# Eureka Math Precalculus Module 1 Lesson 6 Answer Key

## Engage NY Eureka Math Precalculus Module 1 Lesson 6 Answer Key

### Eureka Math Precalculus Module 1 Lesson 6 Exercise Answer Key

Opening Exercise
Perform the indicated arithmetic operations for complex numbers z = -4 + 5i and w = -1-2i.

a. z + w
z + w = -5 + 3i

b. z-w
z-w = -3 + 7i

c.z + 2w
z + 2w = -6 + i

d.z-z
z-z = 0 + 0i

e. Explain how you add and subtract complex numbers.
Add or subtract the real components and the imaginary components separately.

Exercises

Exercise 1.
The length of the vector that represents z1 = 6-8i is 10 because $$\sqrt{6^{2} + (-8)^{2}}$$ = $$\sqrt{100}$$ = 10.
a. Find at least seven other complex numbers that can be represented as vectors that have length 10.
There are an infinite number of complex numbers that meet this criteria; the most obvious are 10, 6 + 8i, 8 + 6i, 10i, -6 + 8i, -8 + 6i, -10, -8-6i, -6-8i, -10i, 8-6i, and -8 + 6i. The associated vectors for these numbers are shown in the sample response for part (b).

b. Draw the vectors on the coordinate axes provided below. c. What do you observe about all of these vectors?
Students should observe that the tips of the vectors lie on the circle of radius 10 centered at the origin.

Exercise 2.
In the Opening Exercise, we computed z + 2w. Calculate this sum using vectors. Exercise 3.
In the Opening Exercise, we also computed z-z. Calculate this sum using vectors. Exercise 4.
For the vectors u and v pictured below, draw the specified sum or difference on the coordinate axes provided.
a. u + v
b. v-u
c. 2u-v
d. -u-3v Exercise 5.
Find the sum of 4 + i and -3 + 2i geometrically.
1 + 3i Exercise 6.
Show that (7 + 2i)-(4-i) = 3 + 3i by representing the complex numbers as vectors. ### Eureka Math Precalculus Module 1 Lesson 6 Problem Set Answer Key

Question 1.
Let z = 1 + i and w = 1-3i. Find the following. Express your answers in a + bi form.
a. z + w
1 + i + 1-3i = 2-2i

b. z-w
1 + i-(1-3i) = 1 + i-1 + 3i
= 0 + 4i

c. 4w
4(1-3i) = 4-12i

d. 3z + w
3(1 + i) + 1-3i = 3 + 3i + 1-3i
= 4 + 0i

e. -w-2z
-(1-3i)-2(1 + i) = -1 + 3i-2-2i
= -3 + i

f. What is the length of the vector representing z?
The length of the vector representing z is $$\sqrt{1^{2} + 1^{2}}$$ = $$\sqrt{2}$$.

g. What is the length of the vector representing w?
The length of the vector representing w is $$\sqrt{1^{2} + (-3)^{2}}$$ = $$\sqrt{10}$$.

Question 2.
Let u = 3 + 2i, v = 1 + i, and w = -2-i. Find the following. Express your answer in a + bi form, and represent the result in the plane. a. u-2v
3 + 2i-2(1 + i) = 3 + 2i-2-2i
= 1 + 0i

b. u-2w
3 + 2i-2(-2-i) = 3 + 2i + 4 + 2i
= 7 + 4i

c. u + v + w
3 + 2i + 1 + i-2-i = 2 + 2i

d. u-v + w
3 + 2i-(1 + i)-2-i = 3 + 2i-1-i-2-i
= 0 + 0i

e. What is the length of the vector representing u?
The length of the vector representing u is $$\sqrt{3^{2} + 2^{2}}$$ = $$\sqrt{13}$$.

f. What is the length of the vector representing u-v + w?
The length of the vector representing u-v + w = $$\sqrt{0^{2} + 0^{2}}$$ = $$\sqrt{0}$$ = 0.

