉ 𝐺. 2. If L is the set of all solutions of the equation 𝑥 2 = 1, then 1 is an element of L, while 2 is not. Generally, there are two ways to describe a set: By listing its elements between curly brackets and separating them by commas, e.g. {0}, {2,67,9}, {𝑥 , 𝑦 , 𝑧}, {0,1,2,3, … } , {2,4,6, … 20} The meaning should be clear from the context. -2- By giving a rule, which determines if a given object is in the set or not; this is also called implicit description; for example 1. {𝑥: 𝑥 is a natural number} 2. {𝑥: 𝑥 is a natural number and 𝑥 > 0} 3. {𝑦: 𝑦 solves (𝑦 + 1). (𝑦 − 3) = 0 } 4. {𝑝: 𝑝 is an even prime number}. In general, we describe sets by {𝑥: 𝑥 ∈ 𝑀 𝑎𝑛𝑑 𝑃(𝑥)}, which also can be written as {𝑥 ∈ 𝑀 ∶ 𝑃(𝑥)}. where 𝑃(𝑥) means that x has the property described by P. We shall use the following conventions in describing certain sets of numbers: ℕ = {1,2,3, … } is the set of natural numbers. ℤ = {… , −3, −2, −1,0,1,2,3 … } is the set of integers. 𝑎 ℚ = {𝑥: 𝑥 = } where 𝑎 ∈ ℤ, 𝑏 ∈ ℤ and 𝑏 ≠ 0 , is the set of 𝑏 rational numbers; observe that each rational number is the ratio of two integers, whence the name. ℝ = {𝑥: 𝑥is a real number}. (see the next section). B) Subsets, Power Sets, Equality of Sets Definition 1.1.2. A set A is a subset of a set B, written as 𝐴 ⊆ 𝐵 if every element of A is also an element of B. So, 𝐴 ⊆ 𝐵 whenever 𝑥 ∈ 𝐴 implies 𝑥 ∈ 𝐵. Observe carefully the difference between ⊆ and ∈. If 𝐵 = {1,2,3}, then 1 is an element of B, but 1 is not a subset of B. The set 𝐴 = {1} which has 1 as its only element, however, is a subset of B, since it fulfils the definition. Note that: 1 is different from {1}, {1} is different from {{1}}. -3- Observe that the unique element of the last set is the set {1}. It is conceivable that a set contains no elements at all; this set is called the empty set, and it is denoted by symbol . It can also be described by a property, e.g 𝜙 = {𝑥: 𝑥 ≠ 𝑥}. Here are some elementary properties of the  relation: 1-  is a subset of every set. 2- For any set 𝐴 , 𝐴 ⊆ 𝐴. 3- If A is a subset of B , and B is a subset of C, then A is a subset of C. Definition 1.1.3. If A is a set, then 𝑃(𝐴) = {𝑋: 𝑋 ⊆ 𝐴} is called the power set of A. It is the set of all subsets of A. Let us look at some power sets: 1- 𝐴 = 𝜙: 𝑃(𝜙) = {𝜙}. Observe that 𝜙 is different from {𝜙}: While 𝜙 has no elements, {𝜙} has exactly one element, namely set 𝜙. 2. A {x }: 𝑃(𝐴) = {𝜙 , {𝑥}}. 3. 𝐴 = {𝑥 , 𝑦}: 𝑃(𝐴) = {𝜙 , {𝑥} , {𝑦} , {𝑥 , 𝑦}} So, we see that 𝑃(𝐴) has four elements. 4. 𝐴 = {𝑥 , 𝑦 , 𝑧}. Observe that 𝑃(𝐴) has eight elements. 𝑃(𝐴) = {𝜙 , {𝑥} , {𝑦} , {𝑧} , {𝑥, 𝑦} , {𝑥, 𝑧} , {𝑦, 𝑧} , {𝑥, 𝑦, 𝑧}}. Definition 1.1.4. Two sets A and B are equal if 𝐴 ⊆ 𝐵and 𝐵 ⊆ 𝐴. If A and B are equal, we write 𝐴 = 𝐵 To prove that two sets A and B are equal, we must show that A is a subset of B, and that B is a subset of A. -4- C) Finite and Infinite Sets Definition 1.1.5. A set M is called finite, if 𝑀 = 𝜙, or if there is natural number n such that the elements of M can be numbered 1, … , 𝑛 in such a way that every element of M appears exactly once in the list. Otherwise, M is called infinite. Example 1.1.6. 1. The set {𝑎 , 𝑏 , 𝑐 , 𝑑} is finite. A possible numbering might look this: Assign 1 to a, assign 2 to b, assign 3 to c, assign 4 to d. Of course, there are other possibilities of listing the set M (Which?) 2. The set of all solutions of the equation 𝑥 2 + 23𝑥 − 17 = 0 is finite, since the number of solutions of a polynomial is at most equal to the degree. 3. The sets ℕ , ℤ , ℚ , ℝ are infinite. 4. The set of all multiples of 5 is infinite. D) Set Operations Definition 1.1.7. Let A and B be given sets; then 1. The intersection 𝐴 ⋂ 𝐵 of A and B is defined by 𝐴 ⋂ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵} If 𝐴 ⋂ 𝐵 = 𝜙 then A and B are called disjoint. 2. The union 𝐴 ⋃ 𝐵 of A and B is defined by 𝐴 ⋃ 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}. 3. The set difference 𝐴 \ 𝐵 of A and B is the set 𝐴\𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵}. -5- 𝐴 \ 𝐵 is also called the relative complement of B with respect to A. If U is given universal set, then 𝑈 \ 𝐴 is just called the complement of A, written as 𝐴𝑐. Example 1.1.8. 1. Let 𝐴 = {𝑥 ∈ ℤ: −1 ≤ 𝑥 < 4} = {−1,0,1,2,3}, and 𝐵 = 𝑥 {𝑥 ∈ ℤ: 1 ≤ 2 ≤ 3} = {2,3,4,5,6}. Then 𝐴⋂𝐵 = {𝑥 ∈ ℤ: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵} 𝑥 = {𝑥 ∈ ℤ ; −1 ≤ 𝑥 < 4 𝑎𝑛𝑑 1 ≤ ≤ 3} = {2,3}. 2 Furthermore, 𝐴⋃𝐵 = {𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵} 𝑥 = {𝑥 ∈ ℤ ∶ −1 ≤ 𝑥 < 4 𝑜𝑟 1 ≤ ≤ 3} 2 = {−1,0,1,2,3,4,5,6} 2. Let ⋃ = ℕ and A be the set of all even natural numbers. Then 𝐴𝑐 is the set of all odd natural numbers. A convenient pictorial representation of the operations defined above are the Venn diagrams, shown in Fig. 1.1.1 − 1.1.3 The shaded area represents the result of the respective operation. Figure 1.1.1: 𝐴⋂𝐵 Figure 1.1.2: 𝐴 ⋃𝐵 Figure 1.1.3: 𝐴/𝐵 A A A B B B -6- Lemma 1.1.9. Let 𝐴 ⊆ ⋃ then 1. 𝐴 ⋂ 𝐴 = 𝐴 , 𝐴 ⋃ 𝐴 = 𝐴; 2. 𝐴 ⋂ 𝜙 = 𝜙, 𝐴 ⋃ 𝜙 = 𝐴; 3. 𝐴 ⋂ ⋃ = 𝐴 , 𝐴 ⋃ 𝑈 = 𝑈; 4. 𝐴 ⋂𝐴𝑐 = 𝜙 , 𝐴 ⋃ 𝐴𝑐 = 𝑈. Lemma 1.1.10. If A and B are sets, then 𝐴 ⋂ 𝐵 = 𝐵 ⋂ 𝐴 and 𝐴 ⋃𝐵 = 𝐵 ⋃ 𝐴. This is called the law of Commutativity for ⋂ and ⋃. Lemma 1.1.11. If 𝐴, 𝐵, 𝐶 are sets, then 1. (𝐴 ⋂ 𝐵) ⋂ 𝐶 = 𝐴 ⋂(𝐵 ⋂ 𝐶) 2. (𝐴 ⋃ 𝐵) ⋃ 𝐶 = 𝐴⋃(𝐵 ⋃ 𝐶) This is called the law of Associativity for ⋂ and ⋃. E) De Morgan Rules, Distributivity, Tables: In the sequel, let U be a universal set, of which all other mentioned sets are subsets. Our first Theorem in this section exhibits a connection between intersection, union, and complementation: Theorem 1.1.12. For all sets 𝐴, 𝐵; (𝐴 ⋃ 𝐵)𝑐 = 𝐴𝑐 ⋂𝐵𝑐 , (𝐴 ⋂ 𝐵)𝑐 = 𝐴𝑐 ⋃ 𝐵𝑐 These are the Rules of De Morgan. -7- Proof: We only show the first part and leave the second as an exercise. " ⊆ " Let 𝑥 ∈ (𝐴 ⋃ 𝐵)𝑐 then x is not in the union of A and B, hence x is neither in A nor in B. Since x is not in A, we have 𝑥 ∈ 𝐴𝑐 and hence x is not in B we have 𝑥 ∈ 𝐵𝑐. Thus, 𝑥 ∈ 𝐴𝑐 and 𝑥 ∈ 𝐵𝑐 which implies that 𝑥 ∈ 𝐴𝑐 ⋂ 𝐵𝑐 "  ": Conversely, let 𝑥 ∈ 𝐴𝑐 ⋂ 𝐵𝑐 then x is not in A and x is not in B. Hence x is not an element of A or B, i.e. 𝑥 ∈ (𝐴 ⋃ 𝐵)𝑐. We shall introduce a new tool for tacking problems of this sort. If A and B are sets, then for an arbitrary element x of our universe of discourse U there are four possibilities. 