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LESSON 3.2 Name Complex Numbers Class 3.2 Date Complex Numbers Essential Question: What is a complex number, and how can you add, subtract, and multiply complex numbers? Common Core Math Standards The student is expected to: COMMON CORE Resource Locker N-CN.A.2 Explore Use the relation i 2 = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. Also N-CN.A.1 In this lesson, you’ll learn to perform operations with complex numbers, which have a form similar to linear binomials such as 3 + 4x and 2 − x. Mathematical Practices COMMON CORE Exploring Operations Involving Complex Numbers A MP.2 Reasoning Add the binomials 3 + 4x and 2 − x. Group like terms. Language Objective Combine like terms. Work with a partner to classify and justify the classification of real, complex, and imaginary numbers. B Subtract 2 − x from 3 + 4x. Rewrite as addition. ENGAGE © Houghton Mifflin Harcourt Publishing Company (3 + 4x) (2 − x) = 6 + (−3x) + 8x Reflect 1. In Step A, you found that (3 + 4x) + (2 − x) = 5 + 3x. Suppose x = i (the imaginary unit). What equation do you get? (3 + 4i) + (2 − i) = 5 + 3i 2. In Step B, you found that (3 + 4x) + (2 − x) = 1 + 5x. Suppose x = i (the imaginary unit). 3. In Step C, you found that (3 + 4x)(2 − x) = 6 + 5x - 4x 2. Suppose x = i (the imaginary unit). What equation do you get? How can you further simplify the right side of this equation? What equation do you get? (3 + 4i) - (2 − i) = 1 + 5i (3 + 4i)(2 − i) = 6 + 5i - 4i 2; because i 2 = -1, the right side of this equation can be simplified to 6 + 5i - 4(-1), or 10 + 5i. Module 3 be ges must EDIT--Chan DO NOT Key=NL-B;CA-B Correction Lesson 2 127 gh “File info” made throu Date Class Name Complex Numbers ly ct, and multip add, subtra can you er, and how ties lex numb utive proper is a comp ion: What tive, and distrib numbers? utative, associa complex and the comm Also N-CN.A.1 2 n i = -1 rs. the relatio ex numbe compl g ly N-CN.A.2 Use lvin ct, and multip to add, subtra rations Invo 3.2 Resource Locker Quest Essential A2_MNLESE385894_U2M03L2 127 operations x. to perform and 2 − , you’ll learn as 3 + 4x In this lesson binomials such linear similar to ials 3 + 4x Add the binom terms. Group like and 2 − x. (3 + 4x) + (2 − x) ( = 3+ = like terms. Combine 3 + 4x. − x from Subtract 2 addition. Rewrite as Group like (3 + 4x) − (2 − x) terms. like terms. Combine ials 3 + 4x the binom Multiply HARDCOVER Ope Exploring s a form Number which have Complex x numbers, with comple Explore Use FOIL. ( ( (3 + 4x) ( + −2 + = 3 + -2 ( -x ) Turn to Lesson 3.2 in the hardcover edition. + 3x = (3 + 4x) = and 2 − x. ) + (4x + ) 2 5 + 1 ) x ) + (4x + 5x ) x ) 2 (2 − x) = 6+ x 8x + -4 (−3x) + 2 =6+ x 5x + -4 Publishin y g Compan like terms. Combine unit). imaginary se x = i (the + 3x. Suppo − x) = 5 4x) + (2 5 + 3i that (3 + unit). 2 − i) = you found imaginary (3 + 4i) + ( se x = i (the get? 5x. Suppo x) = 1 + ion do you 4x) + (2 − What equat 1 + 5i that (3 + ary unit). 2 − i) = you found = i (the imaginon? (3 + 4i) - ( In Step B, 2 Suppose x 5x - 4x . of this equati + do you get? 6 side = ion ) fy the right can be 4x)(2 − x What equat equation that (3 + further simpli side of this C, you found get? How can you Step right In the you 2 3. on do se i = -1, 2 What equati - 4i ; becau = 6 + 5i i ) − 2 ) 5i. (3 + 4i ( ), or 10 + - 4(-1 to 6 + 5i simplified n Mifflin Harcour t Reflect 1. In Step A, © Houghto 2. Lesson 2 127 Module 3 L2 127 4_U2M03 SE38589 A2_MNLE Lesson 3.2 + -4x 2 5x + -4x 2 =6+ Combine like terms. COMMON CORE 127 ) Multiply the binomials 3 + 4x and 2 − x. Use FOIL. