Eureka Math Precalculus Module 3 Lesson 6 Answer Key

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°.   a. Use an ordered pair to write a representation for the points defined by z as they would be represented in the coordinate plane.
(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).
Since x = 3 cos⁡(θ) and y = 3 sin⁡(θ):
x2 + y2 = (3 cos⁡(θ) )2 + (3 sin⁡(θ) )2
x2 + y2 = 9(cos⁡(θ) )2 + 9(sin⁡(θ) )2
x2 + y2 = 9((cos⁡(θ) )2 + (sin⁡(θ) )2 )
We know that (sin⁡(θ) )2 + (cos⁡(θ) )2 = 1, so x2 + y2 = 9.

c. What does the graph of this equation look like in the coordinate plane?
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. b. Rewrite the equation in complex form.
$$\frac{x^{2}}{16} + \frac{y^{2}}{4}$$ = 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.
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 cos2⁡(θ) + sin2⁡(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. 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?
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?
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).
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?
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?
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. 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.
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.
(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).
We have x = 5 cos⁡(θ) and y = 3 sin⁡(θ), so x2 = 25(cos2⁡(θ) ) and y2 = 9(sin2⁡(θ) ). We know
(cos2⁡(θ) + sin2⁡(θ) ) = 1.
Since x2 = 25(cos2⁡(θ) ), then cos2 (θ) = $$\frac{x^{2}}{25}$$. Since y2 = 9(sin2⁡(θ) ), then sin2 (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).
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 x2 = 2 cos2⁡(θ) and y2 = sin2 (θ). We know cos2 (θ) + sin2 (θ) = 1.
Since x2 = 2 cos2⁡(θ), then cos2 (θ) = $$\frac{x^{2}}{2}$$. Then, by substitution, $$\frac{x^{2}}{2}$$ + y2 = 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. b. Rewrite the equation of the ellipse in complex form.
$$\frac{x^{2}}{9} + \frac{y^{2}}{26}$$ = 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)
$$\frac{x^{2}}{16}$$ + $$\frac{y^{2}}{81}$$ = 1

b. z = 6 cos⁡(θ) + i sin⁡(θ)
$$\frac{x^{2}}{36}$$ + y2 = 1

c. z = $$\sqrt{5}$$ cos⁡(θ) + $$\sqrt{10}$$i sin⁡(θ)
$$\frac{x^{2}}{5}$$ + $$\frac{y^{2}}{10}$$ = 1

d. z = 5 – 2i + 4 cos⁡(θ) + 7i sin⁡(θ)
$$\frac{(x – 5)^{2}}{16} + \frac{(y + 2)^{2}}{49}$$ = 1

Question 2.
Sketch the graphs of each equation.
a. z = 3 cos⁡(θ) + i sin⁡(θ) b. z = – 2 + 3i + 4 cos⁡(θ) + i sin⁡(θ) c. $$\frac{(x – 1)^{2}}{9} + \frac{y^{2}}{25}$$ = 1 d. $$\frac{(x – 2)^{2}}{3} + \frac{y^{2}}{15}$$ = 1 Question 3.
Write the complex form of each equation.
a. $$\frac{x^{2}}{16}$$ + $$\frac{y^{2}}{36}$$ = 1
z = 4 cos⁡(θ) + 6i sin⁡(θ)

b. $$\frac{x^{2}}{400}$$ + $$\frac{y^{2}}{169}$$ = 1
z = 20 cos⁡(θ) + 13i sin⁡(θ)

c. $$\frac{x^{2}}{19}$$ + $$\frac{y^{2}}{2}$$ = 1
z = $$\sqrt{19}$$cos⁡(θ) + $$\sqrt{2}$$i sin⁡(θ)

d. $$\frac{(x – 3)^{2}}{100}$$ + $$\frac{(y + 5)^{2}}{16}$$ = 1
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.
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 a2 and b2.

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.
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 ax2 + bx + cy2 + dy + e = 0 with a>0 and c>0 might represent an ellipse. The equation 4x2 + 8x + 3y2 + 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).
4(x2 + 2x) + 3(y2 + 4y) + 4 = 0
4(x2 + 2x + 1) + 3(y2 + 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.
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.
$$\frac{(x + 1)^{2}}{3}$$ + $$\frac{(y + 2)^{2}}{4}$$ = 1
cos2⁡(θ) + sin2⁡(θ) = 1, so cos2⁡(θ) = $$\frac{(x + 1)^{2}}{3}$$ and sin2⁡(θ) = $$\frac{(y + 2)^{2}}{4}$$
3 cos2⁡(θ) = (x + 1)2, so x = $$\sqrt{3}$$ cos⁡(θ) – 1
4 sin2⁡(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. x2 + $$\frac{y^{2}}{9}$$ = 1 Write the complex form of the equation $$\frac{x^{2}}{25} + \frac{y^{2}}{4}$$ = 1. Sketch the graph of the equation.  