## Engage NY Eureka Math Algebra 2 Module 1 Lesson 37 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 37 Example Answer Key

Example 1.

Addition with Complex Numbers

Compute (3 + 4i) + (7 – 20i).

Answer:

(3 + 4i) + (7 – 20i) = 3 + 4i + 7 – 20i = (3 + 7) + (4 – 20)i = 10 – 16i

Example 2.

Subtraction with Complex Numbers

Compute (3 + 4i) – (7 – 20i).

Answer:

(3 + 4i) – (7 – 20i) = 3 + 4i – 7 + 20i = (3 – 7) + (4 + 20)i = – 4 + 20i

Example 3.

Multiplication with Complex Numbers

Compute (1 + 2i) (1 – 2i).

Answer:

(1 + 2i) (1 – 2i) = 1 + 2i – 2i – 4i^{2} – 2 + 4i + 5

= 1 + 0 – 4(-1)

= 1 + 4

= 5

Example 4.

Multiplication with Complex Numbers

Verify that -1 + 2i and -1 – 2i are solutions to x^{2} + 2x + 5 = 0.

Answer:

– 1 + 2i:

(- 1 + 2i)^{2} + 2(-1 + 2i) + 5 = 1 – 4i + 4i^{2} – 2 + 4i + 5

= 4i^{2} – 4i + 4i + 1 – 2 + 5

= – 4 + 0 + 4

= 0

– 1 – 2i:

(- 1 – 2i)^{2} + 2(- 1 – 2i) + 5 = 1 + 4i + 4i^{2} – 2 – 4i + 5

= 4i^{2} + 4i – 4i + 1 – 2 + 5

= – 4 + 0 + 4

= 0

So, both complex numbers – 1 – 2i and – 1 + 21 are solutions to the quadratic equation x^{2} + 2x + 5 = 0.

### Eureka Math Algebra 2 Module 1 Lesson 37 Opening Exercise Answer Key

Solve each equation for x.

a. x – 1 = 0

Answer:

1

b. x + 1 = 0

Answer:

– 1

c. x^{2} – 1 = 0

Answer:

1, – 1

d. x^{2} + 1 = 0

Answer:

No real solution

### Eureka Math Algebra 2 Module 1 Lesson 37 Problem Set Answer Key

Question 1.

Locate the point on the complex plane corresponding to the complex number given in parts (a) – (b). On one set of axes, label each point by its identifying letter. For example, the point corresponding to 5 + 2i should be labeled a.

a. 5 + 2i

b. 3 – 2i â€˜I

c. – 2 – 4i

d. – i

e. \(\frac{1}{2}\) + i

f. âˆš2 – 3i

g. 0

h. \(-\frac{3}{2}+\frac{\sqrt{3}}{2}\)i

Answer:

Question 2.

Express each of the following in a + bi form.

a. (13 + 4i) + (7 + 5i)

Answer:

(13 + 7) + (4 + 5)i = 20 + 9i

b. (5 – i) – 2 (1 – 3i)

Answer:

5 – i – 2 + 6i = 3 + 5i

c. ((5 – i) – 2 (1 – 3i))^{2}

Answer:

(3 + 5i)^{2} = 9 + 30i + 25i^{2}

= 9 + 30i + (- 25)

= – 16 + 30i

d. (3 – i) (4 + 7i)

Answer:

12 – 4i + 21i – 7i^{2} = 12 + 17i – (-7)

= 19 + 17i

e. (3 – i) (4 + 7i) – ((5 – i) – 2(1 – 3i))

Answer:

(19 + 17i) – (3 + 5i) = (19 – 3) + (17 – 5)i

= 16 + 12i

Question 3.

Express each of the following in a + bi form.

a. (2 + 5i) + (4 + 3i)

Answer:

(2 + 5i) + (4 + 3i) = (2 + 4) + (5 + 3)i

= 6 + 8i

b. (- 1 + 2i) – (4 – 3i)

Answer:

(- 1 + 2i) – (4 – 3i) = – 1 + 2i – 4 + 3i

= – 5 + 5i

c. (4 + i) + (2 – i) – (1 – i)

Answer:

(4 + i) + (2 – i) – (1 – i) = 4 + i + 2 – i – 1 + i

= 5 + i

d. (5 + 3i) (3 + 5i)

Answer:

(5 + 3i) (3 + 5i) = 5 âˆ™ 3 + 3 âˆ™ 3i + 5 âˆ™ 5i + 3i âˆ™ 5i

= 15 + 9i + 25i + 15i^{2}

= 15 + 34i – 15

= 0 + 34i

= 34i

e. – i(2 – i) (5 + 6i)

Answer:

– i(2 – i)(5 + 6i) = -i(10 – 5i + 12i – 6i^{2})

= – i(10 + 7i + 6)

= – i(16 + 7i)

= – 16i – 7^{2}

= – 16i + 7

= 7 – 16i

f. (1 + i) (2 – 3i) + 3i(1 – i) – i

Answer:

(1 + i)(2 – 3i) + 3i(1 – L) – i = (2 + 2i – 3i – 3i^{2}) + 3i – 3i^{2} – i

= 2 + 2i – 3i + 3 + 3i + 3 – i

= 8 + i

Question 4.

