CHAPTER 2: BASIC ALGEBRA 44 Simplify each of the following: a. 4(5 + 2) b. 2 (2 − 2 + 6) c. (2 + 3)( − 10) d. e. Answer: a. 4(5 + 2) = (4)(5) + (4)(2) = 20 + 8 b. 2 (2 − 2 + 6) = ( 2 )(2) + ( 2 )(− 2 ) + 2 (6) = 2 3 − 4 + 6 2 c. (2 + 3)( − 10) = (2)() + (2)(−10) + (3)() + (3)(−10) = 2 2 − 20 + 3 − 30 = 2 2 − 17 − 30 d. e. When performing multiplication and division of algebraic expressions, we need to remember the Index Laws and distributive properties. The distributive properties let us multiply a sum by multiplying each addend separately and then add the products. Example 2 3 2 5 × 3 3 7 = (3 ∙ 3)( 1 ∙ 3 )( 2 ∙ 1 ) (5)(7) = 9 4 3 35 2 × 3 = 6 2 × 3 3 2 5 × 3 3 7
CHAPTER 2: BASIC ALGEBRA 45 1. 2 ∙ 3 = 2. 2 2 ∙ 4 −5 = 3. −7 × 9 × 4 × = 4. 10(4 − 3) = 5. (2 + )(4 − 7) = 6. 12(−20 + 11) = 7. 3( + 10)(4 − 3) = 8. −3 2 ( + 1)( − 2) = 9. −5[−3( − 2) + 4] = 10. −[7( − 3) − 5] Exercise 2
CHAPTER 2: BASIC ALGEBRA 46 Simplify each of the following: a. b. c. d. e. Answer: a. b. = 2 4 2 −3 To divide by a fraction, you multiply by the reciprocal of the fraction. Example 3 ÷ = × = 12 7 6 5 5 ÷ 4 2 3 3 5 4 5 3 2 4 3 5 + 10 2 − 5 2 3 5 + 10 2 − 5 2 ( − 7)( + 3) 3 ( + 3) 2 12 7 6 5 5 ÷ 4 2 3 3 5 = 12 7 6 5 5 × 5 4 2 3 3 = 12 7 6 5 × 5 5 × 4 2 3 3 = 60 7−2 6−3 5−3 20 = 3 5 3 2 4 5 3 2 4 = 4 5−1 3−1 1−4 2
CHAPTER 2: BASIC ALGEBRA 47 c. d. e. = ( − 7)( + 3) = + 3 − 7 − 21 1. 2. Exercise 3 3 5 + 10 2 − 5 2 = 3 5 5 2 + 10 2 5 2 − 5 2 = 3 5−2 5 + 10 2−2 5 − 1−2 5 = 3 3 5 + 2 0 1 − −1 5 = 3 3 5 + 2 0 1 − −1 5 = 3 3 5 + 2 − 1 5 3 5 + 10 2 − 5 2 = 3 5 5 2 + 10 2 5 2 − 5 2 = 3 5−2 5 + 10 2−2 5 − 1−2 5 ( − 7)( + 3) 3 ( + 3) 2 = ( − 3)( + 3) 3−2 7 5 = 2 2 4 5 =
CHAPTER 2: BASIC ALGEBRA 48 3. 4. 4. 5. 6. 7. 8. 9. − 2 2 − = (− 60 4 3 102 2 ) = ( 140 3 2 ∙ 7 2 ) ∙ 7 = − 84 2 4 −7 ∙ (−2) ∙ 6 = 40 2 2 −5 ∙ (−4) = 20( + 1) 3 5( + 1) = ( − 3) 2 (2 + 5) 4 ( − 3)(2 + 5) 2 = 11(4 + 7) −2 5(4 + 7) −3 =
CHAPTER 2: BASIC ALGEBRA 49 2 + + = 0 2.2.1 Solve quadratic equations by using factorization and quadratic formula. Quadratic equation is an equation that satisfies both of the following conditions: a) There is only one variable in the equation. b) The word “quadratic” comes from “quadratum”, the Latin word for square. Hence, the highest power of the variable is 2. The general form of quadratic equation is where, a, b and c are constants, ≠ 0 and x as the unknown variable that can also be represented by other letters of alphabet. Examples of quadratic equations: a) 2 + 2 + 1 = 0 b) 2 2 + 4 + 2 = 0 c) 3 2 + 8 = 0 d) 10 2 + 12 = 0 There are two methods of solving quadratic equations, which are: a) by factorization, and b) by using quadratic formula. 2.2 Show quadratic equation
CHAPTER 2: BASIC ALGEBRA 50 Determine whether the following are quadratic equations. a. ( + 2)( + 5) = 0 b. 2( + 3) = 7 c. 3 + 2 2 + 6 = 2 d. 2 + 6 + 9 = 3 Answer: a. ( + 2)( + 5) = 0 2 + 5 + 2 + 10 = 0 2 + 7 + 10 = 0 Quadratic equation b. 2( + 3) = 7 2 2 + 6 = 7 2 2 + 6 − 7 = 0 Quadratic equation c. 3 + 2 2 + 6 = 2 3 + 2 2 + 6 − 2 = 0 3 + 2 2 − 2 + 6 = 0 Not a quadratic equation because the highest power of x is 3. d. 2 + 6 + 9 = 3 2 + 6 + 9 − 3 = 0 2 + 3 + 9 = 0 Quadratic equation Example 4
CHAPTER 2: BASIC ALGEBRA 51 a. Solve the equation 2 = 9 by using factorization. Answer: 2 = 9 2 − 9 = 0 2 3 3 −3 −3 2 −9 0 Therefore; ( + 3)( − 3) = 0 ( + 3) = 0 and ( − 3) = 0 = −3 = 3 Determine whether the following are quadratic equations. 1. ( + 3)( − 2) = 3 2. 3( + 2)( − 1) = 3 3 3. 4. a. Factorization Exercise 4 Example 5 Set the value of Right hand Side to zero. 3 + + 1 = 12 2 + 4 + 4 = 1 Make sure the value of , are the same as in the equation.