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photo and how complex numbers can be used to generate fractal patterns. Then preview the Lesson Performance Task. 5 -x x ) ( = (3 + -2 ) + (4x + x ) = ( 1 + 5x ) Combine like terms. C ) + (4x + + 3x ) 2 (3 + 4x) − (2 − x) = (3 + 4x) + −2 + Group like terms. Essential Question: What is a complex number and how do you add, subtract, and mutiply complex numbers? Possible answer: A complex number has the form a + bi where a and b are real numbers and i is the imaginary unit. You add and subtract complex numbers by combining like terms. You multiply complex numbers by using the distributive property, substituting -1 for i 2 , and combining like terms. ( =( (3 + 4x) + (2 − x) = 3 + 5/22/14 10:28 AM 5/22/14 10:29 AM Explain 1 Defining Complex Numbers EXPLORE A complex number is any number that can be written in the form a + bi, where a and b are real numbers and _ i = √-1 . For a complex number a + bi a is called the real part of the number, and b is called the imaginary part. (Note that “imaginary part” refers to the real multiplier of i; it does not refer to the imaginary number bi.) The Venn diagram shows some examples of complex numbers. 3 + 4i Exploring Operations Involving Imaginary Numbers Complex Numbers 4 - 2i 1-i 5 Real Numbers 17 -5 π 3 √2 3.3 4 INTEGRATE TECHNOLOGY Imaginary Numbers 2i -7i i 11 Students have the option of completing the activity either in the book or online. 6.3i QUESTIONING STRATEGIES Notice that the set of real numbers is a subset of the set of complex numbers. That’s because a real number a can be written in the form a + 0i (whose imaginary part is 0). Likewise, the set of imaginary numbers is also a subset of the set of complex numbers, because an imaginary number bi(where b ≠ 0) can be written in the form 0 + bi(whose real part is 0). Example 1 Do all complex numbers include an imaginary part? Explain. No, a complex number does not always include an imaginary part. All real numbers are also complex numbers. The imaginary unit i is defined as √-1 . You can use the imaginary unit to write the square root of any negative number. ―― Identify the real and imaginary parts of the given number. Then tell which of the following sets the number belongs to: real numbers, imaginary numbers, and complex numbers. 9 + 5i The real part of 9 + 5i is 9, and the imaginary part is 5. Because both the real and imaginary parts of 9 + 5i are nonzero, the number belongs only to the set of complex numbers. -7i the number belongs to these sets: imaginary numbers and complex numbers . Your Turn Identify the real and imaginary parts of the given number. Then tell which of the following sets the number belongs to: real numbers, imaginary numbers, and complex numbers. 11 The real part of 11 is 11, and the imaginary part is 0. Because the imaginary part is 0, the © Houghton Mifflin Harcourt Publishing Company The real part of -7i is 0 , and the imaginary part is -7 . Because the real/imaginary part is 0, 4. How can you tell which part of a complex number is the real part and which is the imaginary part? In a number of the form a + bi, the real part is a, which does not have i as a factor. The imaginary part is bi, where b is a nonzero real number and i is the imaginary unit. EXPLAIN 1 Defining Complex Numbers number belongs to these sets: real numbers and complex numbers. AVOID COMMON ERRORS Module 3 128 Students may write the conjugate of a + bi as -a−bi. Caution them to change the sign only of bi, and not of a, when they write the complex conjugate. Lesson 2 PROFESSIONAL DEVELOPMENT A2_MNLESE385894_U2M03L2 128 Integrate Mathematical Practices This lesson provides an opportunity to address Mathematical Practice MP.2, which calls for students to translate between multiple representations and to “reason abstractly and quantitatively.” Students explore the relationship between operations with complex numbers and operations with binomials. They also describe how complex-number arithmetic operations follow from operations with rational numbers and square roots. 