## Engage NY Eureka Math Precalculus Module 3 Lesson 3 Answer Key

### Eureka Math Precalculus Module 3 Lesson 3 Exploratory Challenge Answer Key

Exploratory Challenge

Consider the equation x^{3} = 1.

a. Use the graph of f(x) = x^{3} – 1 to explain why 1 is the only real number solution to the equation x^{3} = 1.

Answer:

From the graph, you can see that the point (1,0) is the x – intercept of the function. That means that 1 is a zero of the polynomial function and thus is a solution to the equation x^{3} – 1 = 0.

b. Find all of the complex solutions to the equation x^{3} = 1. Come up with as many methods as you can for finding the solutions to this equation.

Answer:

Method 1: Factoring Using a Polynomial Identity and Using the Quadratic Formula

Rewrite the equation in the form x^{3} – 1 = 0, and use the identity a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2}) to factor x^{3} – 1.

x^{3} – 1 = 0

(x – 1)(x^{2} + x + 1) = 0

Then

x – 1 = 0 or x^{2} + x + 1 = 0.

The solution to the equation x – 1 = 0 is 1. The quadratic formula gives the other solutions.

x = \(\frac{ – 1 \pm \sqrt{ – 3}}{2}\) = \(\frac{ – 1 \pm i \sqrt{3}}{2}\)

So, the solution set is

{1, – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i}.

Method 2: Using the Polar Form of a Complex Number

The solutions to the equation x^{3} = 1 are the cube roots of 1.

The number 1 has modulus 1 and argument 0 (or any rotation that terminates along the positive real axis such as 2π or 4π, etc.).

The modulus of the cube roots of 1 is \(\sqrt[3]{1}\) = 1. The arguments are solutions to

3θ = 0

3θ = 2π

3θ = 4π

3θ = 6π.

The solutions to these equations are 0, \(\frac{2\pi}{3}\), \(4\frac{\pi}{3}\), 2π, …. Since the rotations cycle back to the same locations in the complex plane after the first three, we only need to consider 0, \(\frac{2\pi}{3}\), and \(\frac{4\pi}{3}\).

The solutions to the equation x^{3} = 1 are

1(cos(0) + i sin(0) ) = 1

1(cos(\(\frac{2\pi}{3}\)) + i sin(\(\frac{2\pi}{3}\))) = – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i

1(cos(\(\frac{4\pi}{3}\)) + i sin(\(\frac{4\pi}{3}\))) = – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

Method 3: Using the Techniques of Lessons 1 and 2 from This Module

Let x = a + bi; then (a + bi)^{3} = 1 + 0i.

Expand (a + bi)^{3}, and equate the real and imaginary parts with 1 and 0.

(a + bi)^{3} = (a + bi)(a^{2} + 2abi – b^{2}) = a^{3} + 3a^{2} bi – 3ab^{2} – b^{3} i

The real part of (a + bi)^{3} is a^{3} – 3ab^{2}, and the imaginary part is 3a^{3} b – b^{3}. Thus, we need to solve the system

a^{3} – 3ab^{2} = 1

3a^{2} b – b^{3} = 0.

Rewriting the second equation gives us

b(3a^{2} – b^{2}) = 0.

If b = 0, then a^{3} = 1 and a = 1. So, a solution to the equation x^{3} = 1 is 1 + 0i = 1.

If 3a^{2} – b^{2} = 0, then b^{2} = 3a^{2}, and by substitution,

a^{3} – 3a(3a^{2} ) = 1

– 8a^{3} = 1

a^{3} = – \(\frac{1}{8}\) .

This equation has one real solution: – \(\frac{1}{2}\). If = – \(\frac{1}{2}\), then b^{2} = 3 ( – \(\frac{1}{2}\))^{2} = \(\frac{3}{4}\), so b = \(\frac{\sqrt{3}}{2}\) or = – \(\frac{\sqrt{3}}{2}\). The other two solutions to the equation x^{3} = 1 are – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i and – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

### Eureka Math Precalculus Module 3 Lesson 3 Exercise Answer Key

Opening Exercise

Consider the equation x^{n} = 1 for positive integers n.

a. Must an equation of this form have a real solution? Explain your reasoning.

