## Engage NY Eureka Math Precalculus Module 1 Lesson 14 Answer Key

### Eureka Math Precalculus Module 1 Lesson 14 Exercise Answer Key

Exercises

The vertices A(0, 0), B(1, 0), C(1, 1), and D(0, 1) of a unit square can be represented by the complex numbers A = 0, B = 1, C = 1 + i, and D = i.

Exercise 1.

Let L_{1}(z) = – z.

a. Calculate A’ = L_{1}(A), B’ = L_{1}(B), C’ = L_{1}(C), and D’ = L_{1}(D). Plot these four points on the axes.

b. Describe the geometric effect of the linear transformation L_{1}(z) = – z on the square ABCD.

Answer:

Transformation L_{1}rotates the square ABCD by 180Â° about the origin.

Exercise 2.

Let L_{2}(z) = 2z.

a. Calculate A’ = L_{2}(A), B’ = L_{2}(B), C’ = L_{2}(C), and D = L_{2}(D). Plot these four points on the axes.

b. Describe the geometric effect of the linear transformation L_{2}(z) = 2z on the square ABCD.

Answer:

Transformation L_{2}dilates the square ABCD by a factor of 2.

Exercise 3.

Let L_{3}(z) = iz.

a. Calculate A’ = L_{3}(A), B’ = L_{3}(B), C’ = L_{3}(C), and D’ = L_{3}(D). Plot these four points on the axes.

b. Describe the geometric effect of the linear transformation L_{3}(z) = iz on the square ABCD.

Answer:

Transformation L_{3}rotates the square ABCD by 90Â° counterclockwise about the origin.

Exercise 4.

Let L_{4}(z) = (2i)z.

a. Calculate A’ = L_{4}(A), B’ = L_{4}(B), C’ = L_{4}(C), and D’ = L_{4}(D). Plot these four points on the axes.

b. Describe the geometric effect of the linear transformation L_{4}(z) = (2i)z on the square ABCD.

Answer:

Transformation L_{4}rotates the square ABCD by 90Â° counterclockwise about the origin and dilates by a factor of 2.

Exercise 5.

Explain how transformations L_{2}, L_{3}, and L_{4}are related.

Answer:

Transformation L_{4}is the result of doing transformations L_{2}and L_{3}(in either order).

Discussion

â†’ What is the geometric effect of the transformation L(z) = az for a real number a>0?

â†’ The effect of L is dilation by the factor a.

â†’ What happens to a unit square in this case?

â†’ The orientation of the square does not change; it is not reflected or rotated, but the sides of the square are dilated by a.

â†’ What is the effect on the square if a>1?

â†’ The sides of the square will get larger.

â†’ What is the effect on the square if 0<a<1?

â†’ The sides of the square will get smaller.

â†’ What is the geometric effect of the transformation L(z) = az if a = 0?

â†’ If a = 0, then L(z) = 0 for every complex number z. This transformation essentially shrinks the square down to the point at the origin.

â†’ What is the geometric effect of the transformation L(z) = az for a real number a<0?

â†’ If a<0, then L(z) = az = – |a|z , so L is a dilation by |a| and a rotation by 180Â°. This transformation will dilate the original unit square and then rotate it about point A into the third quadrant.

â†’ What is the geometric effect of the transformation L(z) = (bi)z for a real number b>0?

â†’ The transformation L dilates by b and rotates by 90Â° counterclockwise.

â†’ What is the effect on the unit square if b>1?

â†’ The sides of the square will get larger.

â†’ What is the effect on the unit square if 0<b<1?

â†’ The sides of the square will get smaller.

â†’ What is the effect on the unit square if b<0?

â†’ If b<0, then L(z) = (bi)z = i(bz), so L is a dilation by |b| and a rotation by 180Â°, followed by a rotation by 90Â°. This transformation will rotate and dilate the original unit square and then rotate it about point A to the fourth quadrant.

Exercise 6.

We will continue to use the unit square ABCD with A = 0, B = 1, C = 1 + i, D = i for this exercise.

a. What is the geometric effect of the transformation L(z) = 5z on the unit square?

Answer:

By our work in the first five exercises and the previous discussion, we know that this transformation dilates the unit square by a factor of 5.

b. What is the geometric effect of the transformation L(z) = (5i)z on the unit square?