CHAPTER 2: BASIC ALGEBRA 52 b. Solve the equation 4 2 − 25 + 25 = 0 using factorization. Answer: 4 2 − 25 + 25 = 0 2 −5 −20 4 −5 −5 4 2 25 −25 Therefore; ( − 5)(4 − 5) = 0 ( − 5) = 0 and (4 − 5) = 0 = 5 Solve the following quadratic equations by using factorization. 1. 4 2 + 9 − 9 = 0 2. 3 2 − 20 − 32 = 0 3. 2 2 + 23 + 60 = −3 4. 3 2 + 5 = 2 Exercise 5 = 5 4
CHAPTER 2: BASIC ALGEBRA 53 a. Solve the equation 4 2 − 25 = −25 using quadratic formula. Answer: 4 2 − 25 = −25 4 2 − 25 + 25 = 0 = 4 = −25 = 25 By using quadratic formula, = − ± √ 2 − 4 2 = −(−25) ± √(−25) 2 − 4(4)(25) 2(4) = 25 ± √225 8 = 25 ± 15 8 1 = 40 8 or 2 = 10 8 = 5 = 5 4 b. Quadratic Formula The quadratic formula is where a, b and c are constants, ≠ 0 and x as the unknown variable. Identify the value of a, b and c. = − ± √ 2 − 4 2 Example 6
CHAPTER 2: BASIC ALGEBRA 54 Solve the following quadratic equations by using quadratic formula. 1. 2 − − 2 = 4 2. 2 2 + 23 + 63 = 0 3. 2 + 14 + 48 = 0 4. 2 + 6 − 7 = 0 5. 6. Exercise 6 −4 2 − 36 − 56 = 0 12 2 − 40 − 32
CHAPTER 2: BASIC ALGEBRA 55 7. −8 2 + 26 − 10 = 8 8. 48 + 70 = −8 2 9. −25 + 8 = −3 2 10. − 2 + 2 + 15 = −9 11. 2 + 1 = 12. 5 − 3 2 + 1 = 3 − 2
56 3.1 Describe the concepts of real number 3.1.1 Identify type of real numbers 3.1.2 Define real number lines 3.1.3 Define graph of inequalities 3.2 Describe the concepts of complex number 3.2.1 State that complex numbers that derived by combining the real parts with the imaginary parts 3.2.2 Identify that product of two imaginary number is real number 3.2.3 State the conjugate of complex numbers 3.2.4 Describe the operations such as addition, subtraction, multiplication, division, conjugate and equivalent complex numbers 3.3 Solve the complex numbers using Argand's Diagram 3.3.1 Explain graphical representation of complex number through Argand’s Diagrams 3.3.2 Draw a straight line in an Argand’s Diagram to represent a complex number 3.3.3 Use Argand’s Diagrams to find the modulus and argument 3.4 Apply the concepts of complex numbers in other forms 3.4.1 Describe complex numbers in the form of polar, trigonometric and exponential CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM The 16th century Italian mathematician Gerolamo Cardano is credited with introducing complex numbers in his attempts to find solutions to cubic equations. Formally, the complex number system can be defined as the algebraic extension of the ordinary real numbers by an imaginary number i. (https://en.wikipedia.org/wiki/Complex_number)
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 57 3.1.1 Identify type of real numbers There are many types of real numbers. Here are some of them: A natural number, N is a counting number. It starts from 1 onwards. They are located at the right side of the number line (after 0). Examples: 1, 2, 3, 7, 10, 15, 60, 108, 2020 A whole number, W is either a counting number or zero (0). They are located at the right side of the number line and 0 is the smallest whole number. Examples: 0, 1, 2, 3, 5, 13, 16, 40, 217, 3030 An integer, Z is either a whole number or its negative. Positive integers, Z are integers starting from 1 and so on. They are located at the right side of the number line (after 0). They are also natural or counting numbers. Negative integers, Z are integers starting from -1 and so on. They are located at the left side of the number line (before 0). They are negative whole numbers. Examples: -2019, -432, -45, -3, 0, 15, 37, 527, 1010 3.1 Describe the concepts of real number
