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Lesson 3: Roots of Unity NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M3 PRECALCULUS AND ADVANCED TOPICS Lesson 3: Roots of Unity Student Outcomes . Students determine the complex roots of polynomial equations of the form 푥푛 = 1 and, more generally, equations of the form 푥푛 = 푘 positive integers 푛 and positive real numbers 푘. Students plot the 푛th roots of unity in the complex plane. Lesson Notes This lesson ties together work from Algebra II Module 1, where students studied the nature of the roots of polynomial equations and worked with polynomial identities and their recent work with the polar form of a complex number to find the 푛th roots of a complex number in Precalculus Module 1 Lessons 18 and 19. The goal here is to connect work within the algebra strand of solving polynomial equations to the geometry and arithmetic of complex numbers in the complex plane. Students determine solutions to polynomial equations using various methods and interpreting the results in the real and complex plane. Students need to extend factoring to the complex numbers (N-CN.C.8) and more fully understand the conclusion of the fundamental theorem of algebra (N-CN.C.9) by seeing that a polynomial of degree 푛 has 푛 complex roots when they consider the roots of unity graphed in the complex plane. This lesson helps students cement the claim of the fundamental theorem of algebra that the equation 푥푛 = 1 should have 푛 solutions as students find all 푛 solutions of the equation, not just the obvious solution 푥 = 1. Students plot the solutions in the plane and discover their symmetry. GeoGebra can be a powerful tool to explore these types of problems and really helps students to see that the roots of unity correspond to the vertices of a polygon inscribed in the unit circle with one vertex along the positive real axis. Classwork Opening Exercise (3 minutes) Scaffolding: Form students into small groups of 3–5 students each depending on the size of the At this level of mathematics, classroom. Much of this activity is exploration. Start the conversation by having students students may struggle to discuss and respond to the exercises in the opening. Part (c) is an important connection to remember formulas or make for students. More information on the fundamental theorem of algebra can be theorems from previous found in Algebra II Module 1 Lessons 38–40. This information is also reviewed in grades. A quick reference Lesson 1 of this module. The amount that students recall as they work on these exercises sheet or anchor chart can come with their groups can inform decisions about how much scaffolding is necessary as in handy. The Lesson Summary students work through the rest of this lesson. If students are struggling to make sense of boxes from the following the first two problems, ask them to quickly find real number solutions to these equations Algebra II and Precalculus by inspection: 푥 = 1, 푥2 = 1, 푥3 = 1, and 푥4 = 1. lessons would be helpful to have handy during this lesson: . Precalculus Module 1 Lessons 13, 18, 19 . Algebra II Module 1 Lessons 6, 40 Lesson 3: Roots of Unity 38 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M3-TE-1.3.0-08.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M3 PRECALCULUS AND ADVANCED TOPICS Opening Exercise Consider the equation 풙풏 = ퟏ for positive integers 풏. a. Must an equation of this form have a real solution? Explain your reasoning. The number ퟏ will always be a solution to 풙풏 = ퟏ because ퟏ풏 = ퟏ for any positive integer 풏. b. Could an equation of this form have two real solutions? Explain your reasoning. When 풏 is an even number, both ퟏ and −ퟏ are solutions. The number ퟏ is a solution because ퟏ풏 = ퟏ for any MP.3 positive integer. The number −ퟏ is a solution because (−ퟏ)풏 = (−ퟏ)ퟐ풌 where 풌 is a positive integer if 풏 is even and (−ퟏ)ퟐ풌 = ((−ퟏ)ퟐ)풌 = ퟏ풌 = ퟏ. c. How many complex solutions are there for an equation of this form? Explain how you know. We can rewrite the equation in the form 풙풏 − ퟏ = ퟎ. The solutions to this polynomial equation are the roots of the polynomial 풑(풙) = 풙풏 − ퟏ. The fundamental theorem of algebra says that the polynomial 풑(풙) = 풙풏 − ퟏ factors