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵 , 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵 , 𝑥 ∉ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵 , 𝑥 ∉ 𝐴 and 𝑥 ∈ 𝐵. For every one of these cases let us consider if x is in the intersection of A and B: If 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵, then 𝑥 ∈ 𝐴 ⋂ 𝐵. In all three other cases, we have 𝑥 ∉ 𝐴 ⋂ 𝐵. This observation can be put in the form of a table which looks like this: A B 𝐴 ∩ 𝐵 𝐴∪𝐵 1 1 1 1 1 0 0 1 0 1 0 1 0 0 0 0 Here, an entry 1 means that x is in the set pointed to by the column head, and 0 tells us that it is not. -8- Regarding complementation, we have the following table: A 𝐴𝑐 1 0 0 1 With these three tables we can find the table for every set which is formed from A and B by the operations ⋂, ⋃ and 𝐴𝑐. Let us look at the table for the first De Morgan rule: A B 𝐴𝑐 𝐵𝑐 𝐴∪𝐵 (𝐴 ∪ 𝐵)𝑐 𝐴𝑐 ∩ 𝐵𝑐 1 1 0 0 1 0 0 1 0 0 1 1 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 1 Observe that the right columns of the table have the same entries. Theorem 1.1.13. For all sets 𝐴, 𝐵: 𝐴 ⋃ (𝐴 ⋂ 𝐵) = 𝐴 , 𝐴 ⋂(𝐴 ⋃ 𝐵) = 𝐴. These are the absorption laws. Proof: We shall only prove the first equation, and use a table: A B 𝐴⋂𝐵 𝐴 ⋃ (𝐴 ⋂ 𝐵) 1 1 1 1 1 0 0 1 0 1 0 0 0 0 0 0 This proof our claim. The method of tables can also be used with more than just two sets. In general, if there are n sets, then a column in a table for each of these n sets will have 2𝑛 entries. -9- Theorem 1.1.14: For all sets 𝐴, 𝐵, 𝐶 𝐴 ⋂ (𝐵 ⋃ 𝐶) = (𝐴 ⋂ 𝐵)⋃(𝐴 ⋂ 𝐶) , 𝐴 ⋃(𝐵 ⋂ 𝐶) = (𝐴 ⋃ 𝐵)⋂(𝐴 ⋃ 𝐶) These are the distributive Law for and. Exercise set (1) 1- Describe the following sets explicitly: 𝐴 = {𝑥 ∈ ℤ ∶ 𝑥 ≤ 5}, 𝐵 = {𝑥 ∈ ℕ ∶ 𝑥 𝑑𝑖𝑣𝑖𝑑𝑒 24}, 𝐶 = {𝑥 ∈ ℝ ∶ 𝑥 = 2}. 2- Find a property P and a set M such that you can write the following sets in the form {𝑥 ∈ 𝑀 ∶ 𝑃(𝑥)}: 𝐴 = {6, 4 ,8 ,2 ,0}, 𝐵 = {1 ,3 , 5 ,7, … }, 𝐶 = {3 , 5, 11, 2 ,7 ,13}. 3- Find the power set of {𝑎 , 𝑏 , 𝑐 , 𝑑}. How many elements does it have? 4- Given that a set A has n elements, where n is a fixed natural number, how many elements does 𝑃(𝐴) have? You need not prove your answer, but explain your answer. 5- Prove: If 𝐴 = 𝐵 and 𝐵 = 𝐶, then 𝐴 = 𝐶. Prove this exercise for the general case; an example is not a proof! 6- a) Explain why ℕ is infinite. b) Explain why ℤ , ℚ , ℝ are infinite. 7- Are the following sets finite or infinite? a) {𝑥 ∈ ℝ ∶ 𝑥 2 + 2𝑥 − 1 = 0}; b) {𝑥 ∈ ℕ ∶ 𝑥 < 0} c) {𝑥 ∈ ℚ ∶ 0 < 𝑥 < 1} 8- Using the method of tables, prove The second De Morgan rule, the second Absorption Law and the first Distributive Law. - 10 - 1.2. Numbers Number is one of the basic concepts of mathematics. Whole numbers and fractions, both positive and negative, together with the number zero are called rational numbers. Every rational number may be represented in the form a ratio, 𝑝 5 5 , of two integers p and q, for example, , 1.25 =. In 𝑞 7 4 particular, the integer p may be regarded as a ratio of the 𝑝 6 0 integers for example 6 = , 0 =. Rational numbers may 1 1 1 be represented in the form of periodic terminating or no terminating fractions. Numbers represented by no terminating, but no periodic, decimal fractions are called irrational numbers; such are the numbers √2 , √3 , 𝜋 , 2 − √5, etc. Now, we will prove that √2 is irrational number: let √2 is rational number, that is 𝑝 √2 = , 𝑞 ≠ 0 𝑞 𝑝2 Then, 2= 𝑎𝑛𝑑 2𝑞 2 = 𝑝2 𝑞2 - 11 - So, 𝑝2 is even which implies p is even. Therefore, we can choose 𝑝 = 2𝑘 , 𝑘 is integer. Then, 2𝑞 2 = 4𝑘 2 ⇒ 𝑞 2 = 2𝑘 2. Hence, 𝑞 2 is even and so q is even. We have both 𝑝, 𝑞 is even, contradiction. Thus √2 is irrational. The collection of all rational and irrational numbers makes up the set of real number. The real numbers are ordered in magnitude, that is to say, for each pair of real numbers x and y there is one, and only one, of the following relations: 𝑥𝑦 Real numbers may be depicted as points on a number scale. A number scale is as infinite line on which are chosen: 1) a certain point 0 called the origin, 2) a positive direction indicated by an arrow, and 3) a suitable unit of length. Usually, the number scale is made horizontal and takes the positive direction to be from left to right. We denote by ℝ the set of real numbers. - 12 - Definition 1.2.1. The absolute value (or modulus) of a real number x (written |𝑥 |) is a nonnegative real number that satisfies the conditions: |𝑥 | = 𝑥 𝑖𝑓 𝑥 ≥ 0 , |𝑥 | = −𝑥 , 𝑖𝑓 𝑥 < 0 From the definition, it follows 𝑥 ≤ |𝑥 | for any x. Theorem 1.2.2. (i) ∀𝑥 ∈ℝ (|𝑥 | ≥ 0), (ii) ∀𝑥 ∈ ℝ (−|𝑥 | ≤ 𝑥 ≤ |𝑥 |), (iii) ∀ 𝑥 ,𝑦 ∈ ℝ (|𝑥. 𝑦| = |𝑥 | |𝑦|), (iv) ∀ 𝑥 ,𝑦 ∈ ℝ ,𝑦 ≠ 0 (|𝑥/𝑦| = |𝑥 |/|𝑦|), (v) ∀ 𝑥 ,𝑦 ∈ ℝ (𝑥 2 < 𝑦 2 ⟺ |𝑥 | < |𝑦|), (vi) ∀𝑥 ∈ℝ (√𝑥 2 = |𝑥 |), (vii) ∀ 𝑘 > 0 , 𝑥 ∈ ℝ (|𝑥 | < 𝑘 ⟺ −𝑘 < 𝑥 < 𝑘), (viii) ∀ 𝑘 > 0 , 𝑥 ∈ ℝ (|𝑥 | ≤ 𝑘 ⟺ −𝑘 ≤ 𝑥 ≤ 𝑘), (ix) ∀𝑘 < 0 ,𝑥 ∈ ℝ (|𝑥 | > 𝑘 ⟺ 𝑥 < −𝑘 𝑜𝑟 𝑥 > 𝑘), (x) ∀ 𝑥 ,𝑦 ∈ ℝ (|𝑥 + 𝑦| ≤ |𝑥 | + |𝑦|), (xi) ∀ 𝑥 ,𝑦 ∈ ℝ (|𝑥 − 𝑦| ≥ |𝑥 | − |𝑦|). - 13 - Proof: (i) If −𝑥 ≥ 0, |𝑥 | = 𝑥 ≥ 0. If 𝑥 < 0, |𝑥 | = −𝑥 ≥ 0. (ii) Exercise. (iii) Depending on the signs of x and y. We study |𝑥. 𝑦| 𝑎𝑛𝑑 |𝑥 |. |𝑦| 𝑥 𝑦 𝑥𝑦 |𝑥| |𝑦| |𝑥 𝑦| |𝑥|. |𝑦| + + + 𝑥 𝑦 𝑥𝑦 𝑥𝑦 + - - 𝑥 −𝑦 −𝑥 𝑦 −𝑥 𝑦 - + - −𝑥 𝑦 −𝑥 𝑦 −𝑥 𝑦 - - + −𝑥 −𝑦 𝑥𝑦 𝑥𝑦 So,|𝑥 𝑦| = |𝑥 | |𝑦| (iv) The proof is similar to the proof of (iii). (v) First, we suppose that |𝑥 | < |𝑦|. So, 𝑥 2 = |𝑥 |. |𝑥 | < |𝑥 |. |𝑦| < |𝑦|. |𝑦| = 𝑦 2 Now, we suppose that 𝑥 2 < 𝑦 2. If |𝑥 | ≥ |𝑦| then 𝑥 2 = |𝑥 ||𝑥 | ≥ |𝑥 ||𝑦| ≥ |𝑦||𝑦| = 𝑦 2 a contradiction. Hence, |𝑥 | < |𝑦| (vi) Since |𝑥 | is nonnegative and |𝑥 |2 = 𝑥 2 then √𝑥 2 = |𝑥 |. - 14 - (vii) Note that |𝑥 | < 𝑘 ⟺ 𝑥 2 < 𝑘 2 ⟺ 𝑥 2 − 𝑘 2 < 0 ⟺ (𝑥 − 𝑘)(𝑥 + 𝑘) < 0. From the following table we have |𝑥 | < 𝑘 ⟺ 𝑥 ∈ (−𝑘 , 𝑘) -k 0 k --- --- +++ 𝑥−𝑘 --- +++ +++ 𝑥+𝑘 + - + 𝑥2 − 𝑘 (viii) Since |𝑥 | = 𝑘 if and only if 𝑥 = 𝑘 or 𝑥 = −𝑘, then, we have by (vii) that 𝑥 ∈ [𝑘 , 𝑘] ⟺ |𝑥 | ≤ 𝑘. (ix) |𝑥 | > 𝑘 ⟺ 𝑥 2 > 𝑘 2 ⟺ 𝑥 2 − 𝑘 2 > 0 ,and from the above table, we have |𝑥 | > 𝑘 ⟺ 𝑥 > 𝑘 𝑜𝑟 𝑥 < −𝑘 (x) Since (𝑥 + 𝑦)2 = 𝑥 2 + 𝑦 2 + 2𝑥 𝑦 ≤ 𝑥 2 + 𝑦 2 + 2|𝑥 𝑦| = |𝑥 |2 + |𝑦|2 + 2|𝑥 ||𝑦| = (|𝑥 | + |𝑦|)2 Then from (v) |𝑥 + 𝑦| ≤ ||𝑥 | + |𝑦|| = |𝑥 | + |𝑦| (xi) From (x) we have |𝑥 | = |𝑥 − 𝑦 + 𝑦| ≤ |𝑥 − 𝑦| + |𝑦| So, |𝑥 | − |𝑦| ≤ |𝑥 − 𝑦| - 15 - Similarly, |𝑦| − |𝑥 | ≤ |𝑥 − 𝑦| But, |𝑥 − 𝑦| = |𝑦 − 𝑥 | then |𝑥 − 𝑦| ≥ −(|𝑥 | − |𝑦|) and |𝑥 − 𝑦| ≥ ||𝑥 | − |𝑦|| Hence |𝑥 − 𝑦| ≥ ||𝑦| − |𝑥 || Example 1.2.3. Solve each inequality, and then sketch the graph of its solution: (a) |3𝑥 − 2| < 9 (b) |5 − 2𝑥 | > 2 (c) |𝑥 + 4| < |2𝑥 − 1| (d) |𝑥 | + |𝑥 − 2| < 3 𝑥 (e) | | 2 - 16 - ⟺ 7 < 2𝑥 𝑜𝑟 3 > 2𝑥 7 3 ⟺ < 𝑥 𝑜𝑟 > 𝑥 2 2 The solution is the real numbers in the union 3 7 (−∞ , ) ⋃ ( , ∞) 2 2 0 3 2 7 2 (c) |𝑥 + 4| < |2𝑥 − 1| ⟺ (𝑥 + 4)2 < (2𝑥 − 1)2 ⟺ 𝑥 2 − 8𝑥 + 16 < 4𝑥 2 − 4𝑥 + 1 ⟺ 0 < (𝑥 − 5)(𝑥 + 1) -1 5 - - + 𝑥−5 - + + 𝑥+1 + - + (𝑥 − 5)(𝑥 + 1) The solution is the real numbers in the union (−∞ , −1) ⋃(5 , ∞) -1 0 5 7 - 17 - (d) Let 𝑃(𝑥 ) = |𝑥 | + |𝑥 − 2|. Then the values of 𝑃(𝑥) are as follows: 0 2 2  2x 2 2x  2 First: In the interval [2 , ∞) ∶ 5 We have that 𝑃(𝑥 ) < 3 ⟺ 2𝑥 − 2 < 3 ⟺ 𝑥 < , and the 2 5 solution is [2 , ∞)⋂ (−∞ , ) = [2 , 5/2). 