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 58 A fraction is a part of a whole. In the number line, they are located between integers. A fraction can be converted into decimal form. Examples: −4 2 5 , −1 3 , 3 7 , 13 8 , 2 1 6 A rational number, Q is a number that can be written as a ratio of two integers. It can be a fraction, an integer, a whole number, or even a natural number. Examples: −4 2 5 , −1 6 , 0, 3 7 , 13 8 , 3 1 9 , 7 For decimal numbers, it can be a rational number if it satisfies either one of the following: It has finite number of digits. 71.25 → 100 7125 Its digits are repeating in a continuous manner. 3.1111… → 9 1 3 → 9 28 An irrational number, Q is a number that cannot be written as a ratio of two integers. It is a non-terminating and non-repeating decimal. Examples: ,e, 2, 5, 11, 2.3476364982...... The different types of real numbers can be represented by the Venn diagram. N W Z Q R and QQ R Q Z W N R Q
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 59 3.1.2 Define real number lines A real number is a number that can be found on the number line. These are the numbers that we normally use and apply in real-world applications. 3.1.3 Define graph of inequalities When we have an equation such as x = 4 we have a specific value for our variable. With inequalities we will give a range of values for our variable. To do this we will not use equals, but one of the following symbols: > Greater than Greater than or equal to < Less than Less than or equal to Intervals of inequalities can be divided by: a) Open intervals Set Interval Notation Graph {x | a x} or {x | x a} (a,) {x | a x b} (a,b) {x | x b} (,b) {x | x is a real number} (,) b) Closed interval Set Interval Notation Graph {x | a x b} [a,b] a a b b a b
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 60 Formulae: c) Half open intervals Set Interval Notation Graph {x | a x} or {x | x a} [a,) {x | a x b} (a,b] {x | a x b} [a,b) {x | x b} (,b] 3.2.1 State that complex numbers that derived by combining the real parts with the imaginary parts A complex number, C can be written as z a bi , where a and b are real and i 1. The number a is called the real part and b is the imaginary part. When a = 0, z = bi is called an imaginary number. When b = 0, z = a is called a real number. The real numbers set, R is a subset of complex number set, C. R C . \ 3.2 Describe the concepts of complex number Example 1 Classify the real parts and imaginary parts for the following complex numbers: a) x yi b) 4 5i c) 39i d) 7i Answer: a) Real part: x Imaginary part: y b) Real part: 4 Imaginary part: 5 c) Real part: -3 Imaginary part: 9 d) Real part: 0 Imaginary part: -7 a a b b a b
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 61 Remember that: 3.2.2 Identify that product of two imaginary number is real number Express the following numbers in terms of i. a) 49 16 Answer: 11i Therefore: Exercise 1 Example 2 Write in terms of i. a) 4 b) 16 5 c) 81 1 3 9 25 Answer: a) 4 4(1) 4 1 2i b) 16 5 16(1) 5(1) 16 1 5 1 4i 5i 2 4 5 i 4 5 c) 1 81 1 1 3 9 25 81 1 3 9 25 i i 9 3 3 5 2i
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 62 b) 11 4 Answer: 2 11 c) 9 100 2 4 81 Answer: i 6 13 Example 3 Simplify 20 10 30 23 i i i i Answer: Method 1: Method 2: 20 10 30 23 i i i i 20 10 30 23 i i i i i i i i i 11 2 15 2 5 2 10 2 20103023 i i 10 5 15 11 1 1 1 1 83 i 1111i i i 41 2 i i 41 1 i Example 4 Show that 2 35 40 5 4 i i i i Answer: 35 40 5 4 i i i i 2 2 2 2 20 2 17 2 i i i i i i 17 20 2 2 1 i 1 1 i 1 1i 1 1i 1 i 1i 1 2