5/22/14 10:29 AM QUESTIONING STRATEGIES What are the components of a complex number? A complex number has the form a + bi, where a and b are real numbers. Each term of a + bi is given a name: a is called the real part and bi is called the imaginary part. Complex Numbers 128 -1 + i The real part of -1 + i is -1, and the imaginary part is 1. Because both the real and 5. EXPLAIN 2 imaginary parts of -1 + i are nonzero, the number belongs only to the set of complex numbers. Adding and Subtracting Complex Numbers Explain 2 Adding and Subtracting Complex Numbers To add or subtract complex numbers, add or subtract the real parts and the imaginary parts separately. QUESTIONING STRATEGIES Example 2 Can the sum of two imaginary numbers be 0? yes, if both the real parts and the imaginary parts are opposites of each other Add or subtract the complex numbers. ( -7 + 2i ) + ( 5 - 11i ) Group like terms. A pure imaginary number has no real part. When is the sum of two imaginary numbers a pure imaginary number? when the real parts are opposites, for example, (6 + 4i) + (−6 − 8i) = −4i (−7 + 2i) + (5 − 11i) = (−7 + 5)+ (2i + ( −11i )) Combine like terms. = −2 + (−9i) Write addition as subtraction. = −2 − 9i ( 18 + 27i ) - ( 2 + 3i ) Group like terms. ( (18 + 27i) − (2 + 3i) = 18 − AVOID COMMON ERRORS Some students may try to simplify a complex number by combining the real part and the imaginary part. Emphasize that just as unlike terms in an algebraic expression cannot be combined, neither can the real and imaginary parts of a complex number. − 3i ) Reflect 6. Is the sum (a + bi) + (a - bi) where a and b are real numbers, a real number or an imaginary number? Explain. (a + bi) + (a - bi) = (a + a) + (bi + (-bi)) = 2a Since a is a real number, 2a is a real number. So, the sum (a + bi) + (a - bi) is a © Houghton Mifflin Harcourt Publishing Company 2, and therefore have a maximum of two real solutions. Sometimes the two real solutions are a double root, but many quadratic equations have zero real solutions. It is only by introducing complex numbers that we are able to find two solutions for all quadratic equations. ) + ( 27i = 16 + 24 i Combine like terms. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Quadratic equations are equations of degree 2 real number. Your Turn Add or subtract the complex numbers. 7. (17 - 6i) - (9 + 10i) (17 - 6i) - (9 + 10i) = (17 - 9) + (-6i - 10i) = 8 + (-16i) = 8 - 16i 8. (16 + 17i) + (-8 - 12i) (16 + 17i) + (-8 - 12i) = (16 + (-8)) + (17i + (-12i)) = 8 + 5i Module 3 129 Lesson 2 COLLABORATIVE LEARNING A2_MNLESE385894_U2M03L2 129 Peer-to-Peer Activity Have students work in pairs. Each student writes an addition and subtraction problem for an imaginary number (a + bi), a pure imaginary number (bi), and a real number (a). The partners exchange papers and solve all six problems, then exchange papers again to check their answers. 129 Lesson 3.2 5/22/14 10:29 AM Explain 3 Multiplying Complex Numbers EXPLAIN 3 To multiply two complex numbers, use the distributive property to multiply each part of one number by each part of the other. Use the fact that i 2 = -1 to simplify the result. Example 3 Multiplying Complex Numbers Multiply the complex numbers. (4 + 9i)(6 − 2i) Substitute −1 for i . = 24 − 8i + 54i − 18(-1) Combine like terms. = 42 + 46i 2 QUESTIONING STRATEGIES (4 + 9i)(6 − 2i) = 24 − 8i + 54i − 18i 2 Use the distributive property. How is multiplying two imaginary numbers similar to the FOIL method? The same steps are used to multiply two imaginary numbers as are used to multiply two binomials. (−3 + 12i)(7 + 4i) Use the distributive property. (−3 + 12i)(7 + 4i) = -21 −12i + 84i + 48i 2 Substitute −1 for i 2. = -21 − 12i + 84i + 48(−1) Combine like terms. = -69 + 72 i If you use the FOIL method to multiply two imaginary numbers, which of the products, F, O, I, or L, are real? Which are pure imaginary? F and L are real; O and I are pure imaginary. Reflect 9. Is the product of (a + bi)(a - bi), where a and b are real numbers, a real number or an imaginary number? Explain. (a + bi)(a - bi) = a 2 - abi + abi - b 2i 2 AVOID COMMON ERRORS = a 2 - b 2(-1) = a2 + b2 Some students may try to simplify a complex number by combining the real part and the imaginary part. Emphasize that just as unlike terms in an algebraic expression cannot be combined, neither can the real and imaginary parts of a complex number. Since a and b are real numbers, a 2 + b 2 is a real number. So the product (a + bi)(a - bi) is a real number. Your Turn © Houghton Mifflin Harcourt Publishing Company Multiply the complex numbers. 