2 Second: In[0 ,2): We have that 𝑃(𝑥 ) < 3 ⟺ 2 < 3 and this is always true. So, [0 ,2) is the solution. Third: In (−∞, 0): 1 𝑃(𝑥 ) < 3 ⟺ 2 − 2𝑥 < 3 ⟺ −1 < 2𝑥 ⟺ 𝑥 > − 2 1 1 So, the solution is (−∞, 0)⋂ (− , ∞) = (− ,0), 2 2 Hence, the whole solution is the real number in the union 1 5 1 5 (− ,0) ⋃ [0 ,2)⋃ [2 , ) = (− , ). 2 2 2 2 𝑥 𝑥 2 (e) We have 𝑥 ≠ −2 and | | < 1 ⟺ ( ) < 1 2+𝑥 2+𝑥 2 2 ⟺ 𝑥 < (2 + 𝑥 ) ⟺ 0 < 4 + 4𝑥 ⟺ 𝑥 > −1 ⟺ 𝑥 ∈ (−1 , ∞) So, the solution is (−1, ∞)\{−2} = (−1, ∞). - 18 - Example 1.2.4. |𝑥−3| |𝑥|+3 Prove that ≤ , ∀𝑥 ∈ ℝ. 𝑥 2 +4 4 Solution: 1 1 Since 𝑥 2 + 4 ≥ 4, for every 𝑥 ∈ ℝ, then ≤. 𝑥 2 +4 4 |𝑥−3| |𝑥−3| Thus, ≤. Also, |𝑥 − 3| ≤ |𝑥 | + 3. 𝑥 2 +4 4 |𝑥−3| |𝑥|+3 So, ≤. 𝑥 2 +4 4 Exercise Set (1.2) In the following, solve each inequality, and then sketch the graph of its solution: (1) 2𝑥 + 7 > 7 ; (2) 1−𝑥 ≤2; 1 (3) 4 + 5𝑥 ≤ 3𝑥 − 7; (4) 𝑥 + 6 ≥ 14 5 (5) 3𝑥 + 2 < 5𝑥 − 8; (6) −4 < 2 − 9𝑥 < 5; 3−7𝑥 5 (7) −1 < ≤ 6; (8) > 0; 4 7−2𝑥 1 (9) ≤ −2; (10) 𝑥 (𝑥 − 2) ≤ 0 ; 𝑥−2 1 2 (11) 2𝑥 2 + 𝑥 ≤ 1; (12) > ; 𝑥−2 𝑥+3 (13) 𝑥 2 + 𝑥 > 1; (14) 𝑥 3 > 𝑥; - 19 - 2 3 (15) 𝑥 2 + 3𝑥 < 4𝑥 2 ; (16) < ; 𝑥 𝑥−4 1 3 1 1 (17) ≥ ; (18) > ; 𝑥+1 𝑥−2 𝑥 𝑥−1 2 1 3𝑥+2 (19) > ; (20) ≤0; 𝑥+1 𝑥 2𝑥−7 (21) |2𝑥 − 3| ≤ 0.4 ; (22) |5𝑥 − 2| < 6 ; 7−3𝑥 (23) |7𝑥 + 1| > 3; (24) | | ≤ 1; 2 3𝑥+17 (25) | |>9 (26) |6 − 4𝑥 | ≥ |𝑥 − 2| 4 (27) |2𝑥 − 5| ≤ |𝑥 + 4| ; (28) |𝑥 + 1| + |𝑥 − 2| < 7 ; 1 𝑥−2 (29) |𝑥 | > |𝑥 − 1| ; (30) | 1 | ≤1 ; 𝑥+2 2−3𝑥 (31) | |≤4. 1+2𝑥 𝑥+𝑦 (32) Prove that √𝑥𝑦 ≤ , for each 𝑥 , 𝑦 ∈ ℝ+ , and 2 then prove that 𝑎𝑐 + 𝑏𝑑 ≤ √𝑎2 + 𝑏 2 √𝑐 2 + 𝑑2 , ∀𝑎 , 𝑏 , 𝑐 , 𝑑 ∈ ℝ (33) Use the mathematical induction to prove that (1 + 𝑥 )𝑛 ≥ 1 + 𝑛𝑥 for 𝑛 ≥ 1 and 𝑥 > −1, n is integer. 𝑎 𝑎+2𝑏 (34) Prove that √2 is between and for 𝑎, 𝑏 ∈ ℝ+ 𝑏 𝑎+𝑏 - 20 - Chapter 3 Complex Numbers 3 COMPLEX NUMBERS Objectives After studying this chapter you should understand how quadratic equations lead to complex numbers and how to plot complex numbers on an Argand diagram; be able to relate graphs of polynomials to complex numbers; be able to do basic arithmetic operations on complex numbers of the form a + ib ; understand the polar form [ r, θ ] of a complex number and its algebra; understand Euler's relation and the exponential form of a complex number re i θ ; be able to use de Moivre's theorem; be able to interpret relationships of complex numbers as loci in the complex plane. 3.0 Introduction The history of complex numbers goes back to the ancient Greeks who decided (but were perplexed) that no number existed that satisfies 2 x = −1 For example, Diophantus (about 275 AD) attempted to solve what seems a reasonable problem, namely 'Find the sides of a right-angled triangle of perimeter 12 units and area 7 squared units.' C Letting AB = x, AC = h as shown, h then a rea = 1 2 xh 2 2 and perimeter = x + h + x + h A x B 55 Chapter 3 Complex Numbers Activity 1 Show that the two equations above reduce to 2 6x − 43x + 84 = 0 when perimeter = 12 and area = 7. Does this have real solutions? A similar problem was posed by Cardan in 1545. He tried to solve the problem of finding two numbers, a and b, whose sum is 10 and whose product is 40; i.e. a + b = 10 (1) ab = 40 (2) Eliminating b gives a(10 − a) = 40 2 or a − 10a + 40 = 0. Solving this quadratic gives 1 a= (10 ± −60 ) = 5 ± −15 2 This shows that there are no real solutions, but if it is agreed to continue using the numbers a = 5 + −15, b = 5 − −15 then equations (1) and (2) are satisfied. Show that equations (1) and (2) are satisfied by these values of x and y. So these are solutions of the original problem but they are not real numbers. Surprisingly, it was not until the nineteenth century that such solutions were fully understood. The square root of −1 is denoted by i, so that i= −1 and a = 5 + 15 i , b = 5 − 15 i are examples of complex numbers. 56 Chapter 3 Complex Numbers Activity 2 The need for complex numbers Solve if possible, the following quadratic equations by factorising or by using the quadratic formula. If a solution is not possible explain why. (a) x 2 − 1 = 0 (b) x 2 − x − 6 = 0 (c) x 2 − 2x − 2 = 0 (d) x 2 − 2x + 2 = 0 You should have found (a), (b) and (c) straightforward to solve but in (d) a term appears in the solution which includes the square root of a negative number and to obtain solutions you need to use the symbol i = −1 , or i 2 = −1 It is then possible to obtain a solution to (d) in Activity 2. Example Solve x 2 − 2x + 2 = 0. Solution Using the quadratic formula 2 −b ± b − 4ac x = 2a − ( −2 ) ± ( −2 )2 − 4 (1) (2 ) ⇒ x = 2 (1) 2± −4 ⇒ x = 2 But −4 = 4 ( −1) = 4 −1 = 2 −1 = 2i (using the definition of i). 2 ± 2i Therefore x = 2 ⇒ x = 1±i Therefore the two solutions are x = 1 + i and x = 1 − i 57 Chapter 3 Complex Numbers Activity 3 Solve the following equations, leaving your answers in terms of i: 2 2 (a) x + x + 1 = 0 (b) 3x − 4x + 2 = 0 2 2 (c) x + 1 = 0 (d) 2x − 7 = 4x The set of solutions to a quadratic equation such as ax + bx + c = 0 2 can be related to the intercepts on the x-axis when the graph of the function f ( x ) = ax + bx + c 2 is drawn. Activity 4 Quadratic graphs Using a graphics calculator, a graph drawing program on a computer, a spreadsheet or otherwise, draw the graphs of the following functions and find a connection between the existence or not of real solutions to the related quadratic equations. (a) f ( x ) = x − 1 (b) f ( x ) = x − x − 6 2 2 (c) f ( x ) = x − 2x − 2 (d) f ( x ) = x + x + 1 2 2 (e) f ( x ) = 3x − 4x + 2 (f) f ( x ) = x + 1 2 2 You should have noted that if the graph of the function either intercepts the x-axis in two places or touches it in one place then the solutions of the related quadratic equation are real, but if the graph does not intercept the x-axis then the solutions are complex. 