1. Five-Minute Check (over Lesson 3–2) Mathematical Practices Then/Now
New Vocabulary
Example 1: Square Roots of Negative Numbers
Example 2: Products of Pure Imaginary Numbers
Example 3: Equation with Pure Imaginary Solutions
Key Concept: Complex Numbers
Example 4: Equate Complex Numbers
Example 5: Add and Subtract Complex Numbers
Example 6: Real-World Example: Multiply Complex Numbers
Example 7: Divide Complex Numbers
Lesson Menu

2. Solve x 2 – x = 2 by factoring.
A. x = 2 and x = –1
B. x = 2 and x = 1
C. x = 1
D. x = 1 and x = –1
5-Minute Check 1

3. Solve x 2 – x = 2 by factoring.
A. x = 2 and x = –1
B. x = 2 and x = 1
C. x = 1
D. x = 1 and x = –1
5-Minute Check 1

4. Solve c 2 – 16c + 64 = 0 by factoring.
B. c = 4
C. c = 8
D. c = 12
5-Minute Check 2

5. Solve c 2 – 16c + 64 = 0 by factoring.
B. c = 4
C. c = 8
D. c = 12
5-Minute Check 2

6. Solve z 2 = 16z by factoring. A. z = 1 and z = 4 B. z = 0 and z = 16
C. z =–1 and z = 4
D. z = –16
5-Minute Check 3

7. Solve z 2 = 16z by factoring. A. z = 1 and z = 4 B. z = 0 and z = 16
C. z =–1 and z = 4
D. z =–16
5-Minute Check 3

8. Solve 2x 2 + 5x + 3 = 0 by factoring.
A. x = and z = 2
B. x = 0
C. x = –1
D. x = and x = –1
5-Minute Check 4

9. Solve 2x 2 + 5x + 3 = 0 by factoring.
A. x = and z = 2
B. x = 0
C. x = –1
D. x = and x = –1
5-Minute Check 4

10. Write a quadratic equation with the roots –1 and 6 in the form ax 2 + bx + c, where a, b, and c are integers.
A. x2 – x + 6 = 0
B. x2 + x + 6 = 0
C. x2 – 5x – 6 = 0
D. x2 – 6x + 1 = 0
5-Minute Check 5

11. Write a quadratic equation with the roots –1 and 6 in the form ax 2 + bx + c, where a, b, and c are integers.
A. x2 – x + 6 = 0
B. x2 + x + 6 = 0
C. x2 – 5x – 6 = 0
D. x2 – 6x + 1 = 0
5-Minute Check 5

12. In a rectangle, the length is three inches greater than the width
In a rectangle, the length is three inches greater than the width. The area of the rectangle is 108 square inches. Find the width of the rectangle.
A. 6 in.
B. 8 in.
C. 9 in.
D. 12 in.
5-Minute Check 6

13. In a rectangle, the length is three inches greater than the width
In a rectangle, the length is three inches greater than the width. The area of the rectangle is 108 square inches. Find the width of the rectangle.
A. 6 in.
B. 8 in.
C. 9 in.
D. 12 in.
5-Minute Check 6

14. Mathematical Practices 6 Attend to precision. Content Standards
N.CN.1 Know there is a complex number i such that i 2 = –1, and every complex number has the form a + bi with a and b real.
N.CN.2 Use the relation i 2 = –1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. MP

15. You simplified square roots.
Perform operations with pure imaginary numbers.
Perform operations with complex numbers.
Then/Now

16. imaginary unit pure imaginary number complex number complex conjugates
Vocabulary

17. Square Roots of Negative Numbers
Answer:
Example 1

18. Square Roots of Negative Numbers
Answer:
Example 1

19. Square Roots of Negative Numbers
Answer:
Example 1

20. Square Roots of Negative Numbers
Answer:
Example 1

21. A. A. B. C. D.
Example 1

22. A. A. B. C. D.
Example 1

23. B. A. B. C. D.
Example 1

24. B. A. B. C. D.
Example 1

25. A. Simplify –3i ● 2i. –3i ● 2i = –6i 2 = –6(–1) i 2 = –1 = 6 Answer:
Products of Pure Imaginary Numbers
A. Simplify –3i ● 2i.
–3i ● 2i = –6i 2
= –6(–1) i 2 = –1
= 6
Answer:
Example 2