Question 3.
Find the sum of -2-4i and 5 + 3i geometrically.
3-i Question 4.
Show that (-5-6i)-(-8-4i) = 3-2i by representing the complex numbers as vectors. Question 5.
Let z1 = a1 + b1 i, z2 = a2 + b2 i, and z3 = a3 + b3 i. Prove the following using algebra or by showing with vectors.
z1 + z2 = z2 + z1
z1 + z2 = (a1 + b1 i) + (a2 + b2 i)
= (a2 + b2 i) + (a1 + b1 i)
= z2 + z1

b. z1 + (z2 + z3 ) = (z1 + z2 ) + z3
z1 + (z2 + z3 ) = (a1 + b1 i) + ((a2 + b2 i) + (a3 + b3 i))
= ((a1 + b1 i) + (a2 + b2 i)) + (a3 + b3 i)
= (z1 + z2 ) + z3

Question 6.
Let z = -3-4i and w = -3 + 4i.
a. Draw vectors representing z and w on the same set of axes. b. What are the lengths of the vectors representing z and w?
The length of the vector representing z is $$\sqrt{(-3)^{2} + (-4)^{2}}$$ = $$\sqrt{25}$$ = 5.
The length of the vector representing w is $$\sqrt{(-3)^{2} + 4^{2}}$$ = $$\sqrt{25}$$ = 5.

c. Find a new vector, uz, such that uz is equal to z divided by the length of the vector representing z.
uz = $$\frac{-3-4 i}{5}$$ = $$\frac{-3}{5}$$ – $$\frac{4}{5}$$i

d. Find uw, such that uw is equal to w divided by the length of the vector representing w.
uw = $$\frac{-3 + 4 i}{5}$$ = $$\frac{-3}{5}$$ + $$\frac{4}{5}$$i

e. Draw vectors representing uz and uw on the same set of axes as part (a). f. What are the lengths of the vectors representing uz and uw?
The length of the vector representing uz is $$\sqrt{\left(-\frac{3}{5}\right)^{2} + \left(-\frac{4}{5}\right)^{2}}$$ = $$\sqrt{\frac{9}{25} + \frac{16}{25}}$$ = $$\sqrt{\frac{25}{25}}$$ = $$\sqrt{1}$$ = 1.
The length of the vector representing uw is $$\sqrt{\left(-\frac{3}{5}\right)^{2} + \left(\frac{4}{5}\right)^{2}}$$ = $$\sqrt{\frac{9}{25} + \frac{16}{25}}$$ = $$\sqrt{\frac{25}{25}}$$ = $$\sqrt{1}$$ = 1.

g. Compare the vectors representing uz to z and uw to w. What do you notice?
The vectors representing uz and uw are in the same direction as z and w, respectively, but their lengths are only 1.

h. What is the value of uz times uw?
($$\frac{3}{5}$$–$$\frac{4}{5}$$i)($$\frac{3}{5}$$)2-($$\frac{4}{5}$$i)2 = ($$\frac{9}{25}$$) + ($$\frac{16}{25}$$) = 1

i. What does your answer to part (h) tell you about the relationship between uz and uw?
Since their product is 1, we know that uz and uw are reciprocals of each other.

Question 7.
Let z = a + bi.
a. Let uz be represented by the vector in the direction of z with length 1. How can you find uz? What is the value of uz?
Find the length of z, and then divide z by its length.
uz = $$\frac{a + b i}{\sqrt{a^{2} + b^{2}}}$$

b. Let uw be the complex number that when multiplied by uz, the product is 1. What is the value of uw?
From Problem 4, we expect uw = $$\frac{a-b i}{\sqrt{a^{2} + b^{2}}}$$. Multiplying, we get
$$\frac{a + b i}{\sqrt{a^{2} + b^{2}}}$$ ∙ $$\frac{a-b i}{\sqrt{a^{2} + b^{2}}}$$ = $$\frac{a^{2}-(b i)^{2}}{a^{2} + b^{2}}$$
= $$\frac{a^{2} + b^{2}}{a^{2} + b^{2}}$$
= 1

c. What number could we multiply z by to get a product of 1?
Since we know that uz is equal to z divided by the length of z and that uz ∙ uw = 1, we get
z ∙ $$\frac{1}{\sqrt{a^{2} + b^{2}}}$$∙$$\frac{a-b i}{\sqrt{a^{2} + b^{2}}}$$ = z∙$$\frac{a-b i}{a^{2} + b^{2}}$$ = 1
So, multiplying z by $$\frac{a-b i}{a^{2} + b^{2}}$$ will result in a product of 1.

Question 8.
Let z = -3 + 5i.
a. Draw a picture representing z + w = 8 + 2i. b. What is the value of w?
w = 11 – 3i

### Eureka Math Precalculus Module 1 Lesson 6 Exit Ticket Answer Key  Let z = -1 + 2i and w = 2 + i. Find the following, and verify each geometrically by graphing z, w, and each result.

a. z + w
1 + 3i

b. z-w