5-6 Complex Numbers.

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Presentation transcript:

5-6 Complex Numbers

Objectives Identifying Complex Numbers Operations with Complex Numbers

Vocabulary The imaginary number i is defined as a number whose square is -1. So i² = -1 and i = . Square Root of a Negative Real Number For any positive real number a,

Simplifying Numbers Using i Simplify –54 by using the imaginary number i. –54 = –1 • 54 = –1 • 54 = i • 54 = i • 3 6 = 3i 6

Vocabulary A complex number can be written in the form a + bi, where a and b are real numbers, including 0. a + bi Real Part Imaginary Part

Simplifying Imaginary Numbers Write –121 – 7 in a + bi form. –121 – 7 = 11i – 7 Simplify the radical expression. = –7 + 11i Write in the form a + bi.

Vocabulary The absolute value of a complex number is its distance from the origin on the complex number plane. 4i 2i You can plot the points on the graph use the Pythagorean Theorem to find the distance. -4 -2 2 4 -2i -4i

Finding Absolute Value 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

Additive Inverse of a Complex Number Find the additive inverse of –7 – 9i. –7 – 9i –(–7 – 9i) Find the opposite. 7 + 9i Simplify.

Adding Complex Numbers Simplify the expression (3 + 6i) – (4 – 8i). (3 + 6i) – (4 – 8i) = 3 + (–4) + 6i + 8i Use commutative and associative properties. = –1 + 14i Simplify.

Multiplying Complex Numbers 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.

Finding Complex Solutions 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

Real-World Connection 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.

Homework 5-6 Pg 278 #1, 2, 11, 12, 19, 20, 24, 25, 29, 30, 41, 42