10. (6 - 5i)(3 - 10i) (6 - 5i) (3 - 10i) = 18 - 60i - 15i + 50i 2 = 18 - 60i - 15i + 50(-1) = -32 - 75i 11. (8 + 15i)(11 + i) (8 + 15i)(11 + i) = 88 + 8i + 165i + 15i 2 = 88 + 8i + 165i + 15(-1) = 73 + 173i Module 3 INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Reinforce how FOIL is used to multiply two binomials, such as the product of 2 + 3x and 1 − 4x, before asking students to use FOIL to multiply two imaginary numbers, such as the product of 2 + 3i and 1 − 4i. Show the work for the binomials side by side with the work for the imaginary numbers. Ask students to point out the similarities and differences in the work steps. Lesson 2 130 DIFFERENTIATE INSTRUCTION A2_MNLESE385894_U2M03L2.indd 130 5/29/14 9:21 PM Visual Learners To help visual learners see the relationships among the various types of numbers, show them the following diagram. Ask them to give examples of each type of number. Rational Real Complex Numbers Irrational Imaginary Complex Numbers 130 Explain 4 EXPLAIN 4 Solving a Real-World Problem Using Complex Numbers Electrical engineers use complex numbers when analyzing electric circuits. An electric circuit can contain three types of components: resistors, inductors, and capacitors. As shown in the table, each type of component has a different symbol in a circuit diagram, and each is represented by a different type of complex number based on the phase angle of the current passing through it. Solving a Real-World Problem Using Complex Numbers QUESTIONING STRATEGIES If two real-world quantities have values of 12 – 15i and 16 – 20i, do you know which quantity is larger? Explain. No; you can compare the real part only to the real part, and the imaginary part only to the imaginary part. Circuit Component Phase Angle Representation as a Complex Number Resistor 0° Inductor 90° An imaginary number bi where b > 0 -90° An imaginary number bi where b < 0 Capacitor INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Ask students to compare the product of A real number a A diagram of an alternating current (AC) electric circuit is shown along with the impedance (measured in ohms, Ω) of each component in the circuit. An AC power source, which is shown on the left in the diagram and labeled 120 V (for volts), causes electrons to flow through the circuit. Impedance is a measure of each component’s opposition to the electron flow. Example 4 © Houghton Mifflin Harcourt Publishing Company 2 + 3i and 2 – 3i to the product of 3 + 2i and 3 – 2i. Have them explain the relationship between the products. Both have a value of 13. The first product is (2 + 3i)(2 – 3i) = 2 2 – 6i + 6i – 9i 2 = 4 + 9 = 13. The second has the same middle terms and has a value of 9 + 4 = 13. Symbol in Circuit Diagram 4Ω 3Ω 120 V 5Ω Use the diagram of the electric circuit to answer the following questions. The total impedance in the circuit is the sum of the impedances for the individual components. What is the total impedance for the given circuit? Write the impedance for each component as a complex number. • Impedance for the resistor: 4 • Impedance for the inductor: 3i • Impedance for the capacitor: -5i Then find the sum of the impedances. Total impedance = 4 + 3i + (−5i) = 4 − 2i Ohm’s law for AC electric circuits says that the voltage V (measured in volts) is the product of the current I (measured in amps) and the impedance Z (measured in ohms): V = I ∙ Z. For the given circuit, the current I is 24 + 12i amps. What is the voltage V for each component in the circuit? Use Ohm’s law, V = I ∙ Z, to find the voltage for each component. Remember that Z is the impedance from Part A. Module 3 A2_MNLESE385894_U2M03L2.indd 131 131 Lesson 3.2 131 Lesson 2 19/03/14 3:00 PM ( ) ( ) ( ) = I ∙ Z = ( 24 + 