# Complex Numbers - Basic Definitions

by M. Bourne

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Let's first consider what we learned before in Quadratic Equations and Equations of Higher Degree, so we can better understand where complex numbers are coming from.

## Quadratic Equations

Examples of quadratic equations:

• 2x^2 + 3x − 5 = 0
• x^2 − x − 6 = 0
• x^2 = 4

The roots of an equation are the x-values that make it "work" We can find the roots of a quadratic equation either by using the quadratic formula or by factoring.

We can have 3 situations when solving quadratic equations.

### Case 1: Two roots

Example: 2x^2 + 3x − 5 = 0

We proceed to solve this equation using the quadratic formula as we did earlier:

x=(-b+-sqrt(b^2-4ac))/(2a)

=(-3+-sqrt(9+40))/4

=(-3-sqrt(49))/4 or (-3+sqrt49)/4

=-2.5 or 1

We have found 2 roots.

The graph of the quadratic equation  y = 2x^2 + 3x − 5 cuts the x-axis at x = -2.5 and x = 1, as expected, showing our 2 roots:

The curve y = 2x2 + 3x − 5, showing x-intercepts at (−2.5, 0) and (1, 0).

More examples of quadratic equations with 2 roots:

x^2 = 4 has 2 solutions, x = -2 and x = 2.

x^2 − x − 6 = 0 has 2 solutions, x = -2 and x = 3.

2x^2 + 13x − 7 = 0 has 2 solutions, x = -7 and x = 1/2.

### Case 2: One Root

Example: 4x^2 − 12x + 9 = 0

Notice what happens when we use the quadratic formula this time. Under the square root we get 144 − 144 = 0.

x=(-b+-sqrt(b^2-4ac))/(2a)

 =(12+-sqrt(144-144))/8

=12/8

=1.5

So it means we only have one root. We can also say that this is a repeated root, since we are expecting 2 roots.

On the graph of y = 4x^2 −12x + 9, we can see that the graph cuts the x-axis in one place only, at x = 1.5.

The curve y = 4x2 − 12x + 9, showing x-intercept at (1.5, 0).

### Case 3: No Real Roots

Example: x^2 −4x + 20 = 0

x =(-b+-sqrt(b^2-4ac))/(2a)

=(4+-sqrt(16-80))/2

=(4+-sqrt(-64))/2

This example gives us a problem. Under the square root, we get (-64), and we have been told repeatedly by our teachers that we cannot have the square root of a negative number. Can we find such a root?

The curve y = x2 − 4x + 20, which has no x-intercepts.

Continues below

## Summary

A quadratic equation has degree 2 (the highest power of x is 2) and we can have either 2 real roots, one real repeated root or something that involves the square root of a negative number.

## Cubic Equations

Cubic equations are polynomials which have degree 3 (this highest power of x is 3).

In the case of a cubic equation, we expect (up to) 3 real solutions:

Example 1: x^3 − 2x^2 − 5x + 6 = 0 has solutions x = -2, 1 and 3.

The curve y = x3 − 2x2 − 5x + 6, which has 3 x-intercepts.

Example 2: If x^3 = 8, we know the solution x = 2, but we expect 2 other solutions. What are they?

The curve y = x3 − 8, which only has one x-intercept.

## Imaginary Numbers

To allow for these "hidden roots", around the year 1800, the concept of

sqrt(-1)

was proposed and is now accepted as an extension of the real number system. The symbol used is

j = sqrt(-1)

and j is called an imaginary number.

### Why Not i for Imaginary Numbers?

Many textbooks use i as the symbol for imaginary numbers. We use j, because the main application of imaginary numbers is in electricity and electronics, so there is less confusion with i (which is used for current).

Your calculator or computer algebra system will probably use i.

## Powers of j

You may need to look at this reminder example about multiplying square roots before you go any further.

Reminder example

### Reminder Example about Repeated Multiplying of Square Roots

If x = sqrt10, then

x^2 = (sqrt10)^2 = 10

[This because square and square root are inverse processes.]

Let's multiply our previous answer by sqrt10.

x^3 = (sqrt10)^3 = (sqrt10)^2sqrt10  = 10sqrt10

The next step in the pattern is:

x^4 = (sqrt10)^4 = 10 *(sqrt10)^2  = 10 × 10 = 100

[Once again, we are multiplying our previous answer by sqrt10.]

