1. 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 •
= i •
= i •
= 3i 6

2. –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.

3. 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. | i|
| i| =
= = 26

4. 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.

5. 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.

6. (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.

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

8. 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
= ( i – 16) – 4i
= i
The first three output values are –4i, –16 – 4i, i.