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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

imaginary unit pure imaginary number complex number complex conjugates Vocabulary

Square Roots of Negative Numbers Answer: Example 1

Square Roots of Negative Numbers Answer: Example 1

Square Roots of Negative Numbers Answer: Example 1

Square Roots of Negative Numbers Answer: Example 1

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

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

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

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

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

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

Products of Pure Imaginary Numbers Answer: Example 2

Products of Pure Imaginary Numbers Answer: Example 2

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

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

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

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

5y 2 = –20 Subtract 20 from each side. y 2 = –4 Divide each side by 5. Equation with Pure Imaginary Solutions Solve 5y 2 + 20 = 0. 5y 2 + 20 = 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

5y 2 = –20 Subtract 20 from each side. y 2 = –4 Divide each side by 5. Equation with Pure Imaginary Solutions Solve 5y 2 + 20 = 0. 5y 2 + 20 = 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

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

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

Concept

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

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

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

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

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

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

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

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

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

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

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

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

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 = 27 + 6j Add. Answer: Example 6

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

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

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

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

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

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

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

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

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

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

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