2 If the quadratic equation is expressed as ax + bx + c = 0 , then the 2 expression that determines the type of solution is b − 4ac , called the discriminant. In a quadratic equation ax 2 + bx + c = 0, if: b 2 − 4ac > 0 then solutions are real and different b 2 − 4ac = 0 then solutions are real and equal b 2 − 4ac < 0 then solutions are complex 58 Chapter 3 Complex Numbers 3.1 Complex number algebra A number such as 3 + 4i is called a complex number. It is the sum of two terms (each of which may be zero). The real term (not containing i) is called the real part and the coefficient of i is the imaginary part. Therefore the real part of 3 + 4i is 3 and the imaginary part is 4. A number is real when the coefficient of i is zero and is imaginary when the real part is zero. e.g. 3 + 0 i = 3 is real and 0 + 4i = 4i is imaginary. Having introduced a complex number, the ways in which they can be combined, i.e. addition, multiplication, division etc., need to be defined. This is termed the algebra of complex numbers. You will see that, in general, you proceed as in real numbers, but using i = −1 2 where appropriate. But first equality of complex numbers must be defined. If two complex numbers, say a + bi, c + di are equal, then both their real and imaginary parts are equal; a + bi = c + di ⇒ a = c and b = d Addition and subtraction Addition of complex numbers is defined by separately adding real and imaginary parts; so if z = a + bi, w = c + di then z + w = (a + c) + (b + d)i. Similarly for subtraction. Example Express each of the following in the form x + yi. (a) (3 + 5i ) + (2 − 3i ) (b) (3 + 5i ) + 6 (c) 7i − ( 4 + 5i ) 59 Chapter 3 Complex Numbers Solution (a) (3 + 5i ) + (2 − 3i ) = 3 + 2 + (5 − 3)i = 5 + 2i (b) (3 + 5i ) + 6 = 9 + 5i (c) 7i − ( 4 + 5i ) = 7i − 4 − 5i = −4 + 2i Multiplication Multiplication is straightforward provided you remember that i 2 = −1. Example Simplify in the form x + yi : (a) 3(2 + 4i ) (b) (5 + 3i )i (c) (2 − 7i )(3 + 4i ) Solution (a) 3 ( 2 + 4i ) = 3 ( 2 ) + 3 ( 4i ) = 6 + 12i (b) (5 + 3i ) i = ( 5) i + (3i ) i = 5i + 3 i( ) = 5i + (−1) 3 = − 3 + 5i 2 (c) (2 − 7i )(3 + 4i ) = (2 ) (3) − ( 7i ) (3) + (2 ) ( 4i ) − ( 7i ) ( 4i ) = 6 − 21i + 8i − ( −28) = 6 − 21i + 8i + 28 = 34 − 13i In general, if z = a + bi , w = c + di , then z w = (a + bi)(c + di) = a c − b d + (a d + b c)i 60 Chapter 3 Complex Numbers Activity 5 Simplify the following expressions: (a) ( 2 + 6i ) + (9 − 2i ) (b) (8 − 3i ) − (1 + 5i ) (c) 3 ( 7 − 3i ) + i ( 2 + 2i ) (d) (3 + 5i )(1 − 4i ) (e) (5 + 12i )(6 + 7i ) (f) ( 2 + i )2 3 4 (g) i (h) i (i) (1 − i)3 (j) (1 + i)2 + (1 − i)2 (k) (2 + i) + (2 − i) 4 4 (l) ( a + ib) ( a − ib ) Division The complex conjugate of a complex number is obtained by changing the sign of the imaginary part. So if z = a + bi , its Note: an alternative notation complex conjugate, z , is defined by often used for the complex z = a − bi conjugate is z*. Any complex number a + bi has a complex conjugate a − bi and from Activity 5 it can be seen that ( a + bi )( a − bi ) is a real number. This fact is used in simplifying expressions where the denominator of a quotient is complex. Example Simplify the expressions: 1 3 4 + 7i (a) (b) (c) i 1+ i 2 + 5i Solution To simplify these expressions you multiply the numerator and denominator of the quotient by the complex conjugate of the denominator. (a) The complex conjugate of i is −i , therefore 1 1 −i (1)( −i ) −i = × = = = −i i i −i (i )( −i ) −( −1) (b) The complex conjugate of 1 + i is 1 − i , therefore 3 3 1− i 3 (1 − i ) 3 − 3i 3 3 = × = = = − i 1+ i 1+ i 1− i (1 + i ) (1 − i ) 2 2 2 61 Chapter 3 Complex Numbers (c) The complex conjugate of 2 + 5i is 2 − 5i therefore 4 + 7i 4 + 7i 2 − 5i 43 − 6i 43 6 = × = = − i 2 + 5i 2 + 5i 2 − 5i 29 29 29 Activity 6 Division Simplify to the form a + ib 4 1− i 4 + 5i 4i (a) (b) (c) (d) i 1+ i 6 − 5i (1 + 2i )2 3.2 Solving equations Just as you can have equations with real numbers, you can have equations with complex numbers, as illustrated in the example below. Example Solve each of the following equations for the complex number z. (a) 4 + 5i = z − (1 − i ) (b) (1 + 2 i ) z = 2 + 5i Solution (a) Writing z = x + iy , 4 + 5i = ( x + y i ) − (1 − i ) 4 + 5i = x − 1 + ( y + 1)i Comparing real parts ⇒ 4 = x − 1, x = 5 Comparing imaginary parts ⇒ 5 = y +1, y = 4 So z = 5 + 4i. In fact there is no need to introduce the real and imaginary parts of z, since 4 + 5i = z − (1 − i ) ⇒ z = 4 + 5i + (1 − i ) ⇒ z = 5 + 4i (b) (1 + 2i ) z = 2 + 5i 2 + 5i z= 1 + 2i 62 Chapter 3 Complex Numbers 2 + 5i 1 − 2i z= × 1 + 2i 1 − 2i 12 + i 12 1 z= = + i 5 5 5 Activity 7 (a) Solve the following equations for real x and y (i) 3 + 5i + x − yi = 6 − 2i (ii) x + yi = (1 − i)(2 + 8i). (b) Determine the complex number z which satisfies z(3 + 3i) = 2 − i. Exercise 3A 1. Solve the equations: 7. Write in the form x + yi : (a) x + 9 = 0 2 (b) 9x + 25 = 0 2 2 + 3i −4 + 3i 4i (a) (b) (c) 1+ i −2 − i 2−i (c) x + 2 x + 2 = 0 (d) x + x + 1 = 0 2 2 1 3 − 2i p + qi (e) 2 x + 3x + 2 = 0 2 (d) (e) (f) 2 + 3i i r + si 2. Find the quadratic equation which has roots 8. Simplify: 2 ± 3i. (2 + i)(3 − 2i) (1 − i) 3 3. Write the following complex numbers in the (a) (b) 1+ i (2 + i)2 form x + yi. (a) (3 + 2i) + (2 + 4i) (b) (4 + 3i) − (2 + 5i) 1 1 (c) − (c) (4 + 3i) + (4 − 3i) (d) (2 + 7i) − (2 − 7i) 3+i 3−i 9. Solve for z when (e) (3 + 2i)(4 − 3i) (f) (3 + 2i)2 (a) z(2 + i) = 3 − 2i (b) (z + i)(1 − i) = 2 + 3i (g) (1 + i)(1 − i)(2 + i) 1 1 3 4. Find the value of the real number y such that (c) + = (3 + 2i)(1 + iy) z 2−i 1+ i is (a) real (b) imaginary. 10. Find the values of the real numbers x and y in each of the following: 5. Simplify: x y 1 1 1 (a) + =1 (a) i 3 (b) i 4 (c) (d) 2 (e) 3 1+ i 1 − 2i i i i x yi 2 6. If z = 1 + 2i , find (b) + = 2−i i+3 1+ i 2 1 1 (a) z (b) (c) 2 z z 63 Chapter 3 Complex Numbers 11. Given that p and q are real and that 1 + 2i is a 12. The complex numbers u, v and w are related by root of the equation 1 1 1 = +. z 2 + ( p + 5i)z + q(2 − i) = 0 u v w determine: Given that v = 3 + 4i, w = 4 − 3i , find u in the (a) the values of p and q; form x + iy. (b) the other root of the equation. 