12i) 4 = 96 + 48 i Voltage for the inductor = I ∙ Z = (24 + 12i) 3i = -36 + 72 i Voltage for the resistor Voltage for the capacitor = I ∙ Z = (24 + 12i) -5i CONNECT VOCABULARY Have students complete a Venn diagram in which one circle contains real numbers and the other circle imaginary numbers. Emphasize that the overlap shows complex numbers. Have students write three examples of each kind of number in their diagrams. = 60 − 120i Reflect 12. Find the sum of the voltages for the three components in Part B. What do you notice? Sum of voltages = (96 + 48i) + (-36 + 72i) + (60 - 120i) ⌉ ⌈ ⌉ =⌈ ⌊96 + (-36) + 60⌋ + ⌊48i + 72i + (-120i)⌋ ELABORATE = 120 + 0i = 120 The sum of the voltages equals the voltage supplied by the power source. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Relate the real and imaginary parts of a Your Turn 13. Suppose the circuit analyzed in Example 4 has a second resistor with an impedance of 2 Ω added to it. Find the total impedance. Given that the circuit now has a current of 18 + 6i amps, also find the voltage for each component in the circuit. complex number to the real (horizontal) and imaginary (vertical) axes. Students should realize that points on the horizontal axis represent real numbers, points on the vertical axis represent pure imaginary numbers, and points in the quadrants represent complex numbers. Total impedance = 2 + 4 + 3i + (-5i) = 6 - 2i Voltage for the first resistor = I ∙ Z = (18 + 6i)(4) = 72 + 24i Voltage for the second resistor = I ∙ Z = (18 + 6i)(2) = 36 + 12i Voltage for the inductor = I ∙ Z = (18 + 6i)(3i) = -18 + 54i Voltage for the capacitor = I ∙ Z = (18 + 6i)(-5i) = 30 - 90i 14. What kind of number is the sum, difference, or product of two complex numbers? The sum, difference, or product of two complex numbers is always a complex number. 15. When is the sum of two complex numbers a real number? When is the sum of two complex numbers an imaginary number? The sum of two complex numbers is a real number when the imaginary parts of the numbers are additive inverses or both 0. The sum of two complex numbers is an imaginary number when the real parts of the numbers are additive inverses or both 0. © Houghton Mifflin Harcourt Publishing Company Elaborate QUESTIONING STRATEGIES If you multiply a nonzero real number and an imaginary number, is the product real or imaginary? Why? Imaginary; c(bi)= cbi, which is imaginary since cb is real and neither b nor c equals 0. SUMMARIZE THE LESSON Module 3 132 Lesson 2 LANGUAGE SUPPORT A2_MNLESE385894_U2M03L2 132 Connect Vocabulary 5/22/14 10:29 AM How do you add, subtract, and multiply complex numbers? To add or subtract complex numbers, add or subtract their real parts and their imaginary parts separately. To multiply complex numbers, use the distributive property or the FOIL method. Provide pairs of students with 6 to 8 “number cards” or index cards on which are different complex numbers, imaginary numbers, and real numbers. Ask them to sort the cards into those categories. Pairs must agree on the classifications and justify their decisions by writing a short explanation on each card. Students then identify the real and imaginary parts of the complex numbers by labeling. Complex Numbers 132 16. Discussion What are the similarities and differences between multiplying two complex numbers and multiplying two binomial linear expressions in the same variable? The distributive property is used to multiply both complex numbers and binomial linear EVALUATE expressions. When two binomial linear expressions in the same variable are multiplied, the result is a trinomial quadratic expression. When two complex numbers are multiplied, the result is another complex number. 