Answer:

The number 1 will always be a solution to x^{n} = 1 because 1^{n} = 1 for any positive integer n.

b. Could an equation of this form have two real solutions? Explain your reasoning.

Answer:

When n is an even number, both 1 and – 1 are solutions. The number 1 is a solution because 1^{n} = 1 for any positive integer. The number – 1 is a solution because ( – 1)^{n} = ( – 1)^{2k} where k is a positive integer if n is even and

( – 1)^{2k} = (( – 1)^{2})^{k} = 1^{k} = 1.

c. How many complex solutions are there for an equation of this form? Explain how you know.

Answer:

We can rewrite the equation in the form x^{n} – 1 = 0. The solutions to this polynomial equation are the roots of the polynomial p(x) = x^{n} – 1. The fundamental theorem of algebra says that the polynomial

p(x) = x^{n} – 1 factors over the complex numbers into the product of n linear terms. Each term identifies a complex root of the polynomial. Thus, a polynomial equation of degree n has at most n solutions.

Exercises

Solutions to the equation x^{n} = 1 for positive integers n are called the nth roots of unity.

Exercise 1.

What are the square roots of unity in rectangular and polar form?

Answer:

The square roots of unity in rectangular form are the real numbers 1 and – 1.

In polar form, 1(cos(0) + i sin(0) ) and 1(cos(π) + i sin(π) ).

Exercise 2.

What are the fourth roots of unity in rectangular and polar form? Solve this problem by creating and solving a polynomial equation. Show work to support your answer.

Answer:

The fourth roots of unity in rectangular form are 1, – 1, i, – i.

x^{4} = 1

x^{4} – 1 = 0

(x^{2} – 1)(x^{2} + 1) = 0

The solutions to x^{2} – 1 = 0 are 1 and – 1. The solutions to x^{2} + 1 = 0 are i and – i.

In polar form, 1(cos(0) + i sin(0) ), 1(cos(\(\frac{\pi}{2}\)) + i sin(\(\frac{\pi}{2}\)) ), 1(cos(π) + i sin(π) ), 1(cos(\(\frac{3\pi}{2}\)) + i sin(\(\frac{3\pi}{2}\)) ).

Exercise 3.

Find the sixth roots of unity in rectangular form by creating and solving a polynomial equation. Show work to support your answer. Find the sixth roots of unity in polar form.

Answer:

x^{6} – 1 = 0

(x^{3} + 1)(x^{3} – 1) = 0

(x + 1)(x^{2} – x + 1)(x – 1)(x^{2} + x + 1) = 0

By inspection, 1 and – 1 are sixth roots. Using the quadratic formula to find the solutions to x^{2} – x + 1 = 0 and x^{2} + x + 1 = 0 gives the other four roots: \(\frac{1}{2}\) + \(\frac{i \sqrt{3}}{2}\), \(\frac{1}{2}\) – \(\frac{i \sqrt{3}}{2}\), – \(\frac{1}{2}\) + \(\frac{i \sqrt{3}}{2}\), and – \(\frac{1}{2}\) – \(\frac{i \sqrt{3}}{2}\).

In polar form, 1(cos(0) + i sin(0) ), 1(cos(\(\frac{\pi}{3}\)) + i sin(\(\frac{\pi}{3}\)) ), (cos(\(\frac{2\pi}{3}\)) + i sin(\(\frac{2\pi}{3}\)) ), 1(cos(π) + i sin(π) ), 1(cos(\(\frac{4\pi}{3}\)) + i sin(\(\frac{4\pi}{3}\)) ), 1(cos(\(\frac{5\pi}{3}\)) + i sin(\(\frac{5\pi}{3}\)) ).

Exercise 4.

Without using a formula, what would be the polar forms of the fifth roots of unity? Explain using the geometric effect of multiplication complex numbers.

Answer:

The modulus would be 1 because dividing 1 into the product of six equal numbers still means each number must be 1. The arguments would be fifths of 2π, so 0, \(\frac{\pi}{5}\), \(\frac{2\pi}{5}\), \(3\frac{\pi}{5}\), and \(\frac{4\pi}{5}\). The fifth roots of z, when multiplied together, must equal z^{1} = z^{\(\frac{1}{5}\)}∙z^{\(\frac{1}{5}\)}∙z^{\(\frac{1}{5}\)}∙z^{\(\frac{1}{5}\)}∙z^{\(\frac{1}{5}\)} . That would be like starting with the real number 1 and rotating it by \(\frac{1}{5}\) of 2π and dilating it by a factor of 1 so that you ended up back at the real number 1 after 5 repeated multiplications.

Discussion

What is the modulus of each root of unity regardless of the value of n? Explain how you know.

Answer:

The modulus is always 1 because the n^{th/sup> root of 1 is equal to 1. The points are on the unit circle, and the radius is always 1.}

How could you describe the location of the roots of unity in the complex plane?

Answer:

They are points on a unit circle, evenly spaced every \(\frac{2\pi}{n}\) units starting from 1 along the positive real axis.

The diagram below shows the solutions to the equation x^{3} = 27. How do these numbers compare to the cube roots of unity (e.g., the solutions to x^{3} = 1)?

Answer:

They are points on a circle of radius 3 since the cube root of 27 is 3. Each one is a scalar multiple (by a factor of 3) of the cube roots of unity. Thus, they have the same arguments but a different modulus.

### Eureka Math Precalculus Module 3 Lesson 3 Problem Set Answer Key

Question 1.

Graph the n^{th} roots of unity in the complex plane for the specified value of n.

a. n = 3

Answer:

b. n = 4

Answer:

c. n = 5

Answer:

d. n = 6

Answer:

Question 2.

Find the cube roots of unity by using each method stated.

a. Solve the polynomial equation x^{3} = 1 algebraically.

Answer:

x^{3} – 1 = 0, (x – 1)(x^{2} + x + 1) = 0, x = 1, x^{2} + x + 1 = 0, x = \(\frac{ – 1 \pm \sqrt{1 – 4}}{2}\) = – \(\frac{1}{2}\)±\(\frac{\sqrt{3} i}{2}\)

The roots of unity are 1, – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

b. Use the polar form z^{3} = r(cos(θ) + sin(θ) ), and find the modulus and argument of z.

Answer:

z^{3} = 1, r^{3} = 1, r = 1

3θ = 0, 3θ = 2π, 3θ = 4π, …; therefore, θ = 0, \(\frac{2\pi}{3}\), \(\frac{4\pi}{3}\), 2π, …

z = \(\sqrt[3]{r}\)(cos(θ) + i sin(θ) )

z_{1} = 1(cos(0) + i sin(0) ) = 1

z_{2} = 1(cos(\(\frac{2\pi}{3}\)) + i sin(\(\frac{2\pi}{3}\)) ) = – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i

z_{3} = 1(cos(\(\frac{4\pi}{3}\)) + i sin(\(\frac{4\pi}{3}\)) ) = – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i

The roots of unity are 1, – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

c. Solve (a + bi)^{3} = 1 by expanding (a + bi)^{3} and setting it equal to 1 + 0i.

Answer:

(a + bi)^{3} = 1, a^{3} + 3a^{2} bi – 3ab^{2} – b^{3} i = 1; therefore, a^{3} – 3ab^{2} = 1 and 3a^{2} b – b^{3} = 0.

For 3a^{2} b – b^{3} = 0, b(3a^{2} – b^{2} ) = 0, we have either b = 0 or a^{2} – b^{2} = 0.

For b = 0, we substitute it in a^{3} – 3ab^{2} = 1, a^{3} = 1, a = 1; therefore, we have 1 + 0i.

For 3a^{2} – b^{2} = 0, b^{2} = 3a^{2}, we substitute it in a^{3} – 3ab^{2} = 1, a^{3} – 9a^{3} = 1, a^{3} = – \(\frac{1}{8}\), a = – \(\frac{1}{2}\).

For = – \(\frac{1}{2}\), we substitute it in b^{2} = 3a^{2}, and we get b = ±\(\frac{\sqrt{3}}{2}\). Therefore, we have \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i and \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

The roots of unity are 1, – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i.

Question 3.

Find the fourth roots of unity by using the method stated.

a. Solve the polynomial equation x^{4} = 1 algebraically.

Answer:

x^{4} – 1 = 0, (x^{2} + 1)(x + 1)(x – 1) = 0, x = ±i, x = ±1

The roots of unity are 1, i, – 1, – i.

b. Use the polar form z^{4} = r(cos(θ) + sin(θ) ), and find the modulus and argument of z.