Answer:

By our work in the first five exercises, this transformation will dilate the unit square by a factor of 5 and rotate it 90Â° counterclockwise about the origin.

c. What is the geometric effect of the transformation L(z) = (5i^{2})z on the unit square?

Answer:

Since i^{2} = – 1, this transformation is L(z) = – 5z, which will dilate the unit square by 5 and rotate it 180Â° about the origin.

d. What is the geometric effect of the transformation L(z) = (5i^{3})z on the unit square?

Answer:

Since i^{3} = – i, this transformation is L(z) = ( – 5i)z, which will dilate the unit square by a factor of 5 and rotate it 270Â° counterclockwise about the origin.

e. What is the geometric effect of the transformation L(z) = (5i^{4})z on the unit square?

Answer:

Since i^{4} = (i^{2})^{2} = ( – 1)^{2} = 1, this transformation is L(z) = 5z, which is the same transformation as in part (a). Thus, this transformation dilates the unit square by a factor of 5.

f. What is the geometric effect of the transformation L(z) = (5i^{5})z on the unit square?

Answer:

Since i^{5} = i^{4} âˆ™ i = i, this is the same transformation as in part (b). This transformation will dilate the unit square by a factor of 5 and rotate it 90Â° counterclockwise about the origin.

g. What is the geometric effect of the transformation L(z) = (5i^{n})z on the unit square, for some integer

nâ‰¥0?

Answer:

If n is a multiple of 4, then L(z) = (5i^{n})z = 5z will dilate the unit square by a factor of 5.

If n is one more than a multiple of 4, then L(z) = (5i^{n})z = (5i)z will dilate the unit square by a factor of 5 and rotate it 90Â° counterclockwise about the origin.

If n is two more than a multiple of 4, then L(z) = (5i^{n})z = – 5z will dilate the unit square by 5 and rotate it 180Â° about the origin.

If n is three more than a multiple of 4, then L(z) = (5i^{n})z = ( – 5i)z will dilate the unit square by a factor of 5 and rotate it 270Â° counterclockwise about the origin.

### Eureka Math Precalculus Module 1 Lesson 14 Exploratory Challenge Answer Key

Exploratory Challenge

Your group has been assigned either to the 1 – team, 2 – team, 3 – team, or 4 – team. Each team will answer the questions below for the transformation that corresponds to their team number:

L_{1}(z) = (3 + 4i)z

L_{2}(z) = ( – 3 + 4i)z

L_{3}(z) = ( – 3 – 4i)z

L_{4}(z) = (3 – 4i)z.

The unit square ABCD with A = 0, B = 1, C = 1 + i, D = i is shown below. Apply your transformation to the vertices of the square ABCD, and plot the transformed points A’, B’, C’, and D’ on the same coordinate axes.

The solution shown below is for transformation L_{1}. The transformed square for L_{2}, L_{3}, and L_{4}will be rotated 90Â°, 180Â°, and 270Â° counterclockwise about the origin from the one shown, respectively.

For the 1 – team:

a. Why is B’ = 3 + 4i?

Answer:

Because B = 1, we have B’ = L_{1}(B) = (3 + 4i)(1) = 3 + 4i.

b. What is the argument of 3 + 4i?

Answer:

The argument of 3 + 4i is the amount of counterclockwise rotation between the positive x – axis and the ray connecting the origin and the point (3, 4).

c. What is the modulus of 3 + 4i?

Answer:

The modulus of 3 + 4i is |3 + 4i| = \(\sqrt{3^{2} + 4^{2}}\) = \(\sqrt{25}\) = 5.

For the 2 – team:

a. Why is B’ = – 3 + 4i?

Answer:

Because B = 1, we have B’ = L_{2}(B) = ( – 3 + 4i)(1) = – 3 + 4i.

b. What is the argument of – 3 + 4i?

Answer:

The argument of – 3 + 4i is the amount of counterclockwise rotation between the positive x – axis and the ray connecting the origin and the point ( – 3, 4).

c. What is the modulus of – 3 + 4i?

Answer:

The modulus of – 3 + 4i is | – 3 + 4i| = âˆš(( – 3)^{2} + 4^{2}) = \(\sqrt{2}\)5 = 5.

