## Engage NY Eureka Math Precalculus Module 3 Lesson 6 Answer Key

### Eureka Math Precalculus Module 3 Lesson 6 Example Answer Key

Example 1.

Consider again the set of complex numbers represented by z = 3(cos(θ) + i sin(θ) ) for 0° ≤ θ < 360°.

Answer:

a. Use an ordered pair to write a representation for the points defined by z as they would be represented in the coordinate plane.

Answer:

(3 cos(θ),3 sin(θ) )

b. Write an equation that is true for all the points represented by the ordered pair you wrote in part (a).

Answer:

Since x = 3 cos(θ) and y = 3 sin(θ):

x^{2} + y^{2} = (3 cos(θ) )^{2} + (3 sin(θ) )^{2}

x^{2} + y^{2} = 9(cos(θ) )^{2} + 9(sin(θ) )^{2}

x^{2} + y^{2} = 9((cos(θ) )^{2} + (sin(θ) )^{2} )

We know that (sin(θ) )^{2} + (cos(θ) )^{2} = 1, so x^{2} + y^{2} = 9.

c. What does the graph of this equation look like in the coordinate plane?

Answer:

The graph is a circle centered at the origin with radius 3 units.

Example 2.

The equation of an ellipse is given by \(\frac{x^{2}}{16} + \frac{y^{2}}{4}\) = 1.

a. Sketch the graph of the ellipse.

Answer:

b. Rewrite the equation in complex form.

Answer:

\(\frac{x^{2}}{16} + \frac{y^{2}}{4}\) = 1

Since \(\frac{x^{2}}{a^{2}}\) + \(\frac{y^{2}}{b^{2}}\) = 1, we have a = 4 and b = 2.

The complex form of the ellipse is z = a cos(θ) + bi sin(θ) = 4 cos(θ) + 2i sin(θ).

Example 3.

A set of points in the complex plane can be represented in the complex plane as z = 2 + i + 7 cos(θ) + i sin(θ) as θ varies.

a. Find an algebraic equation for the points described.

Answer:

z = 2 + i + 7 cos(θ) + i sin(θ) = (2 + 7 cos(θ) ) + i(1 + sin(θ) )

Since z = x + iy, then x = 2 + 7 cos(θ) and y = 1 + sin(θ).

So cos(θ) = \(\frac{x – 2}{7}\) and sin(θ) = (y – 1).

Since cos^{2}(θ) + sin^{2}(θ) = 1, we have (\(\frac{x – 2}{7}\))^{2} + (y – 1)^{2} = 1, which is equivalent to

\(\frac{(x – 2)^{2}}{49}\) + (y – 1)^{2} = 1.

b. Sketch the graph of the ellipse.

Answer:

### Eureka Math Precalculus Module 3 Lesson 6 Exercise Answer Key

Opening Exercise

a. Consider the complex number z = a + bi.

i. Write z in polar form. What do the variables represent?

Answer:

z = r(cos(θ) + i sin(θ) ), where r is the modulus of the complex number and θ is the argument.

ii. If r = 3 and θ = 90°, where would z be plotted in the complex plane?

Answer:

The point z is located 3 units above the origin on the imaginary axis.

iii. Use the conditions in part (ii) to write z in rectangular form. Explain how this representation corresponds to the location of z that you found in part (ii).

Answer:

z = a + bi, where a = r cos(θ) and b = r sin(θ)

a = 3 cos(90°) = 0; b = 3 sin(90°) = 3

Then z = 3i, which is located three units above the origin on the imaginary axis.

b. Recall the set of points defined by z = 3(cos(θ) + i sin(θ) ) for 0° ≤ θ < 360°, where θ is measured in degrees.

i. What does z represent graphically? Why?

Answer:

It is the set of points that are 3 units from the origin in the complex plane. This is because the modulus is 3, which indicates that for any given value of θ, z is located a distance of 3 units from the origin.

ii. What does z represent geometrically?

Answer:

A circle with radius 3 units centered at the origin

c. Consider the set of points defined by z = 5 cos(θ) + 3i sin(θ).

i. Plot z for θ = 0°, 90°, 180°, 270°, 360°. Based on your plot, form a conjecture about the graph of the set of complex numbers.

Answer:

For θ = 0°, z = 5 cos(0°) + 3i sin(0°) = 5 + 0i ↔ (5,0).

For θ = 90°, z = 5 cos(90°) + 3i sin(90°) = 3i ↔ (0,3i).