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 63 a) Simplify 11 15 40 35 i i i i b) Show that 0 25 30 8 3 i i i i 3.2.3 State the conjugate of complex numbers The conjugate of z a bi , is written as z* a bi or z a bi . It has the same real part as z but an imaginary part of the opposite sign. We will use conjugate to solve the division of complex numbers. When we multiply a complex number by its conjugate, we will get a real number. Complete the following table. z Conjugate, z Real part, R(z) Imaginary part, I(z) 3 4i 7 5i 98i 13i 6 Exercise 2 Example 5 Multiply the following complex numbers with its conjugate. a) z 35i b) z 2 4i Answer: a) zz (35i)(35i) b) zz (2 4i)(2 4i) 2 9 15i 15i 25i 2 48i 8i 16i 9 25(1) 416(1) 34 20 Exercise 3
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 64 3.2.4 Describe the operations such as addition, subtraction, multiplication, division, conjugate and equivalent complex numbers a) Addition of complex numbers. ( ) ( ) z1 z2 a bi c di (a c) (b d)i b) Subtraction of complex numbers. ( ) ( ) z1 z2 a bi c di (a c) (b d)i Simplify the following complex numbers. 1. (3 4i) (68i) Answer: 9 4i 2. (3 2i) (2i) Answer: 1i 3. (5 2i) (7 i) Answer: 12 i 4. (3i) (53i) Answer: 2 2i 5. (2 3i) (5 2i) Answer: 7 i 6. (3i) (2i) Answer: 1 2i Example 6 Solve the following problems. a) (3 2i) (4 6i) b) (7 3i) (5 2i) Answer: a) (3 2i) (46i) b) (7 3i) (5 2i) (3 4) (26)i (7 5) (32)i 7 4i 2 5i Exercise 4
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 65 c) Multiplication of complex numbers. ( )( ) z1 z2 a bi c di 2 ac adi bci bdi (ac bd) (ad bc)i Simplify the following complex numbers and write in the form of a bi . 1. 5(25i) Answer: 10 25i 2. 3i(7 4i) Answer: 12 21i 3. (2 i)(3 2i) Answer: 47i 4. 2 (2 i) Answer: 3 4i Remember that: 1 2 i Exercise 5 Example 7 Solve the following problems and give answer in the standard form, a bi . a) 2(36i) b) 4i(15i) c) (2 4i)(7 3i) 2 d) (56i) Answer: a) 2(36i) b) 4i(15i) 612i 2 4i 20i 20 4i c) (2 4i)(7 3i) 2 d) (56i) 2 146i 28i 12i (56i)(56i) 14 22i 12(1) 2 2530i 30i 36i 26 22i 2560i 36(1) 1160i
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 66 5. (5 2i)(5 2i) Answer: 29 6. (3 4i)(6 i) Answer: 14 27i d) Division of complex numbers. c di a bi z z 2 1 c di a bi c di c di 2 2 2 c (di) ac adi bci bdi 2 2 ( ) ( ) c d ac bd bc ad i Remember that: 1 2 i Conjugate: z c di 2 Example 8 Find the following complex numbers in the form of a bi . i i 3 2 a) i i 1 2 3 4 b) Answer: i i 3 2 a) i i 1 2 3 4 b) (3 ) (3 ) (3 ) (2 ) i i i i ( 1 2 ) ( 1 2 ) ( 1 2 ) (3 4 ) i i i i 2 2 2 (3) ( ) 6 2 3 i i i i 2 2 2 ( 1) (2 ) 3 6 4 8 i i i i 2 9 7 i i 2 2 1 4 3 6 4 8 i i i i 10 7 i 1 4( 1) 3 2 8( 1) i i 10 1 10 7 5 11 2i i 5 2 5 11
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 67 Simplify the following complex numbers and write in the form of a bi . 1. i i 1 7 3 Answer: 2 5i 2. 1 i 5 Answer: i 2 5 2 5 3. i i 5 5 Answer: i 13 5 13 12 4. i i 6 7 2 3 Answer: i 85 32 85 9 5. i i 3 2 Answer: i 10 1 10 7 6. i i 2 6 4 3 Answer: i 20 9 20 13 7. i i 4 2 3 Answer: i 2 1 4 3 8. i i 1 2 3 5 Answer: i 5 11 5 7 Exercise 6