3.3 Argand diagram Any complex number z = a + bi can be represented by an ordered pair (a, b) and hence plotted on xy-axes with the real part measured along the x-axis and the imaginary part along the y-axis. This graphical representation of the complex number field is called an Argand diagram, named after the Swiss mathematician Jean Argand (1768-1822). Example Imaginary imaginary z = 3 + 2i 2 Represent the following complex numbers on an Argand 1 diagram: –3 –2 –1 1 2 3 4 5 (a) z = 3 + 2i (b) z = 4 − 5i (c) z = −2 − i –1 real Real z = –2 – i –2 Solution –3 The Argand diagram is shown opposite. –4 –5 Activity 8 z = 4 – 5i Let z 1= 5 + 2i , z 2 = 1 + 3i , z 3 = 2 − 3i , z 4 = −4 − 7i. (a) Plot the complex numbers z 1 , z 2 , z 3 , z 4 on an Argand diagram and label them. (b) Plot the complex numbers z 1+ z 2 and z 1− z 2 on the same Argand diagram. Geometrically, how do the positions of the numbers z 1+ z 2 and z 1− z 2 relate to z 1 and z 2 ? 64 Chapter 3 Complex Numbers 3.4 Polar coordinates Consider the complex number z = 3 + 4 i as represented on an Imaginary Argand diagram. The position of A can be expressed as coordinates (3, 4), the cartesian form, or in terms of the length 4 A and direction of OA. 3 2 Using Pythagoras' theorem, the length of OA = 3 + 4 = 5. 2 2 1 This is written as z = r = 5. z is read as the modulus or Real O 1 2 3 4 absolute value of z. The angle that OA makes with the positive real axis is  4 θ = tan −1   = 53.13° (or 0.927 radians).  3 This is written as arg( z ) = 53.13°. You say arg( z ) is the y (a, b) argument or phase of z. The parameters z and arg( z ) are in fact the equivalent of polar r coordinates r, θ as shown opposite. There is a simple b connection between the polar coordinate form and the cartesian or rectangular form (a, b): θ a x a = r cos θ , b = r sin θ. Therefore z = a + bi = r cos θ + r i sin θ = r ( cos θ + i sin θ ) where z = r, and arg( z ) = θ. It is more usual to express the angle θ in radians. Note also that it is convention to write the i before sin θ , i.e. i sin θ is preferable to sin θ i. In the diagram opposite, the point A could be labelled 2 3, 2 ( ) or as 2 3 + 2i. 2.0 A The angle that OA makes with the positive x-axis is given by 1.5 −1  2  −1  1  1.0 θ = tan   = tan  . 0.5  2 3  3 O0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 π π π Therefore θ = or 2 π + or 4 π + or... etc. There is an 6 6 6 infinite number of possible angles. The one you should normally use is in the interval − π < θ ≤ π , and this is called the principal argument. 65 Chapter 3 Complex Numbers Using polar coordinates the point A could be labelled with its  π polar coordinates [ r, θ ] as  4, . Note the use of squared  6 brackets when using polar coordinates. This is to avoid confusion with Cartesian coordinates.  π π  Thus 2 3 + 2i = 4  cos   + i sin   .   6  6  Important note: if you are expressing a + ib in its polar form, b where a and b are both positive, then the formula θ = tan −1 is a quite sufficient. But in other cases you need to think about the position of a + ib in the Argand diagram. Example Write z = −1 − i in polar form. Solution Now z = a + ib where a = −1 and b = −1 and in polar form the 2 2 modulus of z = z = r = 1 + 1 = 2 and the argument is y 2 5π 3π (or 225°) : its principal value is −. 4 4 1 −3π  Hence z =  2 , in polar coordinates. (The formula  4  –2 –1 1 2 x b π tan −1 would have given you.) –1 – i –1 a 4 –2 Activity 9 (a) Write the following numbers in [ r, θ ] form: (i) 7 + 2i (ii) 3 − i (iii) −4 + 6i (iv) − 3 − i (b) Write the following in a + bi form: (remember that the angles are in radians)  π (i) 3,  (ii) [ 5, π ] (iii) [6, 4.2 ]  4 −2 π  (iv)  2 ,  3  66 Chapter 3 Complex Numbers 3.5 Complex number algebra You will now investigate the set of complex numbers in the modulus/argument form, [ r, θ ]. Suppose you wish to combine two complex numbers of the form z1 = [r , θ ] 1 1 z2 = [r , θ ] 2 2 Note that, in a + bi form, z 1 = r1 cos θ1 + i r1 sin θ1 and z 2 = r 2 cos θ 2 + i r 2 sin θ 2 So ( )( z 1 z 2 = r1 cos θ1 + i r1 sin θ1 r2 cos θ 2 + i r2 sin θ 2 ) = r 1 r 2( cos θ1 + i sin θ1 ) ( cos θ 2 + i sin θ 2 ) [ = r 1 r 2 ( cos θ1 cos θ 2 − sin θ1 sin θ 2 ) + (sin θ1 cos θ 2 + cos θ1 sin θ 2 )i. ] Simplify the expressions in the brackets. Using the formulae for angles, [ z 1 z 2 = r 1 r 2 cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) ] or, in polar notation [ z 1 z 2 = r 1 r 2, θ1 + θ 2.] For example, [3, 0.5] × [ 4, 0.3] = [12, 0.8]. That is, the first elements of the ordered pairs are multiplied and the second elements are added. Activity 10 Given that z 1 = [3, 0. 7] , z 2 = [2, 1.2 ] and z 3 = [ 4, − 0.5] , (a) find z1 × z2 and z1 × z3 (b) show that [1, 0 ] × z 1= z 1 67 Chapter 3 Complex Numbers (c) (i) find a complex number z = [ r, θ ] such that z × z 2 = [1, 0 ]. (ii) find a complex number z = [ r, θ ] such that z × z 3 = [1, 0 ]. (d) for any complex number [ r, θ ] show that 1   , − θ  × [ r, θ ] = [1, 0 ] (r > 0).  r  Activity 11 = B5* sin(C5* A5) Use a spreadsheet package to plot numbers on an Argand diagram by entering numbers and formulae into cells A5 to E5 = B5* cos(C5* A5) as shown opposite. A5 B5 C5 D5 E5 Cells D5 and E5 calculate the x and y coordinates respectively of the complex number whose modulus and argument are in π 2 0.25 cells B5 and C5 (the argument is entered as a multiple of π ). A second number can be entered in cells B6 and C6 and its (x, y) coordinates calculated by using appropriate formulae in cells D6 and E6. This can be repeated for further numbers (the spreadsheet facility 'FILL DOWN' is useful here). Use the appropriate facility on your spreadsheet to plot the (x, y) values. 