26. A. Simplify –3i ● 2i. –3i ● 2i = –6i 2 = –6(–1) i 2 = –1 = 6 Answer: 6
Products of Pure Imaginary Numbers
A. Simplify –3i ● 2i.
–3i ● 2i = –6i 2
= –6(–1) i 2 = –1
= 6
Answer: 6
Example 2

27. Products of Pure Imaginary Numbers
Answer:
Example 2

28. Products of Pure Imaginary Numbers
Answer:
Example 2

29. A. Simplify 3i ● 5i.
A. 15
B. –15
C. 15i
D. –8
Example 2

30. A. Simplify 3i ● 5i.
A. 15
B. –15
C. 15i
D. –8
Example 2

31. B. Simplify A. B. C. D.
Example 2

32. B. Simplify A. B. C. D.
Example 2

33. 5y 2 = –20 Subtract 20 from each side. y 2 = –4 Divide each side by 5.
Equation with Pure Imaginary Solutions
Solve 5y = 0.
5y = 0 Original equation
5y 2 = –20 Subtract 20 from each side.
y 2 = –4 Divide each side by 5.
Take the square root of each side.
Answer:
Example 3

34. 5y 2 = –20 Subtract 20 from each side. y 2 = –4 Divide each side by 5.
Equation with Pure Imaginary Solutions
Solve 5y = 0.
5y = 0 Original equation
5y 2 = –20 Subtract 20 from each side.
y 2 = –4 Divide each side by 5.
Take the square root of each side.
Answer: y = ±2i
Example 3

35. Solve 2x 2 + 50 = 0. A. x = ±5i B. x = ±25i C. x = ±5 D. x = ±25
Example 3

36. Solve 2x 2 + 50 = 0. A. x = ±5i B. x = ±25i C. x = ±5 D. x = ±25
Example 3

37. Concept

38. Equate Complex Numbers
Find the values of x and y that make the equation 2x + yi = –14 – 3i true.
Set the real parts equal to each other and the imaginary parts equal to each other.
2x = –14 Real parts
x = –7 Divide each side by 2.
y = –3 Imaginary parts
Answer:
Example 4

39. Equate Complex Numbers
Find the values of x and y that make the equation 2x + yi = –14 – 3i true.
Set the real parts equal to each other and the imaginary parts equal to each other.
2x = –14 Real parts
x = –7 Divide each side by 2.
y = –3 Imaginary parts
Answer: x = –7, y = –3
Example 4

40. Find the values of x and y that make the equation 3x – yi = 15 + 2i true.
A. x = 15 y = 2
B. x = 5 y = 2
C. x = 15 y = –2
D. x = 5 y = –2
Example 4

41. Find the values of x and y that make the equation 3x – yi = 15 + 2i true.
A. x = 15 y = 2
B. x = 5 y = 2
C. x = 15 y = –2
D. x = 5 y = –2
Example 4

42. Add and Subtract Complex Numbers
A. Simplify (3 + 5i) + (2 – 4i).
(3 + 5i) + (2 – 4i) = (3 + 2) + (5 – 4)i Commutative and Associative Properties
= 5 + i Simplify.
Answer:
Example 5

43. Add and Subtract Complex Numbers
A. Simplify (3 + 5i) + (2 – 4i).
(3 + 5i) + (2 – 4i) = (3 + 2) + (5 – 4)i Commutative and Associative Properties
= 5 + i Simplify.
Answer: 5 + i
Example 5