Answer:

z^{4} = 1, r^{4} = 1, r = 1

4θ = 0, 4θ = 2π, 4θ = 4π, 4θ = 6π, 4θ = 8π, …; therefore, θ = 0, \(\frac{\pi}{2}\), π, \(\frac{3\pi}{2}\), 2π, …

z = \(\sqrt[3]{r}\)(cos(θ) + i sin(θ) )

z_{1} = 1(cos(0) + i sin(0) ) = 1

z_{2} = 1(cos(\(\frac{\pi}{2}\)) + i sin(\(\frac{\pi}{2}\))) = i

z_{3} = 1(cos(π) + i sin(π) ) = – 1

z_{4} = 1(cos(\(\frac{3\pi}{2}\)) + i sin(\(\frac{3\pi}{2}\))) = – i

The roots of unity are 1, i, – 1, – i.

c. Solve (a + bi)^{4} = 1 by expanding (a + bi)^{4} and setting it equal to 1 + 0i.

Answer:

(a + bi)^{4} = 1, a^{4} + 4a^{3} bi – 6a^{2} b^{2} – 4ab^{3} i + b^{4} = 1.

Therefore, a^{4} – 6a^{2} b^{2} + b^{4} = 1 and 4a^{3} b – 4ab^{3} = 0.

For 4a^{3} b – 4ab^{3} = 0, 4ab(a^{2} – b^{2} ) = 0, we have either a = 0, b = 0, or a^{2} – b^{2} = 0.

For a = 0, we substitute it in a^{4} – 6a^{2} b^{2} + b^{4} = 1, b^{4} = 1, b = ±1; therefore, we have fourth roots of unity i and – i.

For b = 0, we substitute it in a^{4} – 6a^{2} b^{2} + b^{4} = 1, a^{4} = 1, a = ±1 for a. b∈R; therefore, we have fourth roots of unity 1 and – 1.

For a^{2} – b^{2} = 0, a^{2} = b^{2}, we substitute it in a^{4} – 6a^{2} b^{2} + b^{4} = 1, b^{4} – 6b^{4} + b^{4} = 1, 4b^{4} = – 1; there is no solution for b for a. b∈R;

The roots of unity are 1, i, – 1, – i.

Question 4.

Find the fifth roots of unity by using the method stated.

Use the polar form z^{5} = r(cos(θ) + sin(θ) ), and find the modulus and argument of z.

Answer:

z^{5} = 1, r^{5} = 1, r = 1

5θ = 0, 5θ = 2π, 5θ = 4π, 5θ = 6π, 5θ = 8π, 5θ = 10π, …; therefore, θ = 0, \(\frac{2\pi}{5}\), \(\frac{4\pi}{5}\), \(\frac{6\pi}{5}\), \(\frac{8\pi}{5}\), 2π, …

z_{1} = 1(cos(0) + i sin(0) ) = 1

z_{2} = 1(cos(\(\frac{2\pi}{5}\)) + i sin(\(\frac{2\pi}{5}\)) ) = 0.309 + 0.951i

z_{3} = 1(cos(\(\frac{4\pi}{5}\)) + i sin(\(\frac{4\pi}{5}\)) ) = – 0.809 + 0.588i

z_{4} = 1(cos(\(\frac{6\pi}{5}\)) + i sin(\(\frac{6\pi}{5}\)) ) = – 0.809 – 0.588i

z_{5} = 1(cos(\(\frac{8\pi}{5}\)) + i sin(\(\frac{8\pi}{5}\)) ) = 0.309 – 0.951i

The roots of unity are 1, 0.309 + 0.951i, – 0.809 + 0.588i, – 0.809 – 0.588i, 0.309 – 0.951i, 1.

Question 5.

Find the sixth roots of unity by using the method stated.

a. Solve the polynomial equation x^{6} = 1 algebraically.

Answer:

x^{6} – 1 = 0, (x + 1)(x^{2} – x + 1)(x – 1)(x^{2} + x + 1) = 0, x = ±1, x = \(\frac{1}{2}\) ±\(\frac{\sqrt{3} i}{2}\), x = – \(\frac{1}{2}\) ±\(\frac{\sqrt{3} i}{2}\)

The roots of unity are 1, – 1, \(\frac{1}{2}\) + \(\frac{\sqrt{3} i}{2}\), \(\frac{1}{2}\) – \(\frac{\sqrt{3} i}{2}\), – \(\frac{1}{2}\) + \(\frac{\sqrt{3} i}{2}\), – \(\frac{1}{2}\) – \(\frac{\sqrt{3} i}{2}\).

b. Use the polar form z^{6} = r(cos(θ) + sin(θ) ), and find the modulus and argument of z.

