Simplify –54 by using the imaginary number i.

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Presentation transcript:

Simplify –54 by using the imaginary number i. Complex Numbers Lesson 5-6 Additional Examples Simplify –54 by using the imaginary number i. –54 = –1 • 54 = –1 • 54 = i • 54 = i • 3 6 = 3i 6

–121 – 7 = 11i – 7 Simplify the radical expression. Complex Numbers Lesson 5-6 Additional Examples Write –121 – 7 in a + bi form. –121 – 7 = 11i – 7 Simplify the radical expression. = –7 + 11i Write in the form a + bi.

Find each absolute value. Complex Numbers Lesson 5-6 Additional Examples Find each absolute value. a. |–7i| –7i is seven units from the origin on the imaginary axis. So |–7i| = 7 b. |10 + 24i| |10 + 24i| = 102 + 242 = 100 + 576 = 26

Find the additive inverse of –7 – 9i. Complex Numbers Lesson 5-6 Additional Examples Find the additive inverse of –7 – 9i. –7 – 9i –(–7 – 9i) Find the opposite. 7 + 9i Simplify.

Simplify the expression (3 + 6i) – (4 – 8i). Complex Numbers Lesson 5-6 Additional Examples Simplify the expression (3 + 6i) – (4 – 8i). (3 + 6i) – (4 – 8i) = 3 + (–4) + 6i + 8i Use commutative and associative properties. = –1 + 14i Simplify.

(3i)(8i) = 24i 2 Multiply the real numbers. Complex Numbers Lesson 5-6 Additional Examples Find each product. a. (3i)(8i) (3i)(8i) = 24i 2 Multiply the real numbers. = 24(–1) Substitute –1 for i 2. = –24 Multiply. b. (3 – 7i )(2 – 4i ) (3 – 7i )(2 – 4i ) = 6 – 14i – 12i + 28i 2 Multiply the binomials. = 6 – 26i + 28(–1) Substitute –1 for i 2. = –22 – 26i Simplify.

x = ±i 6 Find the square root of each side. Complex Numbers Lesson 5-6 Additional Examples Solve 9x2 + 54 = 0. 9x2 + 54 = 0 9x2 = –54 Isolate x2. x2 = –6 x = ±i 6 Find the square root of each side. Check: 9x2 + 54 = 0 9x2 + 54 = 0 9(i 6)2 + 54 0 9(i(– 6))2 + 54 0 9(6)i 2 + 54 0 9(6i 2) + 54 0 54(–1) –54 54(–1) –54 54 = 54 –54 = –54

Use z = 0 as the first input value. Complex Numbers Lesson 5-6 Additional Examples Find the first three output values for f(z) = z2 – 4i. Use z = 0 as the first input value. f(0) = 02 – 4i = –4i Use z = 0 as the first input value. f(–4i ) = (–4i )2 – 4i First output becomes second input. Evaluate for z = –4i. = –16 – 4i f(–16 – 4i ) = (–16 – 4i )2 – 4i Second output becomes third input. Evaluate for z = –16 – 4i. = [(–16)2 + (–16)(–4i ) + (–16)(–4i) + (–4i )2] – 4i = (256 + 128i – 16) – 4i = 240 + 124i The first three output values are –4i, –16 – 4i, 240 + 124i.