over the complex numbers into the product of 풏 linear terms. Each term identifies a complex root of the polynomial. Thus, a polynomial equation of degree 풏 has at most 풏 solutions. Exploratory Challenge (10 minutes) In this Exploratory Challenge, students should work to apply multiple methods to find the solutions to the equation 푥3 = 1. Give teams sufficient time to consider more than one method. When debriefing the solution methods students found, make sure to present and discuss the second method below, especially if most groups did not attempt the problem. Exploratory Challenge Consider the equation 풙ퟑ = ퟏ. a. Use the graph of 풇(풙) = 풙ퟑ − ퟏ to explain why ퟏ is the only real number solution to the equation 풙ퟑ = ퟏ. From the graph, you can see that the point (ퟏ, ퟎ) is the 풙-intercept of the function. That means that ퟏ is a zero of the polynomial function and thus is a solution to the equation 풙ퟑ − ퟏ = ퟎ. Lesson 3: Roots of Unity 39 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M3-TE-1.3.0-08.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M3 PRECALCULUS AND ADVANCED TOPICS b. Find all of the complex solutions to the equation 풙ퟑ = ퟏ. Come up with as many methods as you can for finding the solutions to this equation. Method 1: Factoring Using a Polynomial Identity and Using the Quadratic Formula Rewrite the equation in the form 풙ퟑ − ퟏ = ퟎ, and use the identity 풂ퟑ − 풃ퟑ = (풂 − 풃)(풂ퟐ + 풂풃 + 풃ퟐ) to factor 풙ퟑ − ퟏ. 풙ퟑ − ퟏ = ퟎ (풙 − ퟏ)(풙ퟐ + 풙 + ퟏ) = ퟎ Then 풙 − ퟏ = ퟎ or 풙ퟐ + 풙 + ퟏ = ퟎ. The solution to the equation 풙 − ퟏ = ퟎ is ퟏ. The quadratic formula gives the other solutions. −ퟏ ± √−ퟑ −ퟏ ± 풊√ퟑ 풙 = = ퟐ ퟐ So, the solution set is ퟏ √ퟑ ퟏ √ퟑ {ퟏ, − + 풊, − − 풊} . ퟐ ퟐ ퟐ ퟐ Method 2: Using the Polar Form of a Complex Number The solutions to the equation 풙ퟑ = ퟏ are the cube roots of ퟏ. The number ퟏ has modulus ퟏ and argument ퟎ (or any rotation that terminates along the positive real axis such as ퟐ흅 or ퟒ흅, etc.). The modulus of the cube roots of ퟏ is ퟑ√ퟏ = ퟏ. The arguments are solutions to ퟑ휽 = ퟎ ퟑ휽 = ퟐ흅 ퟑ휽 = ퟒ흅 ퟑ휽 = ퟔ흅. ퟐ흅 ퟒ흅 The solutions to these equations are ퟎ, , , ퟐ흅, …. Since the rotations cycle back to the same locations in ퟑ ퟑ ퟐ흅 ퟒ흅 the complex plane after the first three, we only need to consider ퟎ, , and . ퟑ ퟑ The solutions to the equation 풙ퟑ = ퟏ are ퟏ(퐜퐨퐬(ퟎ) + 풊 퐬퐢퐧(ퟎ)) = ퟏ ퟐ흅 ퟐ흅 ퟏ √ퟑ ퟏ (퐜퐨퐬 ( ) + 풊 퐬퐢퐧 ( )) = − + 풊 ퟑ ퟑ ퟐ ퟐ ퟒ흅 ퟒ흅 ퟏ √ퟑ ퟏ (퐜퐨퐬 ( ) + 풊 퐬퐢퐧 ( )) = − − 풊. ퟑ ퟑ ퟐ ퟐ Lesson 3: Roots of Unity 40 This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from PreCal-M3-TE-1.3.0-08.2015 Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 3 M3 PRECALCULUS AND ADVANCED TOPICS Method 3: Using the Techniques of Lessons 1 and 2 from This Module Let 풙 = 풂 + 풃풊; then (풂 + 풃풊)ퟑ = ퟏ + ퟎ풊. Expand (풂 + 풃풊)ퟑ, and equate the real and imaginary parts with ퟏ and ퟎ. (풂 + 풃풊)ퟑ = (풂 + 풃풊)(풂ퟐ + ퟐ풂풃풊 − 풃ퟐ) = 풂ퟑ + ퟑ풂ퟐ풃풊 − ퟑ풂풃ퟐ − 풃ퟑ풊 The real part of (풂 + 풃풊)ퟑ is 풂ퟑ − ퟑ풂풃ퟐ, and the imaginary part is ퟑ풂ퟑ풃 − 풃ퟑ. Thus, we need to solve the system 풂ퟑ − ퟑ풂풃ퟐ = ퟏ ퟑ풂ퟐ풃 − 풃ퟑ = ퟎ. Rewriting the second equation gives us 풃(ퟑ풂ퟐ − 풃ퟐ) = ퟎ. If 풃 = ퟎ, then 풂ퟑ = ퟏ and 풂 = ퟏ. So, a solution to the equation 풙ퟑ = ퟏ is ퟏ + ퟎ풊 = ퟏ. If ퟑ풂ퟐ − 풃ퟐ = ퟎ, then 풃ퟐ = ퟑ풂ퟐ, and by substitution, 풂ퟑ − ퟑ풂(ퟑ풂ퟐ) = ퟏ −ퟖ풂ퟑ = ퟏ ퟏ 풂ퟑ = − . ퟖ ퟏ ퟏ ퟏ ퟐ ퟑ √ퟑ √ퟑ − = − 풃ퟐ = ퟑ (− ) = 풃 = = − This equation has one real solution: ퟐ. If ퟐ, then ퟐ ퟒ, so ퟐ or ퟐ . The other ퟏ √ퟑ ퟏ √ퟑ 풙ퟑ = ퟏ − + 풊 − − 풊 two solutions to the equation are ퟐ ퟐ and ퟐ ퟐ . Have different groups present their solutions. If no groups present Method 2, share that with the class. The teacher may also review the formula derived in Module 1, Lesson 19 that is shown below. Given a complex number 푧 with modulus 푟 and argument 휃, the 푛th roots of 푧 are given by 휃 2휋푘 휃 2휋푘 푛√푟 (cos ( + ) + 푖 sin ( + )) 푛 푛 푛 푛 for integers 푘 and 푛 such that 푛 > 0 and 0 ≤ 푘 < 푛. Present the next few questions giving students time to discuss each one in their groups before asking for whole-class responses. Which methods are you most inclined to use and why? Solving using factoring and the quadratic formula is the easiest way to do this provided you know the polynomial identity for the difference of two cubes.
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