Answer:

z^{6} = 1, r^{6} = 1, r = 1

6θ = 0, 6θ = 2π, 6θ = 4π, 6θ = 6π, 6θ = 8π, 6θ = 10π, 6θ = 12π, 6θ = 14π, …; therefore, θ = 0, \(\frac{\pi}{3}\), \(\frac{2}{3}\), π, \(\frac{4\pi}{3}\), \(\frac{5\pi}{3}\), 2π, \(\frac{7\pi}{3}\), …

z_{1} = 1(cos(0) + i sin(0) ) = 1

z_{2} = 1(cos(\(\frac{\pi}{3}\)) + i sin(\(\frac{\pi}{3}\)) ) = \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i

z_{3} = 1(cos(\(\frac{2\pi}{3}\)) + i sin(\(\frac{2\pi}{3}\)) ) = – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i

z_{4} = 1(cos(π) + i sin(π) ) = – 1

z_{5} = 1(cos(\(\frac{4\pi}{3}\)) + i sin(\(\frac{4\pi}{3}\)) ) = – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i

z_{6} = 1(cos(\(\frac{5\pi}{3}\)) + i sin(\(\frac{5\pi}{3}\)) ) = \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i

The roots of unity are 1, \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i, – 1, – \(\frac{1}{2}\) – \(\frac{\sqrt{3}}{2}\) i, \(\frac{1}{2}\) + \(\frac{\sqrt{3}}{2}\) i.

Question 6.

Consider the equation x^{N} = 1 where N is a positive whole number.

a. For which value of N does x^{N} = 1 have only one solution?

Answer:

N = 1

b. For which value of N does x^{N} = 1 have only ±1 as solutions?

Answer:

N = 2

c. For which value of N does x^{N} = 1 have only ±1 and ±i as solutions?

Answer:

N = 4

d. For which values of N does x^{N} = 1 have ±1 as solutions?

Answer:

Any even number N produces solutions ±1.

Question 7.

Find the equation that has the following solutions.

a.

Answer:

x^{8} = 1

b.

Answer:

x^{3} = – 1

c.

Answer:

x^{2} = 1

Question 8.

Find the equation (a + bi)^{N} = c that has solutions shown in the graph below.

Answer:

( – 1 + \(\sqrt{3}\)i)^{3} = 8

### Eureka Math Precalculus Module 3 Lesson 3 Exit Ticket Answer Key

Question 1.

What is a fourth root of unity? How many fourth roots of unity are there? Explain how you know.

Answer:

The fourth root of unity is a number that multiplied by itself 4 times is equal to 1. There are 4 fourth roots of unity. x^{4} = 1 results in solving the polynomial x^{4} – 1 = 0. The fundamental theorem of algebra guarantees four roots since that is the degree of the polynomial.

Question 2.

Find the polar form of the fourth roots of unity.

Answer:

For the fourth roots of unity, n = 4 and r = 1, so each root has modulus 1, and the arguments are 0, \(\frac{2\pi}{4}\), \(\frac{4\pi}{4}\), and \(\frac{6\pi}{4}\). We can rewrite the arguments as 0, \(\frac{\pi}{2}\), π, and \(\frac{3\pi}{2}\). Then, the fourth roots of unity are

x_{1} = cos(0) + i sin(0) = 1

x_{2} = cos(\(\frac{\pi}{2}\)) + i sin(\(\frac{\pi}{2}\)) = i

x_{3} = cos(π) + i sin(π) = – 1

x_{4} = cos(\(\frac{3\pi}{2}\)) + i sin(\(\frac{3\pi}{2}\)) = – i.

Question 3.

Write x^{4} – 1 as a product of linear factors, and explain how this expression supports your answers to Problems 1 and 2.

Answer:

Since there are four roots of unity, there should be four linear factors.

x^{4} – 1 = (x – 1)(x – i)(x + 1)(x + i)