44. Add and Subtract Complex Numbers
B. Simplify (4 – 6i) – (3 – 7i).
(4 – 6i) – (3 – 7i) = (4 – 3) + (–6 + 7)i Commutative and Associative Properties
= 1 + i Simplify.
Answer:
Example 5

45. Add and Subtract Complex Numbers
B. Simplify (4 – 6i) – (3 – 7i).
(4 – 6i) – (3 – 7i) = (4 – 3) + (–6 + 7)i Commutative and Associative Properties
= 1 + i Simplify.
Answer: 1 + i
Example 5

46. A. Simplify (2 + 6i) + (3 + 4i). A. –1 + 2i B. 8 + 7i C. 6 + 12i
D i
Example 5

47. A. Simplify (2 + 6i) + (3 + 4i). A. –1 + 2i B. 8 + 7i C. 6 + 12i
D i
Example 5

48. B. Simplify (3 + 2i) – (–2 + 5i). A. 1 + 7i B. 5 – 3i C. 5 + 8i
D. 1 – 3i
Example 5

49. B. Simplify (3 + 2i) – (–2 + 5i). A. 1 + 7i B. 5 – 3i C. 5 + 8i
D. 1 – 3i
Example 5

50. E = I ● Z Electricity formula
Multiply Complex Numbers
ELECTRICITY In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I ● Z. Find the voltage in a circuit with current 1 + 4j amps and impedance 3 – 6j ohms.
E = I ● Z Electricity formula
= (1 + 4j)(3 – 6j) I = 1 + 4j, Z = 3 – 6j
= 1(3) + 1(–6j) + 4j(3) + 4j(–6j) FOIL
= 3 – 6j + 12j – 24j 2 Multiply.
= 3 + 6j – 24(–1) j 2 = –1
= j Add.
Answer:
Example 6

51. E = I ● Z Electricity formula
Multiply Complex Numbers
ELECTRICITY In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I ● Z. Find the voltage in a circuit with current 1 + 4j amps and impedance 3 – 6j ohms.
E = I ● Z Electricity formula
= (1 + 4j)(3 – 6j) I = 1 + 4j, Z = 3 – 6j
= 1(3) + 1(–6j) + 4j(3) + 4j(–6j) FOIL
= 3 – 6j + 12j – 24j 2 Multiply.
= 3 + 6j – 24(–1) j 2 = –1
= j Add.
Answer: The voltage is j volts.
Example 6

52. ELECTRICITY In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I ● Z. Find the voltage in a circuit with current 1 – 3j amps and impedance 3 + 2j ohms.
A. 4 – j
B. 9 – 7j
C. –2 – 5j
D. 9 – j
Example 6

53. ELECTRICITY In an AC circuit, the voltage E, current I, and impedance Z are related by the formula E = I ● Z. Find the voltage in a circuit with current 1 – 3j amps and impedance 3 + 2j ohms.
A. 4 – j
B. 9 – 7j
C. –2 – 5j
D. 9 – j
Example 6

54. 3 – 2i and 3 + 2i are conjugates.
Divide Complex Numbers A.
3 – 2i and 3 + 2i are conjugates.
Multiply.
i2 = –1
a + bi form
Answer:
Example 7

55. 3 – 2i and 3 + 2i are conjugates.
Divide Complex Numbers A.
3 – 2i and 3 + 2i are conjugates.
Multiply.
i2 = –1
a + bi form
Answer:
Example 7

56. B. Multiply by . Multiply. i2 = –1 a + bi form Answer:
Divide Complex Numbers B.
Multiply by .
Multiply.
i2 = –1
a + bi form
Answer:
Example 7

57. B. Multiply by . Multiply. i2 = –1 a + bi form Answer:
Divide Complex Numbers B.
Multiply by .
Multiply.
i2 = –1
a + bi form
Answer:
Example 7

58. A. A.
B i
C. 1 + i D.
Example 7

59. A. A.
B i
C. 1 + i D.
Example 7

60. B. A. B. C. D.
Example 7

61. B. A. B. C. D.
Example 7