Find the real values of x and y in each of the following equations using the fact that if a + bi = c + di, then a = c and b = d.

a. 5x + 3yi = 20 + 9i

Answer:

5x = 20

x = 4

3yi = 9i

y = 3

b. 2(5x + 9) = (10 – 3y)i

Answer:

2(5x + 9) + 0i = 0 +(10 – 3y)i

2(5x + 9) = 0

x = \(\frac{9}{5}\)

0i = (10 – 3y)i

10 – 3y = 0

y = \(\frac{10}{3}\)

c. 3(7 – 2x) – 5(4y – 3)i = x – 2(1 + y)i

Answer:

3(7 – 2x) = x

21 – 6x = x

21 = 7x

x = 3

– 5 (4y – 3)i = – 2(1 + y)i

– 5 (4y – 3) = – 2(1 + y)

– 20y + 15 = – 2 – 2y

17 = 18y

y = \(\frac{17}{18}\)

Question 5.

Since i^{2} = – 1, we see that

i^{3} = i^{2} âˆ™ i = – 1 âˆ™ i = – i

i^{4} = i^{2} âˆ™ i^{2} = – 1 âˆ™ – 1 = 1.

Plot i, i^{2}, i^{3}, and i^{4} on the complex plane, and describe how multiplication by each rotates points in the complex plane.

Answer:

Multiplying by i rotates points by 90Â° counterclockwise around (0, 0). Multiplying by i^{2} = – 1 rotates points by 180Â° about (0, 0). Multiplying by t^{3} = – i rotates points counterclockwise by 270Â° about the origin, which is equivalent to rotation by 90Â° clockwise about the origin. Multiplying by i^{4} rotates points counter clockwise by 360Â°, which is equivalent to not rotating at all. The points i, i^{2}, i^{3}, and i^{4} are plotted below on the complex plane.

Question 6.

Express each of the following in a + bi form.

a. i^{5}

Answer:

0 + i

b. i^{6}

Answer:

– 1 + 0i

c. i^{7}

Answer:

0 – i

d. i^{8}

Answer:

1 + 0i

e. i^{102}

Answer:

– 1 + 0i

A simple approach is to notice that every 4 multiplications by i result in four 90Â° rotations, which takes i^{4} back to 1. Therefore, divide 102 by 4, which is 25 with remainder 2. So, 102 90Â° rotations is equivalent to 25 360Â° rotations and a 180Â° rotation, and thus i^{102} = – 1.

Question 7.

Express each of the following in a + bi form.

a. (1 + i)^{2}

Answer/:

(1 + i) (1 + i) = 1 + i + i + i^{2}

= 1 + 2i – 1

= 2i

b. (1 + i)^{4}

Answer:

(1 + i)^{4} = ((1 + i)^{2})^{2}

= (2i)^{2}

= 4i^{2}

= – 4

c. (1 + i)^{6}

Answer:

(1 + i)^{6} = (1 + i)^{2} (1 + i)^{4}

= (2i) (- 4)

= – 8i

Question 8.

Evaluate x^{2} – 6x when x = 3 – i.

Answer:

– 10

Question 9.

Evaluate 4x^{2} – 12x when x = \(\)

Answer:

– 10

Question 10.

Show by substitution that \(\frac{5-i \sqrt{5}}{5}\) is a solution to 5x^{2} – 10x + 6 = 0.

Answer:

\(5\left(\frac{5-i \sqrt{5}}{5}\right)^{2}\) – 10 \(\left(\frac{5-i \sqrt{5}}{5}\right)\) = \(\frac{1}{5}\) (5 – iâˆš5) (5 – iâˆš5) – 2(5 – iâˆš5) + 6

= \(\frac{1}{5}\) (25 – 10iâˆš5 + 5i^{2}) – 2 (5 – iâˆš5) + 6

= \(\frac{1}{5}\) (25 – 10iâˆš5 – 5) – 2 (5 – iâˆš5) + 6

= 5 – 2iâˆš5 – 1 – 10 + 2iâˆš5 + 6

= 0

Question 11.

a. Evaluate the four products below.

Evaluate âˆš9 âˆ™ âˆš4

Answer:

3 âˆ™ 2 = 6

Evaluate âˆš9 âˆ™ âˆš-4

Answer:

3 âˆ™ 2i = 6i

Evaluate âˆš-9 âˆ™ âˆš4.

Answer:

3i âˆ™ 2 = 6i

Evaluate âˆš-9 âˆ™ âˆš-4

Answer:

3i âˆ™ 2i = 6i^{2} = – 6

b. Suppose a and b are positive real numbers. Determine whether the following quantities are equal or not equal.

Answer:

âˆša âˆ™ âˆšb and âˆš-a âˆ™ âˆš-b

Answer:

not equal

âˆš-a âˆ™ âˆšb and âˆša âˆ™ âˆš-b

Answer:

equal

### Eureka Math Algebra 2 Module 1 Lesson 37 Exit Ticket Answer Key

Express the quantities below in a + bi form, and graph the corresponding points on the complex plane. If you use one

set of axes, be sure to label each point appropriately.

(1 + i) – (1 – i)

Answer:

(1 + i) – (1 – i) = 0 + 2i

= 2i

(1 + i) (1 – i)

Answer:

(1 + i) (1 – i) = 1 + i – i – i^{2}

= 1 – i^{2}

= 1 + 1

= 2 + 0i

= 2

i(2 – i) (1 + 2i)

Answer:

i(2 – i) (1 + 2i) = i(2 + 4i – i – 2i^{2})

= i(2 + 3i – 2(-1))

= i(2 + 3i + 2)

= i(4 + 3i)

= 4i + 3i^{2}

= – 3 + 4i