17. Essential Question Check-In How do you add and subtract complex numbers? To add or subtract complex numbers, combine like terms. ASSIGNMENT GUIDE Concepts and Skills Practice Explore Exploring Operations Involving Imaginary Numbers Exercise 1 Example 1 Defining Complex Numbers Exercises 2–5 Example 2 Adding and Subtracting Complex Numbers Exercises 6–9 Example 3 Multiplying Complex Numbers Exercises 10–20 Evaluate: Homework and Practice 1. Find the sum of the binomials 3 + 2x and 4 - 5x. Explain how you can use the result to find the sum of the complex numbers 3 + 2i and 4 - 5i. • Online Homework • Hints and Help • Extra Practice (3 + 2x) + (4 - 5x) = (3 + 4) + (2x - 5x) = 7 - 3x Replacing x with the imaginary unit i gives this result: (3 + 2i) + (4 - 5i) = 7 - 3i. 2. Find the product of the binomials 1 - 3x and 2 + x. Explain how you can use the result to find the product of the complex numbers 1 - 3i and 2 + i. (1 - 3x)(2 + x) = 2 - 6x + x - 3x 2 = 2 - 5x - 3x 2 Replacing x with the imaginary unit i gives this result: (1 - 3i)(2 + i) = 2 - 5i-3i 2. Because i 2 = -1, the result can be further simplified as follows: (1 - 3i)(2 + i) = 2 - 5i - 3i 2 = 2 - 5i - 3(-1) = 5 - 5i © Houghton Mifflin Harcourt Publishing Company Identify the real and imaginary parts of the given number. Then tell which of the following sets the number belongs to: real numbers, imaginary numbers, and complex numbers. 3. 5+i The real part is 5, and the imaginary part is 1. Because both the real and imaginary parts are nonzero, the number belongs only to the set of complex numbers. 4. 7 - 6i The real part is 7, and the imaginary part is -6. Because both the real and imaginary parts are nonzero, the number belongs only to the set of complex numbers. Module 3 Exercise A2_MNLESE385894_U2M03L2 133 COMMON CORE Mathematical Practices 1 Recall of Information MP.2 Reasoning 15–18 2 Skills/Concepts MP.2 Reasoning 19–22 2 Skills/Concepts MP.4 Modeling 23 2 Skills/Concepts MP.2 Reasoning 3 Strategic Thinking MP.2 Reasoning 2 Skills/Concepts MP.3 Logic 26 Lesson 3.2 Depth of Knowledge (D.O.K.) 1–14 24–25 133 Lesson 2 133 5/22/14 10:29 AM 5. 25 QUESTIONING STRATEGIES The real part is 25, and the imaginary part is 0. Because the imaginary part is 0, the number belongs to these sets: real numbers and complex numbers. 6. _ i√ 21 What is a complex conjugate? Two complex numbers of the form a + bi and a − bi are complex conjugates. The product of complex conjugates is always a real number. ― The real part is 0, and the imaginary part is √21 . Because the real part is 0, the number belongs to these sets: imaginary numbers and complex numbers. Add. 7. (3 + 4i) + (7 + 11i) 8. Just as every real number corresponds to a point on the real number line, every complex number corresponds to a point in the complex plane. The complex plane has a horizontal axis called the real axis and a vertical axis called the imaginary axis. Ask students whether they can graph the line y = 3x + 4 on the complex plane. = 8 - 2i = 10 + 15i (-1 - i) + (-10 + 3i) CRITICAL THINKING (2 + 3i) + (6 - 5i) = (2 + 6) + (3i + 5i) (3 + 4i) + (7 + 11i) = (3 + 7) + (4i + 11i) 9. (2 + 3i) + (6 - 5i) 10. (-9 - 7i) + (6 + 5i) (-1 - i) + (-10 + 3i) = (-1 - 10) + (-i + 3i) (-9 - 7i) + (6 + 5i) = (-9 + 6) + (-7i + 5i) = -3 - 2i = -11 + 2i Subtract. 11. (2 + 3i) - (7 + 6i) 12. (4 + 5i) - (14 - i) (2 + 3i) - (7 + 6i) = (2 - 7) + (3i - 6i) (4 + 5i) - (14 - i) = (4 - 14) + (5i + i) = -5 - 3i = -10 + 6i 13. (-8 - 3i) - (-9 - 5i) 14. (5 + 2i) - (5 - 2i) (5 + 2i) - (5 - 2i) = (5 - 5) + (2i + 2i) (-8 - 3i) - (-9 - 5i) = (-8 + 9) + (-3i + 5i) = 4i = 1 + 2i 15. (2 + 3i)(3 + 5i) 16. (7 + i)(6 - 9i) (2 + 3i)(3 + 5i) = 6 + 10i + 9i + 15i 2 (7 + i)(6 - 9i) = 42 - 63i + 6i - 9i 2 = 6 + 10i + 9i + 15(-1) = -9 + 19i 17. (-4 + 11i)(-5 - 8i) (-4 + 11i)(-5 - 8i) = 20 + 32i - 55i - 88i = 42 - 63i + 6i - 9(-1) = 51 - 57i 18. (4 - i)(4 + i) 2 = 20 + 32i - 55i - 88(-1) = 108 - 23i Module 3 A2_MNLESE385894_U2M03L2 134 134 © Houghton Mifflin Harcourt Publishing Company Multiply. (4 - i)(4 + i) = 16 + 4i - 4i - i 2 = 16 + 4i - 4i -(-1) = 17 Lesson 2 5/22/14 10:29 AM Complex Numbers 134 AVOID COMMON ERRORS Use the diagram of the electric circuit and the given current to find the total impedance for the circuit and the voltage for each component. Watch for students who simplify a complex number by combining the real part and the imaginary part. Clarify that just as unlike terms in an algebraic expression cannot be combined, real and imaginary parts of a complex number cannot be combined. 