2 1 Label rows and columns if it makes it easier. –4 –3 –2 –1 1 2 3 4 –1 Experiment with different values of r and θ. –2 An example is shown in the graph opposite and the related –3 spreadsheet below. –4 –5 68 Chapter 3 Complex Numbers Exercise 3B 1. Mark on an Argand diagram the points 8. In this question, angles are in radians. representing the following numbers: (a) (i) Plot the following complex numbers on (a) 2 (b) 3i (c) −i (d) 1 + 2i (e) 3 − i an Argand diagram and label them: (f) −2 + 3i π −π  z1 = [ 4,0 ] , z2 = 3,  , z3 =  2, 2. The points A, B, C and D represent the numbers  2  2  z 1 , z 2 , z 3, and z 4 and O is the origin. π 5π  z 4 = 3,  , z5 =  2, (a) If OABC is a parallelogram, and z 1 = 1 + i ,  3  3  z 2 = 4 + 5i , find z 3. π (ii) Let the complex number z = 1,   2 (b) Find z 2 and z 4 when ABCD is a square and Calculate z × z1 , z × z2 , etc. and plot the points on the same diagram as in (i). (i) z 1 = 2 + i, z 3 = 6 + 7i What do you notice? (ii) z 1 = 6 − 2i, z 3 = 6i π (b) Repeat (a) (ii) using z = 1,   3 3. Find the modulus and argument of (c) In general, what happens when a complex (a) 1 − i (b) 1 + 3i (c) 3 − 3i (d) 3 + 2i number is multiplied by [1, θ ] ? Make up some 4. Show that examples to illustrate your answer. (a) z = z (b) arg z = − arg z π and illustrate these results on an Argand (d) Repeat (a) (ii) using z =  0.5,   2 diagram. (e) In general, what happens when a complex 5. Find the modulus and argument of z 1 , z 2 , z 1 z 2 π z1 number is multiplied by  0.5,  ? Make up and when z 1= 1 + i and z 2 = 3 + i. What do  2 z2 some examples to illustrate your answer. you notice? π 6. Write in the form a + bi (f) Repeat (e) for 3,   3  π  π (g) Describe what happens when a complex (a)  4,  (b)  5,   3  2 π number is multiplied by 3, . Make up  3π   3 (c) 3 2 , −  (d) [ 4, 13 π ] some examples to illustrate your answer.  4  7. Write in polar form (a) 1 + i (b) −2 + i (c) −5 (d) 4i (e) 3 + 4i (f) −3 − 4i (g) 3 − 4i (h) −3 + 4i 3.6 De Moivre's theorem An important theorem in complex numbers is named after the French mathematician, Abraham de Moivre (1667-1754). Although born in France, he came to England where he made the acquaintance of Newton and Halley and became a private teacher of Mathematics. He never obtained the university position he sought but he did produce a considerable amount of research, including his work on complex numbers. 69 Chapter 3 Complex Numbers The derivation of de Moivre's theorem now follows.  π π Consider the complex number z =  cos + i sin .  3 3  π π  π π z =  cos + i sin  ×  cos + i sin  2 Then  3 3   3 3 π π π π = cos 2 − sin 2 + 2i cos sin 3 3 3 3 2π 2π = cos + i sin 3 3 or with the modulus/argument notation  π z = 1,   3 2  π  π  2π  and z = 1,  × 1,  = 1, .  3  3  3  Remember that any complex number z = x + y i can be written in the form of an ordered pair [ r, θ ] where r = x + y 2 2 and  y θ = tan −1  .  x If the modulus of the number is 1, then z = cos θ + i sin θ and z2 = (cos θ + i sin θ )2 = cos 2 θ − sin 2 θ + 2i cos θ sin θ = cos 2θ + i sin 2θ z 2 = [1, θ ] = [1, 2θ ]. 2 i.e. Activity 12 (a) Use the principle that, with the usual notation, [r 1, θ1 ] × [r 2, θ 2 ] = [r 1 r 2, θ1 + θ 2 ] n  π π to investigate  cos + i sin  when n = 0, 1, 2, 3,..., 12.  6 6 70 Chapter 3 Complex Numbers (b) In the same way as in (a), investigate n  π π  3cos +3i sin   6 6 for n = 0, 1, 2,..., 6. You should find from the last activity that (cos θ + i sin θ )n = cos( nθ ) + i sin( nθ ). [ ] In [ r, θ ] form this is [ r, θ ] = r , nθ and de Moivre's theorem n n states that this is true for any rational number n. A more rigorous way of deriving de Moivre's theorem follows. Activity 13 Show that ( cos θ + i sin θ ) = cos nθ + i sin nθ for n = 3 and n = 4. n Activity 14 Show that (cos kθ + i sin kθ )(cos θ + i sin θ ) = cos( k + 1)θ + i sin( k + 1)θ. Hence show that if (cos θ + i sin θ )k = cos kθ + i sin kθ then (cos θ + i sin θ )k +1 = cos(( k + 1)θ ) + i sin(( k + 1)θ ). The principle of mathematical induction will be used to prove that ( cos θ + i sin θ ) = cos( nθ ) + i sin( nθ ) for all positive integers. n Let S(k) be the statement ' ( cos θ + i sin θ ) = cos kθ + i sin kθ '. k As S(1) is true and you have shown in Activity 14 that S(k) implies S( k + 1) then S(2) is also true. But then (again by Activity 14) S(3) is true. But then... Hence S(n) is true for n = 1, 2,3, K. This is the principle of mathematical induction (which you meet more fully later in the book). So for all positive integers n, 71 Chapter 3 Complex Numbers (cos θ + i sin θ )n = cos nθ + i sin nθ If n is a negative integer, then let m = −n 1 (cos θ + i sin θ )n = (cos θ + i sin θ )− m = (cos θ + i sin θ )m where m is positive and, from the work above, (cos θ + i sin θ )m = (cos mθ + i sin mθ ). 1 Therefore (cos θ + i sin θ )n = (cos mθ + i sin mθ ) Activity 15 Show that 1 = cos mθ − i sin mθ (cos mθ + i sin mθ ) and hence that ( cos θ + i sin θ ) = cos nθ + i sin nθ when n is a n negative integer. Hint : multiply top and bottom by ( cos mθ − i sin mθ ) and use the fact that sin( − A) = − sin( A). p When n is a rational number, i.e. n = where p and q are q integers, then as q is an integer q   p   p   cos  θ + i sin   θ  = ( cos pθ + i sin pθ )   q   q   Since p is an integer cos p θ + i sin p θ = ( cos θ + i sin θ ) , p and hence q   p   p   cos  θ + i sin   θ  = ( cos θ + i sin θ ) p   q   q   72 Chapter 3 Complex Numbers   p   p   cos  θ + i sin   θ  = ( cos θ + i sin θ ) q p Thus   q   q   Therefore cos n θ + i sin n θ = ( cos θ + i sin θ ) n for any rational number n and clearly this leads to (r (cos θ + i sin θ ))n = r n (cos nθ + i sin nθ ) 3.7 Applications of de Moivre's theorem There are many applications of de Moivre's theorem, including the proof of trigonometric identities. Example Prove that cos 3θ = cos3 θ − 3cos θ sin 2 θ. Solution By de Moivre's theorem: cos 3 θ + i sin 3 θ = ( cos θ + i sin θ ) 3 = cos θ + 3cos θ (i sin θ ) + 3cos θ (i sin θ ) + (i sin θ ) 3 2 2 3 = cos3 θ + 3i cos 2 θ sin θ − 3cos θ sin 2 θ − i sin 3 θ = cos θ − 3cos θ sin θ + i (3cos θ sin θ − sin θ ) 3 2 2 3 Comparing real parts of the equation above you obtain cos 3θ = cos3 θ − 3cos θ sin 2 θ Example Simplify the following expression: cos 2θ + i sin 2θ cos 3θ + i sin 3θ Solution cos 2θ + i sin 2θ (cos θ + i sin θ )2 1 = = cos 3θ + i sin 3θ (cos θ + i sin θ ) ( cos θ + i sin θ )1 3 73 Chapter 3 Complex Numbers = ( cos θ + i sin θ ) −1 = ( cos( − θ ) + i sin( − θ )) = cos θ − i sin θ Exercise 3C 1. Use de Moivre's theorem to prove the trig 3. Simplify the following expressions: identities: cos5θ + i sin 5θ cos θ − isin θ (a) sin 2 θ = 2sin θ cos θ (a) (b) cos2 θ − i sin 2 θ cos 4 θ − isin 4 θ (b) cos5θ = cos θ −10 cos θ sin θ + 5cos θ sin θ 5 3 2 4 2. If z = cos θ + isin θ then use de Moivre's theorem to show that: 1 1 (a) z + = 2 cos θ (b) z 2 + 2 = 2 cos2 θ z z 1 (c) z n + n = 2 cos nθ z Activity 16 Make an educated guess at a complex solution to the equation = B7 * sin(A5* C7) z = 1 and then use the facilities of the spreadsheet to raise it to 3 the power 3 and plot it on the Argand diagram. If it is a solution = B7 * cos(A5* C7) of the equation then the resultant point will be plotted at distance 1 unit along the real axis. The initial spreadsheet layout from B7 C7 D7 E7 Activity 11 can be adapted. In addition, the cells shown opposite are required. = B5^ 3 What does the long formula in cell C7 do? Is it strictly necessary in this context? ( = C5* 3 - 2 * int ( C5*3 2 ) ) Below are two examples of the output from a spreadsheet using these cells – the first one is not a cube root of 1 but the second is. 1 0.8 0.6 0.4 0.2 –1 –0.5 0.5 1 1 0.8 0.6 0.4 0.2 –1 –0.5 0.5 1 74 Chapter 3 Complex Numbers 3.8 Solutions of z = 1 3 Write down one solution of z 3 = 1. n=1 1 De Moivre's theorem can be used to find all the solutions of z 3 = 1. Let z = [ r, θ ] –1 1 then 3 3 [ z = [ r, θ ] = r ,3θ 3 ] and you can express 1 as 1 = [1, 2nπ ] where n is an integer. n=2 –1 Then [r ,3θ ] = [1, 2nπ ] 3 r = 1 and 3θ = 2nπ 3 Therefore 2nπ i.e. r = 1 and θ= 3 The solutions are then given by letting n = 0, 1, 2,... If n = 0, z1 = [1, 0 ] = 1  2π  2π 2π If n = 1, z2 = 1,  = cos + i sin  3  3 3 1 3 = − + i 2 2  4π  4π 4π If n = 2, z3 = 1,  = cos + i sin  3  3 3 1 3 = − − i 2 2 What happens if n = 3 , 4,... ? Activity 17 Cube roots of unity Plot the three distinct cube roots of unity on an Argand diagram. What do you notice? 75 Chapter 3 Complex Numbers Activity 18 Use de Moivre's theorem to find all solutions to the following equations and plot the results on an Argand diagram. (a) z 4 = 1 (b) z 3 = 8 (c) z 3 = i 3.9 Euler's theorem x You have probably already met the series expansion of e , namely x2 x3 x4 e = 1+ x + x + + +K 2! 3! 4! Also the series expansions for cos θ and sin θ are given by θ2 θ4 θ6 cos θ = 1 − + − +K 2! 4! 6! θ3 θ5 θ7 sin θ = θ − + − +K 3! 5! 7! Activity 19 (a) For each of the following values of θ , use the series for ex iθ with x replaced by iθ to calculate (to 4 d.p.) the value of e. (Write your answer in the form a + bi.) (i) θ = 0 (ii) θ = 1 (iii) θ = 2 (iv) θ = −0. 4 (b) Calculate cos θ and sin θ for each of the values in (a). (c) Find a connection between the values of e i θ , cos θ and sin θ for each of the values of θ given in (a) and make up one other example to test your conjecture. (d) To prove this for all values of θ , write down the series iθ expansions of e , cos θ and sin θ and show that iθ e = cos θ + i sin θ. 76 Chapter 3 Complex Numbers The previous activity has shown that e i θ = cos θ + i sin θ which is sometimes known as Euler's theorem. It is an important result, and can be used to derive de Moivre's theorem in a simple way. If z is any complex number then in polar form z = x + yi = r cos θ + r i sin θ = r ( cos θ + i sin θ ) iθ = re , using Euler's theorem. r y θ ( ) i ( nθ ) n Thus z n = re i θ = r nen i θ = r ne x or (r cos θ + ir sin θ )n = r n (cos(nθ ) + i sin(nθ )) ⇒ (cos θ + i sin θ )n = cos( nθ ) + i sin( nθ ) which is de Moivre's theoem. What assumptions about complex number algebra have been made in the 'proof' above? One interesting result can be obtained from Euler's theorem by putting θ = π. This gives iπ e = cos π + i sin π = −1 + i × 0. So ei π + 1 = 0 This is often referred to as Euler's equation, since it connects the five most 'famous' numbers 0, 1, π , e, i with a ' + ' and ' = ' sign! Try substituting other values of θ in Euler's theorem and see what equation is derived. 77 Chapter 3 Complex Numbers 3.10 Exponential form of a complex number When a complex number z has modulus r, which must be non- negative, and argument θ , which is usually taken such that it satisfies − π < θ ≤ π , you have already shown that it can be represented in the forms (i) r ( cos θ + i sin θ ) (ii) [ r, θ ] (iii) re i θ Expression (iii) is referred to as the exponential form of a complex number. Activity 20 Write each of the following complex numbers in the exponential form. π π 2π  (a) 2  cos + i cos  (b) 5, (c) 1 − i 3  3 3  3  3.11 Solving equations You have already investigated the solutions of the equation z 3 = 1 and similar equations using a spreadsheet and by using de Moivre's theorem. A similar approach will now be used to solve more complicated equations. Example Write down the modulus and argument of the complex number 4 − 4i. Solve the equation z 5 = 4 − 4i , expressing your answers in the exponential form. Solution 4 − 4i = {4 2 + ( −4 ) 2 }=4 2 As before it is often helpful to make a small sketch of an Argand diagram to locate the correct quadrant for the argument. 78 Chapter 3 Complex Numbers −π Imaginary arg( 4 − 4i ) = imaginary So 4 Therefore the complex number 4 − 4i can be expressed as 4 4 2 , − π  real  4  Real It is quite convenient to work using the polar form of a complex –4 4 – 4i number when solving z = 4 − 4i. 5 Let z = [ r, θ ] , then z 5 = r 5 , 5θ. [ ] So as to obtain all five roots of the equation, the argument is π considered to be 2n π − where n is an integer. 4 Equating the results π [r , 5θ ] = 4 5 2 , 2n π − 4  r5 = 4 2 ⇒ r = 2 π π 5θ = 2n π − ⇒ θ = (8n − 1) 4 20 Now choose the five appropriate values of n so that θ lies between − π and π. −17 π n = −2 ⇒ θ = 20 −9 π n = −1 ⇒ θ = 20 −π n=0 ⇒ θ = 20 7π n =1 ⇒ θ = 20 15π 3π n=2 ⇒ θ = or 20 4 The solutions in exponential form are therefore 3π − 17 πi − 920π i π − 20 7π i i 2e 20 , 2e , 2e , 2e 20 and 2e 4. 