## 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.

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

â†’ The point shifts to the left and down the same number of units.

â†’ A right triangle is formed. What are the values of the legs and the hypotenuse?

â†’ 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.

â†’ x^{2}+x^{2}=(\(\sqrt{2}\))^{2}

â†’ 2x^{2}=2, so x=1.

â†’ 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.

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

â†’ Begin by factoring out -1: -1(a-bi).

â†’ 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.

â†’ Did your conjecture match the answer to Exercise 3?

â†’ 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.

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

â†’ The graph of the parabola would shift upward 2 units.

â†’ How does this differ from adding the real number 2 to a complex number?

â†’ 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 90Â° counterclockwise rotation about the origin of w.

Answer:

wâˆ™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=Â±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=Â±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.