ATM - Complex Numbers Lesson

Complex Numbers

Another name for complex numbers is imaginary numbers. These numbers are of the complex form a + bi, where a and b are real numbers and i is the imaginary part.

Why do we need an imaginary number? An imaginary number helps us deal with a negative under a square root sign. For example, how do you solve x² = -5. Without the acceptance of an imaginary number, this is not possible.

We know that two negatives numbers multiply to a positive number, so how can we multiply two negatives numbers and get a negative? This is where the imaginary number comes in.

So what is the i, the imaginary part? The letter i stands for the square root of -1, . Defining i = allows us to calculate the square root of -1.

Example 1: Solve x² = -5

= Take the square root of both sides.

$LaTeX: x=\pm\sqrt{-1\cdot5}$ $x=\pm\sqrt{-1\cdot5}$ A negative sign means times negative 1.

$LaTeX: x=\pm i\sqrt{5}$ $x=\pm i\sqrt{5}$ For ease of reading when you have a square root, the i can be written in front of the root rather than behind.

*Remember anytime you take a square root in a problem, you must include the positive and negative solutions.

Now let's look at the basic rules that you should understand for the imaginary number i.

There are just 4 basic rules to remember with the imaginary number i.

i =

i² = i * i

= * = ^-1

i³ = i * i * i

= * *

= ^-1 * = ^-i

ⁱ⁴= i * i * i * i*

= * * *

= -1 * -1 = 1

When you use the i, it is never written as a higher exponent than i to the first power, it always must be reduced.

So what happens with i⁵ ?

ⁱ⁵= i * i * i * i* i

= * * * *

= -1 * -1 *

= 1 * = i

Yes, the pattern starts over.

To simplify I quickly always divide the exponent by 4 if the exponent is greater than 4 and simplify the remainder.

Example 2: Simplify

a. i⁵ 5/4 has a remainder of 1

i⁵ = i¹ = i Simplify by using the remainder for the exponent

b. i³³ 33/4 has a remainder of 1

i³³ = i¹ = i Simplify by using the remainder for the exponent

c. i¹⁰⁴ 104/4 has a remainder of 0

i¹⁰⁴ = i⁰ = 1 Simplify by using the remainder for the exponent. Any number to the zero power = 1.

d. i²² 22/4 has a remainder of 2

i²² = i² = -1 Simplify by using the remainder for the exponent.

^{As you notice, using the simple method of division of the exponent by 4 since the pattern of i repeats every set of 4 allows us to simplify this imaginary number quickly.}

Imaginary Addition and Subtraction

Adding and subtracting imaginary numbers works just like addition and subtraction of any other variable, you combine like terms.

Example 3:

a. Simplify (8 + 2i) + (3 - 6i)

(8 + 2i) + (3 - 6i) Written in complex number a + bi form.

8 + 3 + 2i - 6i Regroup to combine like parts

11 - 4i

b. Simplify (8 + 2i) - (3 - 6i) Combine like terms

(8 + 2i) - (3 - 6i) Written in complex number a + bi form.

8 - 3 + 2i - (-6i) Regroup to combine like parts. Watch for the negative signs. Use ( ).

5 + 8i Combine like terms

Imaginary Multiplication

Multiplication with the imaginary numbers also follows the rules of multiplication and distribution of factors with one addition. Every time the imaginary i has an exponent greater than 1, it must be simplified.

Example 4:

a. Simplify -3(4 - 7i) - (5 - i)

-3(4 - 7i) - (5 - i) Written in complex number a + bi form.

-12 + 21i - 5 + i Distribute the multipliers of -3 and -1

-17 + 22i Combine like terms and write the result in a + bi form

b. Simplify (8 + 2i)(3 - 6i)

(8 + 2i) (3 - 6i) Written in complex number a + bi form.

8(3 - 6i) + 2i(3 - 6i) Set up to distribute

24 - 48i + 6i - 12i² Distribute

24 - 42i - 12 (-1) Combine like terms and simplify i²

24 - 42i + 12 Simplify

36 - 42i Combine like terms and write the result in a + bi form.

c. Simplify (2 + 3i)(2 - 3i)

(2 + 3i)(2 - 3i) Written in complex number a + bi form.

2(2 - 3i) + 3i(2 - 3i) Regroup to combine like parts. Watch for the negative signs. Use ( ).

4 - 6i + 6i + 9i² Distribute

4 - 9(-1) Combine like terms, simplify i²

4 + 9 Simplify

13 Combine like terms. Note this was the difference of squares 2² - (3i)².

Imaginary Division

Now let's discuss the last section of the imaginary number properties, division. The general rule is that a square root, cube root, etc. are not left in the denominators after an expression is simplified. It must be removed from the denominator. To do this we use the property that we just saw above in Example 4c.

In Example 4c, the two factors could have quickly used the difference of squares technique which eliminates the middle term. Creating the difference of squares to eliminate the middle term in the denominator is called creating the conjugate.

So let's try this creation of a conjugate. If you are given 6 - 4i, what is the other factor that when multiplied with the first, will not create a middle term.

Solution: 6 + 4i

(6-4i)(6+4i) = 6²- (4i)²

= 36 + 16i²

= 36 + 16(-1)

= 36 - 16

= 20

Note that the multiplication of a complex number and its conjugate results in a number value. Watch the video to understand why conjugates are important.

^{Now let's try an example in complex numbers that will require the use of a conjugate.}

Example 5: Simplify

Identify the conjugate and create a multiplication in the form of 1.

Create a single fraction.

3(4+7i) - 5i(4+7i)/4^2 - (7i)^ Numerator set up to distribute, denominator use the difference of squares.

Distribute.

Combine like terms and simplify.

Simplify.

Answer in complex form.

The video that follows provides another view of the division with the complex numbers as well as practice with greatest common factor and fraction simplification.

After reviewing the examples above, try these problems to verify you understand the imaginary i and the complex number form. Remember to use the principal of division by 4 for the exponent of i when simplifying. For detailed solutions, download the Complex Numbers Solution handout Links to an external site..

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