79 Chapter 3 Complex Numbers Activity 21 Show that 1 + i is a root of the equation z 4 = −4 and find each of the other roots in the form a + bi where a and b are real. Plot the roots on an Argand diagram. By considering the diagonals, or otherwise, show that the points are at the vertices of a square. Calculate the area of the square. Activity 22 Given that k ≠ 1 and the roots of the equation z 3 = k are α , β ( x − 2) and γ , use the substitution z = to obtain the roots of the ( x + 1) equation ( x − 2 )3 = k ( x + 1)3 Exercise 3D 1. By using de Moivre's theorem, find all solutions 4. Using the results from Question 1(b), solve the to the following equations, giving your answers equation in polar form. Plot each set of roots on an 1 + 27i ( x + 1) = 0 3 Argand diagram and comment on the symmetry. (a) z 4 = 16 (b) z 3 = −27i (c) z 5 = −1 giving your answers in the form a + bi. 5. Solve the equation z 3 = i ( z −1) giving your 3 2. Find the cube roots of (a) 1 + i (b) 2i − 2 answers in the form a + bi. giving your answers in exponential form. Plot the solutions on an Argand diagram and comment on your results. 3. Using the answers from Question 1(a), determine 6. Determine the four roots of the equation the solutions of the equation ( x + 1)4 = 16( x − 1)4 ( z − 2 )4 + ( z + 1)4 = 0 and plot them on an Argand diagram. giving your answers in the form a + bi. 3.12 Loci in the complex plane Suppose z is allowed to vary in such a way that z − 1 = 2. You could write z = x + i y and obtain {( x − 1) 2 } + y2 = 2 or ( x − 1)2 + y 2 = 4 80 Chapter 3 Complex Numbers You can immediately identify this as the cartesian equation of a Imaginary circle centre (1, 0 ) and radius 2. In terms of the complex plane, the centre is 1 + 0i. This approach could be adopted for most problems and the exercise is simply one in algebra, lacking any geometrical feel for the locus. Real Instead, if ω is a complex number, you can identify z − ω as the distance of z from the point represented by ω on the complex Imaginary plane. The locus z − 1 = 2 can be interpreted as the set of points that are 2 units from the point 1 + 0i; in other words, a circle centre 1 + 0i and radius 2. 1 Real Activity 23 Illustrate the locus of z in the complex plane if z satisfies (a) z − (3 + 2i ) = 5 (b) z − 2 + i = 1 + 3i (c) z + 2i = 2 (d) z−4 =0 Activity 24 Describe the path of a point which moves in a fixed plane so that it is always the same distance from two fixed points A and B. Illustrate the locus of z in the case when z satisfies z + 3 = z − 4i. You would probably have had some difficulty in writing down a cartesian equation of the locus in Activity 24, even though you could describe the locus geometrically. Activity 25 Describe the locus of z in the case where z moves in such a way that z = z + 2 − 2i. Now try to write down the cartesian equation of this locus which should be a straight line. By writing z = x + i y, try to obtain the same result algebraically. 81 Chapter 3 Complex Numbers Activity 26 Investigate the locus of P when P moves in the complex plane and represents the complex number z which satisfies z +1 = k z −1 for different values of the real number k. Why does k = 1 have to be treated as a special case? Example The point P represents the complex number z on an Argand diagram. Describe the locus geometrically and obtain a cartesian equation for the locus in the cases (a) z = z−4 (b) z + z−4 =6 (c) z =2z−4 Solution imaginary Imaginary (a) From your work in Activity 25, you should recognise this as a straight line. In fact, it is the mediator, or perpendicular bisector, of the line segment joining the origin to the point 4 + 0i. It should be immediately obvious that its cartesian equation 2 4 real Real is x = 2 ; however, writing z = x + iy z = x + iy = x − 4 + iy Squaring both sides gives x 2 + y 2 = ( x − 4) + y 2 2 leading to 0 = −8x + 16 or x = 2. imaginary Imaginary (b) You may be aware of a curve that is traced out when the sum of the distances from two fixed points is constant. You could try using a piece of string with its ends fastened to two fixed points. The curve is called an ellipse. –1 4 5 Real real A sketch of the locus is shown opposite. 82 Chapter 3 Complex Numbers You can obtain a cartesian equation by putting z = x + i y x + iy + x − 4 + iy = 6 So x 2 + y2 + (( x − 4 ) 2 + y2 = 6 ) ( x − 4)2 + y2 = 6 − ( x 2 + y2 ) [ ] 2 ⇒ x 2 − 8x + 16 + y 2 = 36 − 12 (x 2 ) + y2 + x 2 + y2 12 (x 2 ) + y 2 = 20 + 8x 3 (x 2 ) + y 2 = 5 + 2x ( ) 9 x 2 + y 2 = 25 + 20x + 4x 2 5x 2 − 20x + 9y 2 = 25 5( x − 2 ) + 9y 2 = 45 2 ( x − 2 )2 y2 + =1 9 5 (c) You should have discovered in Activity 26 that the locus Imaginary imaginary will be a circle when the relationship is of this form. It is called the circle of Apollonius. You could possibly sketch the locus without finding the cartesian equation. 2 4 6 real Real Let z = x + iy x + iy = 2 x − 4 + iy (x 2 + y2 = 2 ) (( x − 4 ) 2 + y2 ) ( x 2 + y 2 = 4 x 2 − 8x + 16 + y 2 ) 0 = 3x 2 + 3y 2 − 32x + 64 In order to find the centre and radius you can complete the square 32 64 x 2 + y2 − x+ =0 3 3 2  x − 16  + y2 = 256 − 64 = 64  3 9 3 9 16 8 Centre of circle is at + 0i and radius is. 3 3 83 Chapter 3 Complex Numbers Activity 27 By recognising the locus z − 2 = 3 z − 10 as the circle of Apollonius, use the idea of simple ratios to determine the coordinates of the centre and the radius of the circle. Check your answer by finding the cartesian equation of the circle. P Discover the locus of the vertex P 45o Activity 28 By folding a piece of paper, create an angle of 45° and cut it fixed Fixed point fixed out. Now mark two fixed points on a piece of paper and point Fixed point point explore the locus of the vertex as you keep the two sides of the cut-out in contact with the fixed points as shown. P2 You should find that P moves on the arc of a circle. P1 Alternatively, when you have a circle and two fixed points A and B, if you choose a sequence of points P1 , P2 , P3 ,... on the P3 circumference, what do you notice about the angles AP1B, AP2 B, AP3B, etc.? A B This is an example of the constant angle locus. Example The point P represents z in the complex plane. Find the locus of P in each of the cases below when z satisfies 5π (a) argz = 6 −π (b) arg( z − 2 + 3i ) = 4 z − 1 π (c) arg =  z + 1 4 imaginary Imaginary Locus locus Solution (a) The locus is a half-line starting at the origin making an 5 angle π with the real axis. real Real 6 84 Chapter 3 Complex Numbers

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