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 68 e) Equivalent of complex numbers. Two complex numbers are equal if and only if their real and imaginary parts are respectively equal: a bi c di then a c and b d Example 9 Solve the following equations for the value of x and y a) 2x i 4 yi b) x 3i2 i 1 yi Answer: a) 2x i 4 yi Compare real part: 2x 4 2 2 4 x Compare imaginary part: i yi y 1 Therefore, x 2 and y 1 b) x 3i2 i 1 yi 2x xi 6i 3i 1 yi 2 2x xi 6i 31 1 yi 2x 3 (x 6)i 1 yi Compare real part: 2x 3 1 2x 2 x 1 Compare imaginary part: (x 6)i yi When x 1 (1) 6 y y 7 Therefore, x 1 and y 7
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 69 Find the value of x and y for the following questions. 1. 5x 7yi x 7i Answer: x 0, y 1 2. (x yi)(23i) i Answer: 13 2 , 13 3 x y 3. 2 2 3 i y i x Answer: x 10, y 20 Exercise 7
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 70 3.3.1 Explain graphical representation of complex number through Argand’s Diagrams Complex number z a bi can be represented by Argand Diagram. The x axis of the diagram represents real part (usually in term of r). The y axis of the diagram represents imaginary part (usually in term of i). Write the complex number in the given Argand’s Diagram 1. 2. 3. 4. 3.3 Solve the complex numbers using Argand’s Diagram b a θ r i Exercise 8 5 3 5 +3i 6 −3 −4 2 −√7 −5 z = a + bi
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 71 3.3.2 Draw a straight line in an Argand’s Diagram to represent a complex number Draw an Argand Diagram for each complex number 1. z 2 4i 2. z 3i 9 3. z 5 i 4. z 6 7i 3.3.3 Use Argand’s Diagrams to find the modulus and argument The line segment represents the modulus, |z| or length of complex number. Modulus 2 2 z a b An angle from the positive x axis to the line segment represents the argument of complex number, . Argument a b tan 1 Arguments are different according to the quadrants. Exercise 9 2 −4 90° 180° 0° 270° 1 st Quadrant bbbququadran 2 nd Quadrant 3 rd Quadrant 4 th Quadrant
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 72 Example 10 Find modulus and argument for the following complex number a) 5 i b) 1 4i c) 2 3i d) 3 6i Answer: a) 5 i Modulus: 5 1 6 2 2 z Argument: (Complex number is in 1st quadrant) 24 09 5 1 1 tan . b) 1 4i Modulus: 1 4 17 2 2 z Argument: (Complex number is in 2nd quadrant) 104.04 1 4 180 tan 1 c) 2 3i Modulus: 2 3 13 2 2 z Argument: (Complex number is in 3rd quadrant) 236.31 2 3 180 tan 1 d) 3 6i Modulus: 3 6 9 36 45 2 2 z Argument: (Complex number is in 4th quadrant) 296.57 3 6 360 tan 1
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 73 Sketch the Argand’s diagram and then find modulus and argument for the following complex number. 1. 1 7i Answer: z 50; 278.13 2. 3i 2 7 Answer: z 4.61; 139.40 3. 3 5i Answer: z 14; 52.24 4. 4 9i Answer: z 97; 293.96 Exercise 10
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 74 5. 33i Answer: z 18; 225 6. 5 2i Answer: z 3; 138.19 7. Given Z 4 3i 1 and Z 1 i 2 . Calculate the modulus and argument of . Z1 Z2 Answer: z 29; 21.80 3.4.1 Describe complex numbers in the form of polar, trigonometric and exponential Complex numbers also can be represented in Polar form, Trigonometric form and Exponential form. If z is the modulus and is the argument of a complex number, then: 3.4 Apply the concepts of complex numbers in other forms