For the 3 – team:

a. Why is B’ = – 3 – 4i?

Answer:

Because B = 1, we have B’ = L_{3}(B) = ( – 3 – 4i)(1) = – 3 – 4i.

b. What is the argument of – 3 – 4i?

Answer:

The argument of – 3 – 4i is the amount of counterclockwise rotation between the positive x – axis and the ray connecting the origin and the point ( – 3, – 4).

c. What is the modulus of – 3 – 4i?

Answer:

The modulus of – 3 – 4i is | – 3 – 4i| = \(\sqrt{( – 3)^{2} + ( – 4)^{2}}\) = \(\sqrt{25}\) = 5.

For the 4 – team:

a. Why is B’ = 3 – 4i?

Answer:

Because B = 1, we have B’ = L_{4}(B) = (3 – 4i)(1) = 3 – 4i.

b. What is the argument of 3 – 4i?

Answer:

The argument of 3 – 4i is the amount of counterclockwise rotation between the positive x – axis and the ray connecting the origin and the point (3, – 4).

c. What is the modulus of 3 – 4i?

Answer:

The modulus of 3 – 4i is |3 – 4i| = \(\sqrt{3^{2} + ( – 4)^{2}}\) = \(\sqrt{25}\) = 5.

All groups should also answer the following:

a. Describe the amount the square has been rotated counterclockwise.

Answer:

The square has been rotated the amount of counterclockwise rotation between the positive x – axis and ray \(\overrightarrow{A B^{\prime}}\).

b. What is the dilation factor of the square? Explain how you know.

Answer:

First, we need to calculate the length of one side of the square. The length AB’ is given by

AB’ = \(\sqrt{(4 – 0)^{2} + (3 – 0)^{2} = 5}\). Then, the dilation factor of the square is 5 because the final square has sides that are five times longer than the sides of the original square.

c. What is the geometric effect of your transformation L_{1}, L_{2}, L_{3}, or L_{4}on the unit square ABCD?

Answer:

(Answered for transformation L_{1}.) The transformation rotates the square counterclockwise by the argument of (3 + 4i) and dilates it by a factor of the modulus of 3 + 4i.

d. Make a conjecture: What do you expect to be the geometric effect of the transformation L(z) = (2 + i)z on the unit square ABCD?

Answer:

This transformation should rotate the square counterclockwise by the argument of 2 + i and dilate it by a factor of |2 + i| = \(\sqrt{2^{2} + 1^{2}}\) = \(\sqrt{5}\).

e. Test your conjecture with the unit square on the axes below.

Answer:

### Eureka Math Precalculus Module 1 Lesson 14 Exit Ticket Answer Key

Question 1.

Identify the linear transformation L that takes square ABCD to square A’B’C’D’ as shown in the figure on the right.

Answer:

The transformation L takes the point B = 1 to the point B’ = – 3 – 2i, so this transformation is given by L(z) = ( – 3 – 2i)z.

Question 2.

Describe the geometric effect of the transformation L(z) = (1 – 3i)z on the unit square ABCD, where A = 0, B = 1, C = 1 + i, and D = i. Sketch the unit square transformed by L on the axes on the right.

Answer:

This transformation dilates by |1 – 3i| = \(\sqrt{1^{2} + 3^{2}}\) = \(\sqrt{10}\) and rotates counterclockwise by arg(1 – 3i).

### Eureka Math Precalculus Module 1 Lesson 14 Problem Set Answer Key

Question 3.

Find the modulus and argument for each of the following complex numbers.

a. z_{1} = \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\)i

Answer:

\(\left|\frac{\sqrt{3}}{2} + \frac{1}{2} i\right|\) = 1, z_{1} is in Quadrant I; thus, arg(z_{1}) = arctan(\(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\)) = 30Â° = \(\frac{\pi}{6}\) rad.

b. z_{2} = 2 + 2\(\sqrt{3}\)i

Answer:

|2 + 2\(\sqrt{3}\)i| = 4, z_{2} is in Quadrant I; thus, arg(z_{2}) = arctan((2\(\sqrt{3}\))/( 2 )) = 60Â° = \(\frac{Ï€}{3}\)rad.