For θ = 180°, z = 5 cos(180°) + 3i sin(180°) = – 5 + 0i ↔ ( – 5,0).

For θ = 270°, z = 5 cos(270°) + 3i sin(270°) = – 3i ↔ (0, – 3i).

This set of points seems to form an oval shape centered at the origin.

ii. Compare this graph to the graph of z = 3(cos(θ) + i sin(θ) ). Form a conjecture about what accounts for the differences between the graphs.

Answer:

The coefficients of cos(θ) and i sin(θ) are equal for z = 3(cos(θ) + i sin(θ) ), which results in a circle, which has a constant radius, while the coefficients are different for z = 5 cos(θ) + 3i sin(θ), which seems to stretch the circle.

Exercises 1–2

Exercise 1.

Recall the set of points defined by z = 5 cos(θ) + 3i sin(θ).

a. Use an ordered pair to write a representation for the points defined by z as they would be represented in the coordinate plane.

Answer:

(5 cos(θ), 3 sin(θ) )

b. Write an equation in the coordinate plane that is true for all the points represented by the ordered pair you wrote in part (a).

Answer:

We have x = 5 cos(θ) and y = 3 sin(θ), so x^{2} = 25(cos^{2}(θ) ) and y^{2} = 9(sin^{2}(θ) ). We know

(cos^{2}(θ) + sin^{2}(θ) ) = 1.

Since x^{2} = 25(cos^{2}(θ) ), then cos^{2} (θ) = \(\frac{x^{2}}{25}\). Since y^{2} = 9(sin^{2}(θ) ), then sin^{2} (θ) = \(\frac{x^{2}}{9}\). By substitution, \(\frac{x^{2}}{25} + \frac{y^{2}}{9}\) = 1.

Exercise 2.

Find an algebraic equation for all the points in the coordinate plane traced by the complex numbers z = \(\sqrt{2}\) cos(θ) + i sin(θ).

Answer:

All the complex numbers represented by z can be written using the ordered pair (\(\sqrt{2}\) cos(θ),sin(θ) ) in the coordinate plane.

We have x = \(\sqrt{2}\) cos(θ) and y = sin(θ), so x^{2} = 2 cos^{2}(θ) and y^{2} = sin^{2} (θ). We know cos^{2} (θ) + sin^{2} (θ) = 1.

Since x^{2} = 2 cos^{2}(θ), then cos^{2} (θ) = \(\frac{x^{2}}{2}\). Then, by substitution, \(\frac{x^{2}}{2}\) + y^{2} = 1.

Exercise 3.

The equation of an ellipse is given by \(\frac{x^{2}}{9} + \frac{y^{2}}{26}\) = 1.

a. Sketch the graph of the ellipse.

Answer:

b. Rewrite the equation of the ellipse in complex form.

Answer:

\(\frac{x^{2}}{9} + \frac{y^{2}}{26}\) = 1

|a| = 3

|b| = \(\sqrt{26}\)

The complex form of the ellipse is z = a cos(θ) + bi sin(θ) = 3 cos(θ) + \(\sqrt{26}\)i sin(θ).

### Eureka Math Precalculus Module 3 Lesson 6 Problem Set Answer Key

Question 1.

Write the real form of each complex equation.

a. z = 4 cos(θ) + 9i sin(θ)

Answer:

\(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{81}\) = 1

b. z = 6 cos(θ) + i sin(θ)

Answer:

\(\frac{x^{2}}{36}\) + y^{2} = 1

c. z = \(\sqrt{5}\) cos(θ) + \(\sqrt{10}\)i sin(θ)

Answer:

\(\frac{x^{2}}{5}\) + \(\frac{y^{2}}{10}\) = 1

d. z = 5 – 2i + 4 cos(θ) + 7i sin(θ)

Answer:

\(\frac{(x – 5)^{2}}{16} + \frac{(y + 2)^{2}}{49}\) = 1

Question 2.

Sketch the graphs of each equation.

a. z = 3 cos(θ) + i sin(θ)

Answer:

b. z = – 2 + 3i + 4 cos(θ) + i sin(θ)

Answer:

c. \(\frac{(x – 1)^{2}}{9} + \frac{y^{2}}{25}\) = 1

Answer:

d. \(\frac{(x – 2)^{2}}{3} + \frac{y^{2}}{15}\) = 1

Answer:

Question 3.