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 75 Form Formulae Cartesian Form a bi Example: z 2 3i Polar Form z ( is in degree) Example: z 1356.13 Trigonometric Form z(cos isin) ( is in degree) Example: z 13(cos56.13 isin56.13) Exponential Form i z e ( is in radian ) Example: i z e 0.9797 13 Recall back: 0 0 0 1 1 180 rad (degree radian) Recall back: radian 180 x x (Degree radian) degree 180 x rad x (Radian degree) Example 11 Convert 3 4i to the polar form, trigonometric form and exponential form. Answer: z 3 4i Modulus: (3) ( 4) 5 2 2 z Argument: (Complex number is in 4th quadrant) 3 4 360 tan 1 306.87 (θ in degree) 5.356 180 306.87 (θ in radian) i) Polar form: z 5306.87 ii) Trigonometric form: z 5(cos306.87 isin306.87) iii) Exponential form: i z e 5.356 5 3 −4
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 76 1. Convert 3 2i to polar form. Answer: 5 320.77 2. Convert 1.11863.43 to Cartesian form. Answer: 0.5i 3. Convert 7 5i to exponential form. Answer: i e 3.762 74 4. Convert 25i to trigonometric form. Answer: 29(cos68.20 isin 68.20) 5. Express 5178 in Cartesian form, trigonometric form and exponential form. Answer: i CF i TF i EF e 3.107 : 5 0.17 , :5(cos178 sin178), 5 Exercise 11
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 77 6. Express 10(cos218 isin218) in Cartesian form, polar form and exponential form. Answer: i CF i PF EF e 3.8 1 :7.886.16 , :10178 , 10 7. Express i e 5.06 14 in Cartesian form, polar form and trigonometric form. Answer: CF :1.283.52i, PF : 14 290 , TF 14 (cos 290isin 290) 8. Convert 5 12i to the polar form and exponential form. Answer: i PF EF e 1.18 :1367.38 , 13 3.4.2 Solve multiplication and division of complex numbers in other forms (Additional subtopic) Formulae for polar form: Given and then, Multiplication : Division :
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 78 Formulae for trigonometric form: Given and then, Multiplication : Division : Example 12 Given that 3.568 1 z and 749 2 z , find the values of 1 2 z z and 2 1 z z and state them in the polar form. Answer: (3.5 68 )(7 49 ) 1 2 z z (3.57)(68 49) 24.5117 7 49 3.5 68 2 1 z z (68 49 ) 7 3.5 0.519 Formulae for exponential form: Given and then, Multiplication : Division :
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 79 Example 13 Given that i z e 4.2 1 3 and i z e 1.3 2 9 . Find 1 2 z z in exponential form and 2 1 z z in trigonometry form Answer: (3 )( 9 ) 4.2 1.3 1 2 i i z z e e i e (4.2 1.3) (3 ( 9)) i e 5.5 27 Exponential form: i z z e 5.5 1 2 27 i i e e z z 1.3 4.2 2 1 9 3 i e (4.2 1.3) 3 1 i e 2.9 3 1 166.16 180 2.9 Trigonometry form: cos166.16 sin166.16 3 1 2 1 i z z
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 80 Example 14 Given that Z 5 6i 1 , 4(cos30 sin30 ) 2 Z i and i Z e 1.13 3 2 . Express Z1 , Z2 and Z3 in polar forms. Then, find 3 1 2 Z Z Z in polar form. Answer: Z 5 6i 1 Modulus: (5) (6) 61 2 2 Z1 Argument: (Complex number is in 1st quadrant) 5 6 tan 1 50.19 (θ in degree) Z 5 6i 1 4(cos30 sin30 ) 2 Z i Polar form: 6150.19 Z1 Polar form: 430 Z2 i Z e 1.13 3 2 64.74 180 1.13 Polar form: 264.74 Z3 Therefore, 2 64.74 61 50.19 4 30 3 1 2 Z Z Z (50.19 30 64.74 ) 2 ( 61 4) 15.6215.45
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 81 1. If 12(cos125 sin125 ) 1 z i and 3(cos75 sin 75 ) 2 z i , find the values of 1 2 z z in Polar form. Answer: 36 200 2. Given i z e 0.35 1 2.5 , 2.2(cos15 sin15 ) 2 z i and z 2 8i 3 find the values of 1 3 2 z z z in exponential form. Answer: i e 1.72 7.26 3. Given, Z 2 3i 1 , 567 Z2 and i Z e 1.6 3 3 . Express Z1 , Z2 and Z3 in trigonometric forms. Then, find 3 1 2 Z Z Z in trigonometric form. Answer: 6(cos31.64 sin 31.64 ) 3 1 2 i Z Z Z Exercise 12
