1. Splash Screen

2. Lesson Menu Five-Minute Check (over Lesson 4–3) CCSS 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

3. Over Lesson 4–3 5-Minute Check 1 A.2, –1 B.1, 2 C.1, 1 D.–1, 1 Solve x 2 – x = 2 by factoring.

4. Over Lesson 4–3 5-Minute Check 2 A.2 B.4 C.8 D.12 Solve c 2 – 16c + 64 = 0 by factoring.

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

6. Over Lesson 4–3 5-Minute Check 4 Solve 2x 2 + 5x + 3 = 0 by factoring. A. B.0 C.–1 D. –1

7. Over Lesson 4–3 5-Minute Check 5 A.x 2 – x + 6 = 0 B.x 2 + x + 6 = 0 C.x 2 – 5x – 6 = 0 D.x 2 – 6x + 1 = 0 Write a quadratic equation with the roots –1 and 6 in the form ax 2 + bx + c, where a, b, and c are integers.

8. Over Lesson 4–3 5-Minute Check 6 A.6 in. B.8 in. C.9 in. D.12 in. 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.

9. CCSS 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. Mathematical Practices 6 Attend to precision.

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

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

12. Example 1 Square Roots of Negative Numbers Answer: A.

13. Example 1 Square Roots of Negative Numbers B. Answer:

14. A. B. C. D. Example 1 A.

15. B. C. D. Example 1 B.

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

17. Example 2 Products of Pure Imaginary Numbers B. Answer:

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

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

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

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

22. Concept

23. Example 4 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. Answer: x = –7, y = –3 2x=–14Real parts x=–7Divide each side by 2. y=–3Imaginary parts

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

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

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

27. Example 5 A.–1 + 2i B.8 + 7i C.6 + 12i D.5 + 10i A. Simplify (2 + 6i) + (3 + 4i).

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

29. Example 6 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 =27 + 6jAdd. Answer: The voltage is 27 + 6j volts.

30. Example 6 A.4 – j B.9 – 7j C.–2 – 5j D.9 – j 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.

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

32. Example 7 Divide Complex Numbers B. Multiply. i 2 = –1 a + bi form Answer: Multiply by.

33. A. Example 7 A. B.3 + 3i C.1 + i D.

34. A. B. C. D. Example 7 B.

35. End of the Lesson