c. z_{3} = – 3 + 5i

Answer:

|3 + 5i| = \(\sqrt{34}\), z_{3}is in Quadrant II; thus, arg(z) = Ï€ – arctan(\(\frac{5}{3}\))â‰ˆÏ€ – 1.030â‰ˆ2.112 rad.

d. z_{4} = – 2 – 2i

Answer:

| – 2 – 2i| = 2\(\sqrt{2}\), z_{4}is in Quadrant III; thus, arg(z_{4}) = Ï€ + arctan(\(\frac{2}{2}\)) = Ï€ + \(\frac{Ï€}{4}\) = \(\frac{5Ï€}{4}\) rad.

e. z_{5} = 4 – 4i

Answer:

|4 + 4i| = 4\(\sqrt{2}\), z_{5} is in Quadrant IV; thus, arg(z_{5} ) = 2Ï€ – arctan(\(\frac{4}{4}\)) = 2Ï€ – ( \(\frac{Ï€}{4}\)) = ( 7\(\frac{Ï€}{4}\)) rad.

f. z_{6} = 3 – 6i

Answer:

|3 – 6i| = 3\(\sqrt{5}\), z_{6} is in Quadrant IV; thus, arg(z_{6} ) = 2Ï€ – arctan(\(\frac{6}{3}\)) = 2Ï€ – 1.107 = 5.176 rad.

Question 4.

For parts (a)â€“(c), determine the geometric effect of the specified transformation.

a. L(z) = – 3z

Answer:

The transformation L dilates by 3 and rotates by 180Â° about the origin.

b. L(z) = – 100z

Answer:

The transformation L dilates by 100 and rotates by 180Â° about the origin.

c. L(z) = – \(\frac{1}{3}\)z

Answer:

The transformation L dilates by \(\frac{1}{3}\)and rotates by 180Â° about the origin.

d. Describe the geometric effect of the transformation L(z) = az for any negative real number a.

Answer:

The transformation L dilates by |a| and rotates by 180Â° about the origin.

Question 5.

For parts (a)â€“(c), determine the geometric effect of the specified transformation.

a. L(z) = ( – 3i)z

Answer:

The transformation L dilates by 3 and rotates counterclockwise by 270Â° about the origin.

b. L(z) = ( – 100i)z

Answer:

The transformation L dilates by 100 and rotates by 270Â° about the origin.

c. L(z) = ( – \(\frac{1}{3}\)i)z

Answer:

The transformation L dilates by \(\frac{1}{3}\)and rotates counterclockwise by 270Â° about the origin.

d. Describe the geometric effect of the transformation L(z) = (bi)z for any negative real number b.

Answer:

The transformation L dilates by |b| and rotates by 270Â° counterclockwise about the origin.

Question 6.

Suppose that we have two linear transformations, L_{1}(z) = 3z and L_{2}(z) = (5i)z.

a. What is the geometric effect of first performing transformation L_{1}and then performing transformation L_{2}?

Answer:

The transformation L_{1}dilates by 3, dilates by 5, and rotates by 90Â° counterclockwise about the origin.

b. What is the geometric effect of first performing transformation L_{2}and then performing transformation L_{1}?

Answer:

The transformation L_{1}dilates by 5, rotates by 90Â° counterclockwise about the origin, and then dilates by 3.

c. Are your answers to parts (a) and (b) the same or different? Explain how you know.

Answer:

The answers are the same.

L_{2}(L_{1}(z)) = (5i) L_{1}(z) = (5i)(3z) = (15i)z For example, let z = 2 – 3i.

L_{1} = 3(2 – 3i) = 6 – 9i

L_{2} = (5i)(2 – 3i) = 15 + 10i

L_{2}(L_{1}) = (5i)(6 – 9i) = 45 + 30i

L_{1}(L_{2}) = 3(15 + 10i) = 45 + 30i

L_{1}(L_{2}(z)) = 3L_{2}(z) = 3((5i)z) = (15i)z

Question 7.

Suppose that we have two linear transformations, L_{1}(z) = (4 + 3i)z and L_{2}(z) = – z. What is the geometric effect of first performing transformation L_{1}and then performing transformation L_{2}?