The next step in the pattern is:

x^5 = (sqrt10)^5 = 100sqrt10

The next step in the pattern is:

x^6 = (sqrt10)^6 = 1000

What we are doing next in Powers of j works in the same way.

Recall:

(sqrta)^2 = a, for any value of a.

and

j = sqrt(-1)

Using these, we can derive the following:

j^2 = (sqrt-1)^2 = -1

Multiplying by j again gives us:

j^3 = j^2(j) = -j

Continuing the process gives us:

j^4 = j^3(j) = -j(j) = -(-1) = 1

j^5 = j^4(j) = 1 × j = j

j^6 = j^5(j) = j × j = -1 etc

### Example 3: Using j

Express the following in terms of the imaginary number j:

a. sqrt(-16)

Answer

sqrt(-16)

sqrt(-1)sqrt(16)=jxx4=4j

b. sqrt(-100)

Answer

sqrt(-100)

sqrt(-1)sqrt(100)=jxx10=10j

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c. sqrt(-7)

Answer

sqrt(-7)

sqrt(-1)sqrt(7)=jxxsqrt(7)=jsqrt(7)

Care: We don't write this as sqrt(7)j, because it could be confused with sqrt((7j)

d. sqrt(-2)sqrt(-18)

Answer

sqrt(-2)sqrt(-18)

=jsqrt(2)xxjsqrt(18)

=j^2xxsqrt(2xx18)

=(-1)sqrt(36)

=-6

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e. sqrt(-2 × -18)

(NOT the same as Number 4! - Note the difference.)

Answer

sqrt((-2)(-18))

sqrt(36)=6

## Complex Numbers

Complex numbers have a real part and an imaginary part.

### Example 4: Complex numbers

a. 5 + 6j

Real part: 5, Imaginary part: 6j

b. −3 + 7j

Real part:  −3, Imaginary part: 7j

## Notation

We can write the complex number 2 + 5j as 2 + j5.

There is no difference in meaning.

## Solving Equations with Complex Numbers

We now return to our problem from above. We didn't know then what to do with sqrt(-64). Now we can write the solution using complex numbers, as follows:

x=(4+-sqrt(-64))/2

=(4+-jsqrt(64))/2

=(4+-8j)/2

=2-4j or 2+4j

## Equivalent Complex Numbers

Two complex numbers x + yj and a + bj are equivalent if:

The real parts are equal (x = a), and
The imaginary parts are equal (y = b).

### Example 5: Equivalent complex numbers

Given that 3 + 2j= a + bj, then

a = 3 and b = 2.

### Exercises

1. Express in terms of j:

-sqrt(-2/5)

Answer

-sqrt(-2/5)

= -sqrt((-1)(2/5))

 = -sqrt(-1)xxsqrt(2/5)

= -jsqrt(2/5)

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2. Simplify each of the following:

a. sqrt(-2)sqrt(-8)

Answer

sqrt(-2)sqrt(-8)

=(sqrt(-1)sqrt2)(sqrt(-1)sqrt8)

=(jsqrt2)(jsqrt8)

=j^2sqrt16

=(-1)(4)

=-4

Easy to understand math videos:
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b. sqrt((-2)(-8))

Answer

sqrt((-2)(-8))=sqrt16=4

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c. j^2 − j^6

Answer

j^2 = -1, and

j^6 = (j^2)^3 = (-1)^3 = -1

So

j^2 − j^6 =(-1) − (-1) = 0

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d. (sqrt(-2))^2+j^4

Answer

(sqrt(-2))^2 = -2, and

j^4 = (j^2)^2 = (-1)^2 = 1, so

(sqrt(-2))^2+j^4 =-2 + 1 = -1

## Forms of Complex Numbers

We can write complex numbers in 3 different ways:

We can write complex numbers in 3 different ways:

Rectangular form: x + yj

Example: 5 + 6j

Polar form: r(cos θ + j sin θ)

Example: 8(cos24° + j sin 24°)

Exponential form: re

Example: 6e2.5j

We will meet polar form and exponential form later in this chapter, but first, let's see how to perform basic operations with complex numbers.