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,

1

aia

Simplify –54 by using the imaginary number i.

= –1 • 54

= i • 54

= i • 3 6

–54 = –1 • 54

= 3i 6

Simplifying Numbers Using i

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

Write –121 – 7 in a + bi form.

–121 – 7 = 11i – 7 Simplify the radical expression.

= –7 + 11i Write in the form a + bi.

Simplifying Imaginary Numbers

Vocabulary

2i

4i

-2i

-4i

2 4-4 -2

The absolute value of a complex number is its distance from the origin on the complex number plane.

You can plot the points on the graph use the Pythagorean Theorem to find the distance.

Find each absolute value.

|10 + 24i| = 102 + 242

a. |–7i|

–7i is seven units from the origin on the imaginary axis.

So |–7i| = 7

b. |10 + 24i|

= 100 + 576 = 26

Finding Absolute Value

Find the additive inverse of –7 – 9i.

–7 – 9i

–(–7 – 9i) Find the opposite.

7 + 9i Simplify.

Additive Inverse of a Complex Number

Simplify the expression (3 + 6i) – (4 – 8i).

(3 + 6i) – (4 – 8i) = 3 + (–4) + 6i + 8i Use commutative and associative properties.

= –1 + 14iSimplify.

Adding 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.

Multiplying Complex Numbers

x = ±i 6 Find the square root of each side.

Solve 9x2 + 54 = 0.

9x2 + 54 = 0

9x2 = –54 Isolate x2.

x2 = –6

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

Finding Complex Solutions

Find the first three output values for f(z) = z2 – 4i. Use z = 0 as

the first input value.

f(–4i ) = (–4i )2 – 4i First

output becomes second input.

Evaluate for z = –4i.= –16 – 4if(–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.

f(0) = 02 – 4i= –4i

Use z = 0 as the first input value.

Real-World Connection

Homework

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