Answer:

We have |4 + 3i| = 5, and the argument of 4 + 3i is arctan(\(\frac{3}{4}\))â‰ˆ0.644 radians, which is about 36.87Â°. Therefore, the transformation L_{1} followed by L_{2}dilates with scale factor 5, rotates by approximately 36.87Â° counter clockwise, and then rotates by 180Â°.

Question 8.

Suppose that we have two linear transformations, L_{1}(z) = (3 – 4i)z and L_{2}(z) = – z. What is the geometric effect of first performing transformation L_{1}and then performing transformation L_{2}?

Answer:

We see that |3 – 4i| = 5, and the argument of 3 – 4i is arctan(\(\frac{4}{3}\))â‰ˆ2Ï€ – 5.356 radians, which is about 306.87Â°. Therefore, the transformation L_{1}followed by L_{2}dilates with scale factor 5, rotates by approximately 306.87Â° counter clockwise, and then rotates by 180Â°.

Question 9.

Explain the geometric effect of the linear transformation L(z) = (a – bi)z, where a and b are positive real numbers.

Answer:

Note that the complex number a – bi is represented by a point in the fourth quadrant. The transformation L dilates with scale factor |a – bi| and rotates counterclockwise by 2Ï€ – arctan(\(\frac{b}{a}\)).

Question 10.

In Geometry, we learned the special angles of a right triangle whose hypotenuse is 1 unit. The figures are shown above. Describe the geometric effect of the following transformations.

a. L_{1}(z) = (\(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\)i)z

|\(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\)i| = 1″, ” arg(z) = 30Â° = Ï€/6 rad

The transformation L_{1} rotates counterclockwise by 30Â°.

b. L_{2}(z) = (2 + 2\(\sqrt{3}\)i)z

Answer:

|2 + 2\(\sqrt{3}\)i| = 4, arg(z) = 60Â° = \(\frac{Ï€}{3}\)rad

The transformation L_{2} dilates with scale factor 4 and rotates counterclockwise by 60Â°.

c. L_{3}(z) = (\(\frac{\sqrt{2}}{2}\) + \(\frac{\sqrt{2}}{2}\)i)z

|\(\frac{\sqrt{2}}{2}\) + \(\frac{\sqrt{2}}{2}\)i| = 1, arg(z) = 45Â° = \(\frac{Ï€}{4}\) rad

The transformation L_{3} dilates by 1 and rotates counterclockwise by 45Â°.

d. L_{4}(z) = (4 + 4i)z

Answer:

|4 + 4i| = 4\(\sqrt{2}\), arg(z) = 45Â° = \(\frac{Ï€}{4}\) rad

The transformation L_{4}dilates with scale factor 4\(\sqrt{2}\) and rotates counterclockwise by 45Â°.

Question 11.

Recall that a function L is a linear transformation if all z and w in the domain of L and all constants a meet the following two conditions:

i. L(z + w) = L(z) + L(w)

ii. L(az) = aL(z)

Show that the following functions meet the definition of a linear transformation.

a. L_{1}(z) = 4z

Answer:

L_{1}(z + w) = 4(z + w) = 4z + 4w = L_{1}(z) + L_{1}(w)

L_{1}(az) = 4(az) = 4az = a(4z) = aL_{1}(z)

b. L_{2}(z) = iz

Answer:

L_{2}(z + w) = i(z + w) = iz + iw = L_{2}(z) + L_{1}(w)

L_{2}(az) = i(az) = iaz = a(iz) = aL_{2}(z)

c. L_{3}(z) = (4 + i)z

Answer:

L_{3}(z + w) = (4 + i)(z + w) = (4 + i)z + (4 + i)w = L_{3}(z) + L_{3}(w)

L_{3}(az) = (4 + i)(az) = (4 + i)az = a((4 + i)z) = aL_{3}(z)

Question 12.

The vertices A(0, 0), B(1, 0), C(1, 1), D(0, 1) of a unit square can be represented by the complex numbers A = 0, B = 1, C = 1 + i, D = i. We learned that multiplication of those complex numbers by i rotates the unit square by 90Â° counterclockwise. What do you need to multiply by so that the unit square will be rotated by 90Â° clockwise?

Answer:

We need to multiply by i^{3} = – i.