## Engage NY Eureka Math Precalculus Module 1 Lesson 9 Answer Key

### Eureka Math Precalculus Module 1 Lesson 9 Example Answer Key

Example

Given the complex number z, find a complex number w such that z+w is shifted \(\sqrt{2}\) units in a southwest direction.

→ Begin by plotting the complex number. What does it mean for the point to be shifted in a southwest direction?

→ The point shifts to the left and down the same number of units.

→ A right triangle is formed. What are the values of the legs and the hypotenuse?

→ The legs are both x, and the hypotenuse is \(\sqrt{2}\).

Give students an opportunity to solve for x on their own and use the information to determine the complex number w.

→ x^{2}+x^{2}=(\(\sqrt{2}\))^{2}

→ 2x^{2}=2, so x=1.

→ Since the point was shifted 1 unit down and 1 unit to the left, the complex number must be -1-i.

### Eureka Math Precalculus Module 1 Lesson 9 Exercise Answer Key

Exercises

Exercise 1.

Taking the conjugate of a complex number corresponds to reflecting a complex number about the real axis. What operation on a complex number induces a reflection across the imaginary axis?

Answer:

For a complex number a+bi, the reflection across the imaginary axis is -a+bi. Alternatively, for a complex number z, the reflection across the imaginary axis is –\(\overline{\boldsymbol{z}}\).

Students may have answered that the reflection of a+bi across the imaginary axis is -a+bi. Discuss as a class how to write this in terms of the conjugate of the complex number.

→ Is it possible to write -a+bi another way? (Recall that the complex number z can be written as a+bi.)

→ Begin by factoring out -1: -1(a-bi).

→ Replace a-bi with \(\overline{\boldsymbol{z}}\): –\(\overline{\boldsymbol{z}}\).

Exercise 2.

Given the complex numbers w=-4+3i and z=2-5i, graph each of the following:

a. w

b. z

c. w+2

d. z+2

e. w-1

f. z-1

Answer:

Exercise 3.

Describe in your own words the geometric effect adding or subtracting a real number has on a complex number.

Answer:

Adding a real number to a complex number shifts the point to the right on the real (horizontal) axis, while subtracting a real number shifts the point to the left.

When students have finished the exercise, confirm as a class the answer to Exercise 3.

→ Did your conjecture match the answer to Exercise 3?

→ Answers will vary.

Some students may no doubt have guessed that adding a positive real value (i.e., w+2) to the complex number would shift the point vertically instead of horizontally. They may be confusing the translation of a function, such as f(x)=x^{2},

with that of a complex number. Make clear that even though comparisons are made between the complex and coordinate planes, the geometric effects are different. Use the following discussion points to clarify.

→ What is the effect of adding a constant to a function like f(x)=x^{2}? (For example, f(x)=x^{2}+2.)

→ The graph of the parabola would shift upward 2 units.

→ How does this differ from adding the real number 2 to a complex number?

→ The point representing the complex number would shift two units to the right, not vertically like the function.

Exercise 4.

Given the complex numbers w=-4+3i and z=2-5i, graph each of the following:

a. w

b. z

c. w+i

d. z+i

e. w-2i

f. z-2i

Answer:

Exercise 5.

Describe in your own words the geometric effect adding or subtracting an imaginary number has on a complex number.

Answer:

Adding an imaginary number to a complex number shifts the point up the imaginary (vertical) axis, while subtracting an imaginary number shifts the point down.

### Eureka Math Precalculus Module 1 Lesson 9 Problem Set Answer Key

Question 1.

Given the complex numbers w=2-3i and z=-3+2i, graph each of the following:

a. w-2

Answer:

w-2=2-3i-2=-3i

b. z+2

Answer:

z+2=-3+2i+2=-1+2i

c. w+2i

Answer:

w+2i=2-3i+2i=2-i

d. z-3i

Answer:

z-3i=-3+2i-3i=-3-i

e. w+z

Answer:

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

f. z-w

Answer:

z-w=-3+2i-(2-3i)=-5+5i

Question 2.

Let z=5-2i. Find w for each case.

a. z is a 90° counterclockwise rotation about the origin of w.

Answer:

w∙i=z; therefore, w=\(\frac{z}{i}\)=\(\frac{5-2 i}{i}\)=\(\frac{2+5 i}{-1}\)=-2-5i.

b. z is reflected about the imaginary axis from w.

Answer:

w=-\(\overline{\boldsymbol{z}}\); therefore, w=-(5+2i)=-5-2i.

c. z is reflected about the real axis from w.

Answer:

w=\(\overline{\boldsymbol{z}}\); therefore, w=5+2i.

Question 3.

Let z=-1+2i, w=4-i. Simplify the following expressions.

a. z+\(\overline{\boldsymbol{w}}\)

Answer:

z+\(\overline{\boldsymbol{w}}\)=-1+2i+4+i=3+3i

b. |w-\(\overline{\boldsymbol{z}}\) |

Answer:

|w-\(\overline{\boldsymbol{z}}\) |=|4-i-(-1-2i)|=|4-i+1+2i|=|5+i|=\(\sqrt{(5)^{2}+(1)^{2}}\)=\(\sqrt{26}\)

c. 2z-3w

Answer:

2z-3w=-2+4i-(12-3i)=-2+4i-12+3i=-14+7i

d. \(\frac{\boldsymbol{z}}{\boldsymbol{w}}\)

Answer:

\(\frac{\boldsymbol{z}}{\boldsymbol{w}}\)=\(\frac{-1+2 i}{4-i}\)=\(\frac{(-1+2 i)(4+i)}{(4-i)(4+i)}\)=\(\frac{-6+7 i}{16+1}\) = \(\frac{-6}{17}\) + \(\frac{7i}{17}\)

Question 4.

Given the complex number z, find a complex number w where z+w is shifted:

a. 2\(\sqrt{2}\) units in a northeast direction.

Answer:

x^{2}+x^{2}=(2\(\sqrt{2}\))^{2}, 2x^{2}=8, x=±2. Therefore, w=2+2i.

b. 5\(\sqrt{2}\) units in a southeast direction.

Answer:

x^{2}+x^{2}=(5\(\sqrt{2}\))^{2}, 2x^{2}=50, x=±5. Therefore, w=5-5i.

### Eureka Math Precalculus Module 1 Lesson 9 Exit Ticket Answer Key

Question 1.

Given z=3+2i and w=-2-i, plot the following in the complex plane:

a. z

b. w

c. z-2

d. w+3i

e. w+z

Answer:

Question 2.

Given z=a+bi, what complex number represents the reflection of z about the imaginary axis? Give one example to show why.

Answer:

–\(\overline{\boldsymbol{z}}\), the negative conjugate of z. For example, z=2+3i,

–\(\overline{\boldsymbol{z}}\)=-(2-3i)=-2+3i, which is reflected about the imaginary axis.

Question 3.

What is the geometric effect of T(z)=z+(4-2i)?

Answer:

T(z) shifts 4 units to the right and the 2 units downward.