19. 20. 1Ω 120 V 4Ω 3Ω 120 V 3Ω The circuit has a current of 12 + 36i amps. The circuit has a current of 19.2 - 14.4i amps. Total impedance = 4 + 3i Total impedance = 1 - 3i Voltage for the resistor = I ∙ Z Voltage for the resistor = I ∙ Z = (19.2 - 14.4i)(4) = (12 + 36i)(1) = 76.8 - 57.6i = 12 + 36i Voltage for the inductor = I ∙ Z Voltage for the capacitor = I ∙ Z = (19.2 - 14.4i)(-3i) = (12 + 36i)(-3i) = 43.2 + 57.6i = 108 - 36i 6Ω 21. 2Ω 120 V 4Ω The circuit has a current of 7.2 + 9.6i amps. The circuit has a current of 16.8 + 2.4i amps. Total impedance = 6 + 2i + (-10i) = 6 - 8i Total impedance = 7 + 3i + (-4i) = 7 - i Voltage for the resistor = I ∙ Z Voltage for the resistor = I ∙ Z © Houghton Mifflin Harcourt Publishing Company = (7.2 + 9.6i)(6) = (16.8 + 2.4i)(7) = 43.2 + 57.6i = 117.6 + 16.8i Voltage for the inductor = I ∙ Z Voltage for the inductor = I ∙ Z = (7.2 + 9.6i)(2i) = (16.8 + 2.4i)(3i) = -19.2 + 14.4i = -7.2 + 50.4i Voltage for the capacitor = I ∙ Z Voltage for the capacitor = I ∙ Z = (7.2 + 9.6i)(-10i) = (16.8 + 2.4i)(-4i) = 96 - 72i Module 3 A2_MNLESE385894_U2M03L2 135 Lesson 3.2 3Ω 120 V 10 Ω 135 7Ω 22. = 9.6 - 67.2i 135 Lesson 2 5/22/14 10:29 AM 23. Match each product on the right with the corresponding expression on the left. B −16 + 30i A. (3 − 5i)(3 + 5i) B. (3 + 5i)(3 + 5i) D −34 C. (−3 − 5i)(3 + 5i) A 34 D. (3 − 5i)(−3 − 5i) C 16 − 30i INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Have students look for a pattern in powers of the imaginary unit i, going beyond i2, the highest power used in the lesson. Students will need to deduce that all the higher powers can be simplified by repeatedly dividing out i2. A. (3 - 5i)(3 + 5i) = 9 + 15i - 15i - 25i 2 = 9 + 15i - 15i - 25(-1) = 34 B. (3 + 5i)(3 + 5i) = 9 + 15i + 15i + 25i 2 = 9 + 15i + 15i + 25(-1) = -16 + 30i C. (-3 - 5i)(3 + 5i) = -9 - 15i - 15i - 25i 2 = -9 - 15i - 15i - 25(-1) = 16 - 30i D. (3 - 5i)(-3 - 5i) = -9 - 15i + 15i + 25i 2 = -9 - 15i + 15i + 25(-1) = -34 H.O.T. Focus on Higher Order Thinking 24. Explain the Error While attempting to multiply the expression (2 - 3i)(3 + 2i), a student made a mistake. Explain and correct the error. (2 - 3i)(3 + 2i) = 6 - 9i + 4i - 6i 2 = 6 - 9(-1) + 4(-1) - 6(1) =5 © Houghton Mifflin Harcourt Publishing Company =6+9-4- 6 _ The student incorrectly defined i as being equal to −1 instead of √ -1 . The student should have written the product as 6 − 9i + 4i − 6(−1) = 12 − 5i. Module 3 A2_MNLESE385894_U2M03L2.indd 136 136 Lesson 2 23/05/14 3:24 PM Complex Numbers 136 _ _ _ _ 25. Critical Thinking Show that √ 3 + i√3 and -√ 3 - i√3 are the square roots of 6i. PEERTOPEER DISCUSSION ― ― ― ― Show that the square of each number is 6i. ( √3 + i √3)( √3 + i √3) = 3 + 3i + 3i + 3i 2 Ask students to work with a partner to find examples of each of the following terms: a complex number, a real number, and an imaginary number. complex: a + bi; real: a; imaginary: bi = 3 + 3i + 3i + 3(-1) = 6i (- √―3 - i √―3)(- √―3 - i √―3) = 3 + 3i + 3i + 3i 2 = 3 + 3i + 3i + 3(-1) = 6i JOURNAL 26. Justify Reasoning What type of number is the product of two complex numbers that differ only in the sign of their imaginary parts? Prove your conjecture. Have students write about how complex numbers can be applied to the real world. How might complex numbers help describe how electric circuits operate? The product of two complex numbers that differ only in the sign of their imaginary parts is a real number. Proof: Let the complex numbers be a + bi and a - bi where a and b are real and b ≠ 0. Multiplying the numbers gives the following result: (a + bi)(a - bi) = a 2 + abi - abi - b 2i 2 = a 2 - b 2(-1) = a2 + b2 © Houghton Mifflin Harcourt Publishing Company Since the set of real numbers is closed under all