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 82 FINAL EXAMINATION QUESTIONS SESSION JUNE 2019 a) Solve the following expressions in the form of a bi . i. (8 7i)(3i) [3 marks] ii. 2 i 5 [4 marks] b) Calculate the modulus and argument of each of the following complex numbers. Then, sketch the Argand diagram. i. 3 4i [4 marks] ii. 4 7i [4 marks] c) Given that, 5.39158.2 1 z and 2.5(cos 189 sin189 ) 2 z i . i. Express 2 z in the form of a bi . [2 marks] ii. Express 1 z in exponential form. [2 marks] iii. Calculate the value of 1 2 z z and 2 1 z z in Polar form. [6 marks] SESSION DECEMBER 2018 a) Determine the following expressions in the form of a bi . i. 3i(5 8i) [2 marks] ii. (2 4i)(3 5i) [3 marks] iii. i i 4 3 5 [5 marks] b) Calculate the modulus and argument for z 3 5i and sketch an Argand Diagram. Express the answer in the trigonometric form. [7 marks] c) Given 3(cos60 sin 60 ) 1 z i and i z e 0.785 2 5 . i. Express to 2 z Trigonometric form. [2 marks] ii. Express to 1 z Exponential form. [2 marks] iii. Calculate the value of 1 2 z z and 2 1 z z in Polar form. [4 marks]
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 83 SESSION JUNE 2018 a) i. Calculate the following complex numbers: a. (4 5i) (6 7i) [2 marks] b. (8 i) (2 3i) [2 marks] ii. Given that a bi (2 i)(5 4i) . Compute the value of 2 (a b) . [6 marks] b) i. Determine the modulus and argument of 7 3i [5 marks] ii. Given Z 2 5i 1 , 7(cos50 sin50 ) 2 Z i and i Z e 2.62 3 7 . Express Z1 , Z2 and Z3 in polar forms. Then, calculate 3 1 2 Z Z Z in polar form. [10 marks] SESSION DECEMBER 2017 a) Determine each of the following complex numbers in a bi form: i. (7 9i) (15 7i) [2 marks] ii. (4 2i) (23i) [2 marks] iii. 5(2 i) [1 mark] iv. i i 3 2 1 [5 marks] b) Calculate the modulus and argument for z 7 4i and sketch an Argand Diagram. Express the answer in the trigonometric form. [7 marks] c) Given 723 1 z and 2040 2 z . i. Express to 2 z Trigonometric form. [2 marks] ii. Express to 1 z Exponential form. [2 marks] iii. Calculate the value of 1 2 z z and 2 1 z z . [4 marks]
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 84 SESSION JUNE 2017 a) Determine the following expressions in the form of a bi . i. 8i(7 2i) [2 marks] ii. (2 6i)(13i) [3 marks] iii. 6 7i 4 [5 marks] b) Given Z 5 5i 1 and Z 1 i 2 : i. Solve Z1 Z2 in terms of a bi form. [3 marks] ii. Illustrate Z1 Z2 in an Argand Diagram. [2 marks] iii. Calculate the modulus and argument of . Z1 Z2 [7 marks] iv. Express . Z1 Z2 to exponential form and trigonometric form. [3 marks] SESSION DECEMBER 2016 a) i. Solve the following complex numbers: a) (1 2i) (3 4i) [2 marks] b) (5 6i) (7 8i) [2 marks] ii. Given that a bi (54i)(3 2i) . Find the value of 2 (a b) . [6 marks] b) i. Determine the modulus and argument of the following complex number 95i . [5 marks] ii. Given, Z 5 6i 1 , 4(cos30 sin30 ) 2 Z i and i Z e 1.11 3 2 . Express Z1 , Z2 and Z3 in polar forms. Then, find 3 1 2 Z Z Z in polar form. [10 marks]
CHAPTER 3: REAL AND COMPLEX NUMBER SYSTEM 85 SESSION JUNE 2016 a) Solve each of the following complex number in the form of a bi . i. 3(3i) [2 marks] ii. (1 2i) (15i) [2 marks] iii. (35i) (2 6i) [2 marks] iv. i i 1 2 2 3 [4 marks] b) i. Change z 5(cos30isin30) in Cartesian and Exponential form. [4 marks] ii. Given that i z e 2.3 1 5 and i z e 1.8 2 3 . Find 1 2 z z and express the answer in exponential form. [3 marks] iii. Given that z 5 2i . Answer the following questions: a) Sketch the Argand’s diagram for z . [2 marks] b) Find the modulus and the argument for z . [5 marks] c) Chage the answer in Question 3(b)(iii)(b) into polar form. [1 mark] SESSION DECEMBER 2015 a) Solve: i. 3(4 i) 3(5 7i) [3 marks] ii. (3 6i)(5 2i) [3 marks] iii. 3 2 5 i i [4 marks] b) Given Z 1 2i 1 and Z 3 4i 2 . i. Find the modulus and the argument for Z1 and sketch a Argand’s diagram. [6 marks] ii. Carry out Z1 Z2 in polar, trigonometric and exponential form. [9 marks]