Write the complex form of each equation.

a. \(\frac{x^{2}}{16}\) + \(\frac{y^{2}}{36}\) = 1

Answer:

z = 4 cos(θ) + 6i sin(θ)

b. \(\frac{x^{2}}{400}\) + \(\frac{y^{2}}{169}\) = 1

Answer:

z = 20 cos(θ) + 13i sin(θ)

c. \(\frac{x^{2}}{19}\) + \(\frac{y^{2}}{2}\) = 1

Answer:

z = \(\sqrt{19}\)cos(θ) + \(\sqrt{2}\)i sin(θ)

d. \(\frac{(x – 3)^{2}}{100}\) + \(\frac{(y + 5)^{2}}{16}\) = 1

Answer:

z = 3 – 5i + 10 cos(θ) + 4i sin(θ)

Question 4.

Carrie converted the equation z = 7 cos(θ) + 4i sin(θ) to the real form \(\frac{x^{2}}{7}\) + \(\frac{y^{2}}{4}\) = 1. Her partner Ginger said that the ellipse must pass through the point (7 cos(0),4 sin(0) ) = (7,0) and this point does not satisfy Carrie’s equation, so the equation must be wrong. Who made the mistake, and what was the error? Explain how you know.

Answer:

Ginger is correct. Carrie set a = 7 and b = 4, which is correct, but then she made an error in converting to the real form of the equation by dividing by a and b instead of a^{2} and b^{2}.

Question 5.

Cody says that the center of the ellipse with complex equation z = 4 – 5i + 2 cos(θ) + 3i sin(θ) is (4, – 5), while his partner, Jarrett, says that the center of this ellipse is ( – 4,5). Which student is correct? Explain how you know.

Answer:

Cody is correct. This ellipse is the translation of the ellipse with equation z = 2 cos(θ) + 3i sin(θ) by the vector 〈4, – 5〉, which moves the center of the ellipse from the origin to the point (4, – 5).

Extension:

Question 6.

Any equation of the form ax^{2} + bx + cy^{2} + dy + e = 0 with a>0 and c>0 might represent an ellipse. The equation 4x^{2} + 8x + 3y^{2} + 12y + 4 = 0 is such an equation of an ellipse.

a. Rewrite the equation \(\frac{(x – h)^{2}}{a^{2}}\) + \(\frac{(y – k)^{2}}{b^{2}}\) = 1 in standard form to locate the center of the ellipse (h,k).

Answer:

4(x^{2} + 2x) + 3(y^{2} + 4y) + 4 = 0

4(x^{2} + 2x + 1) + 3(y^{2} + 4y + 4) = – 4 + 4(1) + 3(4)

4(x + 1)^{2} + 3(y + 2)^{2} = 12

\(\frac{4(x + 1)^{2}}{12}\) + \(\frac{3(y + 2)^{2}}{12}\) = 1

\(\frac{(x + 1)^{2}}{3}\) + \(\frac{(y + 2)^{2}}{4}\) = 1

The center of the ellipse is the point ( – 1, – 2).

b. Describe the graph of the ellipse, and then sketch the graph.

Answer:

The graph of the ellipse is centered at ( – 1, – 2). It is elongated vertically with a semi – major axis of length 2 units and a semi – minor axis of length \(\sqrt{3}\) units.

c. Write the complex form of the equation for this ellipse.

Answer:

\(\frac{(x + 1)^{2}}{3}\) + \(\frac{(y + 2)^{2}}{4}\) = 1

cos^{2}(θ) + sin^{2}(θ) = 1, so cos^{2}(θ) = \(\frac{(x + 1)^{2}}{3}\) and sin^{2}(θ) = \(\frac{(y + 2)^{2}}{4}\)

3 cos^{2}(θ) = (x + 1)^{2}, so x = \(\sqrt{3}\) cos(θ) – 1

4 sin^{2}(θ) = (y + 2)^{2}, so y = 2 sin(θ) – 2

z = x + iy

= \(\sqrt{3}\) cos(θ) – 1 + i(2 sin(θ) – 2)

= – 1 – 2i + \(\sqrt{3}\) cos(θ) + 2i sin(θ)

### Eureka Math Precalculus Module 3 Lesson 6 Exit Ticket Answer Key

Question 1.

Write the real form of the complex equation z = cos(θ) + 3i sin(θ). Sketch the graph of the equation.

Answer:

x^{2} + \(\frac{y^{2}}{9}\) = 1

Question 2.

Write the complex form of the equation \(\frac{x^{2}}{25} + \frac{y^{2}}{4}\) = 1. Sketch the graph of the equation.

Answer:

z = 5 cos(θ) + 2i sin(θ)