Section 7.7

• Previously, when we encountered square roots of negative numbers in solving equations, we would say “no real solution” or “not a real number”.

• That’s still true.• However, we will now introduce a new set

of numbers.• Imaginary numbers which includes the

imaginary unit i.

Real numbers and imaginary numbers are both subsets of a new set of numbers.

Complex numbers

Every complex number can be written as a sum of a real number and an imaginary number.

The imaginary unit i is the number whose square is –1.

We can write the square root of a negative number in terms of i.

1i

12 i

Write the following with the i notation.

25 125

32 132 1216

121 1121

Example

5 i

11 i

24 i 24 i

Complex numbers can be written in the form a + bi (called standard form), with both a and b as real numbers.

a is a real number and bi would be an imaginary number.

If b = 0, a + bi is a real number.

If a = 0, a + bi is an imaginary number.

Example

Write each of the following in the form of a complex number in standard form a + bi.

6 = 6 + 0i

8i = 0 + 8i

24 164 62 i

256 1256 6 + 5i

620 i

Adding and subtracting complex numbers is accomplished by combining their corresponding components.

(a + bi) + (c + di) = (a + c) + (b + d)i

(a + bi) – (c + di) = (a – c) + (b – d)i

Add or subtract the following complex numbers. Write the answer in standard form a + bi.

Example

(4 + 3) + (6 – 2)i = 7 + 4i

(8 + 2i) – (4i) = (8 – 0) + (2 – 4)i = 8 – 2i

(4 + 6i) + (3 – 2i) =

The technique for multiplying complex numbers varies depending on whether the numbers are written as single term (either the real or imaginary component is missing) or two terms.

Note that the product rule for radicals does NOT apply for imaginary numbers.

2516 ii 54 220i )1(20 20

2516 2516 20400

Multiply the following complex numbers.

8i • 7i

56i2

56(-1)

-56

Example

Multiply the following complex numbers. Write the answer in standard form a + bi.

5i(4 – 7i)

20i – 35i2

20i – 35(-1)

20i + 35

35 + 20i

Example

Multiply the following complex numbers. Write the answer in standard form a + bi.

(6 – 3i)(7 + 4i)

42 + 24i – 21i – 12i2

42 + 3i – 12(-1)

42 + 3i + 12

54 + 3i

Example

In the previous chapter, when trying to rationalize the denominator of a rational expression containing radicals, we used the conjugate of the denominator.

Similarly, to divide complex numbers, we need to use the complex conjugate of the number we are dividing by.

The conjugate of a + bi is a – bi.

The conjugate of a – bi is a + bi.

The product of (a + bi) and (a – bi) is

(a + bi)(a – bi)

a2 – abi + abi – b2i2

a2 – b2(-1)

a2 + b2, which is a real number.

Divide the following complex numbers. Write the answer in standard form.

i

i

34

26

i

i

i

i

34

34

34

26

2

2

9121216

681824

iii

iii

)1(916

)1(62624 i

25

2618 ii

25

26

25

18

Example

Divide the following complex numbers.

i6

5

i

i

i 6

6

6

5

236

30

i

i

)1(36

30i

36

30ii

6

5

Example

• So far, we have looked at only two powers of i, i and i2

• There is an interesting pattern within the powers of i.

1i

12 i

iiiii )1(23

1)1)(1(224 iii

iiiii )1(45

1)1)(1(246 iii

iiiii ))(1(347

1)1)(1(448 iii

The powers recycle through each multiple of 4.

14 ki

Simplify each of the following powers.

Example

53i ii52 ii 134 i1 i

17i 17

1

i

ii16

1

i1

1

i

1

i

i

i

1

2i

i

)1(

i

1

ii