operations, a 2 + b 2 is a real number. Module 3 A2_MNLESE385894_U2M03L2.indd 137 137 Lesson 3.2 137 Lesson 2 19/03/14 2:59 PM Lesson Performance Task INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Students should be able to recognize when Just as real numbers can be graphed on a real number line, complex numbers can be graphed on a complex plane, which has a horizontal real axis and a vertical imaginary axis. When a set that involves complex numbers is graphed on a complex plane, the result can be an elaborate self-similar figure called a fractal. Such a set is called a Julia set. iterations for the Julia set result in a constant or two distinct values which switch back and forth. Students can try different values of Z 0 and c to explore this phenomenon, such as Z 0 = i – 1, and c = i + 1, which returns a constant value Z n = –i +1. Consider Julia sets having the quadratic recursive rule 2 ƒ(n + 1) = (ƒ(n)) + c for some complex number ƒ(0) and some complex constant c. For a given value of c, a complex number ƒ(0) either belongs or doesn’t belong to the “’filled-in” Julia set corresponding to c depending on what happens with the sequence of numbers generated by the recursive rule. a. Letting c = i, generate the first few numbers in the sequence defined by ƒ(0) = 1 and 2 ƒ(n + 1) = (ƒ(n)) + i . Record your results in the table. AVOID COMMON ERRORS f(n + 1) =( f(n)) + i 2 f (n) n 0 f (0) = 1 2 f (1) = (f (0)) + i = (1) + i = 1 + i 1 f (1) = 1 + i f (2) = (f (1)) + i = (1 + i) + i = 3i 2 f (2) = 3i f (3) = (f (2)) + i = 3i 3 f (3) = -9 + i f (4) = (f (3)) + i = Students may have difficulty with the notation 2 f(n + 1) = (f(n)) + c. Have students read this aloud as: “f(n + 1) equals the quantity f(n) squared plus c.” Note that the nested parentheses indicate that the entire quantity f(n) is squared. 2 2 2 ( 2 2 2 -9 + i -9 + i ) +i= 2 80 -17i _ b. The magnitude of a complex number a + bi is the real number √a 2 + b 2 . In the complex plane, the magnitude of a complex number is the number’s distance from the origin. If the magnitudes of the numbers in the sequence generated by a Julia set’s recursive rule, where ƒ(0) is the starting value, remain bounded, then ƒ(0) belongs to the “filled-in” Julia set. If the magnitudes increase without bound, then ƒ(0) doesn’t belong to the “filled-in” Julia set. Based on your completed table for ƒ(0) = 1, would you say that the number belongs to the “filled-in” Julia set corresponding to c = i? Explain. c. Would you say that ƒ(0) = i belongs to the “filled-in” Julia set corresponding to c = i? Explain. __ _ b. The magnitude of f(0) = 1 is √1 2 + 02 = √1 = 1. The magnitude of f(1) = 1 + i is __ __ _ _ √1 2 + 12 = √2 . The magnitude of f(2) = 3i is √0 2 + 32 = √9 = 3. The magnitude __ _ of f(3) = -9 + i is √(-9 2) + 12 = √82 . The magnitude of f(4) = 80 - 17i __ _ is √80 2 + ( -17)2 = √6689 . The magnitudes appear to be increasing without bound, so f(0) = 1 does not belong to the “filled-in” Julia set. c. For f(0) = i , the sequence of numbers generated by the recursive rule is f (0) = i, f (1) = -1 + i, f (2) = -i, f (3) = -1 + i, f (3) = -1 + i, f (4) = -i and so on. Since the © Houghton Mifflin Harcourt Publishing Company ⋅ Image Credits: ©koi88/ Shutterstock ( ) +i= _ magnitudes of the numbers never exceed √2 , f(0) = i belongs to the “filled-in” Julia set. Module 3 Lesson 2 138 EXTENSION ACTIVITY A2_MNLESE385894_U2M03L2 138 Students can use the Julia set to generate a visual pattern. Each set of ordered pairs (a, b) in the grid represents a complex number a + bi. Use these complex numbers to calculate up to three iterations for the Julia set, using c = 0. Then find the absolute value of the result, which is √a 2 + b 2 . 5/22/14 10:29 AM ――― If |Z 1|>2, stop, and color that square red. If |Z 2|>2, stop, and color that square green. If |Z 3|>2, stop, and color that square blue. Have students experiment with larger grids and with c ≠ 0. Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. Complex Numbers 138