86 4.1 Describe Differentiation 4.1.1 Limits and differentiation 4.1.2 Techniques of Differentiation 4.1.3 Rule of Differentiation: a. Chain Rule b. Product Rule c. Quotient Rule 4.1.4 Calculate gradients of tangent to a curve by applying the differentiation techniques 4.1.5 Compute the tangent and normal equations 4.2 Clarify Integration 4.2.1 Describe the integration as the inverse of differentiation. 4.2.2 Indefinite integral of algebraic function 4.2.3 Indefinite integral of a algebraic function with initial conditions given 4.2.4 Definite integrals 4.2.5 Solve the problems related to definite integrals CHAPTER 4: DIFFERENTIATION AND INTEGRATION Calculus is developed by Leibniz and Newton. The idea behind differentiation on the other hand ... (Source: https://www.archimedes-lab.org/numeral.html)
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 87 Find derivative of the following functions. a) () = 2 b) () = −4 Answer: a) ′() = →0 (+ )−() = →0 2(+ )−2 = 2+2−2 = 2 = 2 b) ′() = →0 (+ )−() = →0 −4− (−4) = 0 = 0 4.1.1 Identify limits and differentiation - Let f (x) be a function and x is small change at variables x. - It is defined as - Derivative of a function f (x) is a process to measure rate of change of function with respect to the change in variable x. - The process of finding derivative of a function is called differentiation. 4.1 Describe differentiation Example 1 ′() = →0 (+ )−()
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 88 1. Find ′() if () = 2 2 2. Find ′() if () = ( − 1) 2 3. Find ′() if () = 1 3 −1 Exercise 1
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 89 Derive the following equations a) () = 10 b) = − 1 2 c) = 2 d) = √45 Answer: 4.1.2 Techniques of Differentiation - Derivative of function f (x) is called ‘f prime’ or f ’(x). - Other derivative notation of any function (f, y etc.): a) Constant Rule a) () = 10 ′() = 0 b) = − 1 2 = 0 c) = 2 = 0 d) = √45 = 0 = = 0 Example 2 ′ () = ′ () =
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 90 Derive the following functions a) = 10 b) = 5 c) = −9 Answer: a) = 10 = 10( 1−1 ) = 10 b) = 5 = 5( 5−1 ) = 5 4 c) = −9 = −9( −9−1 ) = −9 −10 = − 9 10 b) Power Rule Differentiate: a) = 2 3 b) = −1 000 c) = 2 d) = √10 Exercise 2 = = (−1 ) Example 3
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 91 Differentiate the following functions a) = √ b) = 3 8 4 c) = 4 3 2 Answer: a) = √ = 1 2 = 1 2 ( 1 2 −1 ) = − 1 2 2 = 1 2√ b) = 3 8 4 = 8 ( 3 8−1 4 ) = 6 7 c) = 4 3 2 = 4 −2 3 = −2(4 −2−1) 3 = −8 3 3 Example 4
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 92 Derive the following functions a) = 2 b) = 1 4 c) = 4 √ 3 d) = 1 6 5 e) () = −3 3 f) () = −4 −6 g) = 4 3 3 Exercise 3
CHAPTER 4: DIFFERENTIATION AND INTEGRATION 93 Find differentiation of the following functions a) = 4 3 + 2 b) () = 3 − 5 c) () = 2 2 + 7 − 9 d) = ( 4 + 2 ) 2 Answer: a) = 4 3 + 2 = 3(4 3−1 ) + 2( 2−1 ) = 12 2 + 2 b) () = 3 − 5 ′ () = 3( 3−1 ) − 5 = 3 2 − 5 c) () = 2 2 + 7 − 9 ′() = 2( 2−1 ) − 7 = 4 − 7 d) = ( 4 + 2 ) 2 = ( 4 + 2 )( 4 + 2 ) = 8 + 2 6 + 4 = 8( 8−1 ) + 6(2 6−1 ) + 4( 4−1 ) = 8 7 + 12 5 + 4 3 c) Sum and Difference Rule = () ± () = ′() ± ′() Example 5