# Complex Numbers - Basic Definitions

by M. Bourne

### On this page:

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* = 2*x*^{2} + 3*x* − 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* = 4*x*^{2} − 12*x* + 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* = *x*^{2} − 4*x* + 20, which has no *x*-intercepts.

## 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* = *x*^{3} − 2*x*^{2} − 5*x* + 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* = *x*^{3} − 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.

Please support IntMath!

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`

**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)`

Please support IntMath!

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

Answer

`sqrt(-2)sqrt(-18)`

`=jsqrt(2)xxjsqrt(18)`

`=j^2xxsqrt(2xx18)`

`=(-1)sqrt(36)`

`=-6`

Please support IntMath!

**e. ** `sqrt(-2 × -18)`

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

Answer

## 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)`

Get the Daily Math Tweet!

IntMath on Twitter

**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`

**b.** `sqrt((-2)(-8))`

Answer

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

**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`

**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`

Get the Daily Math Tweet!

IntMath on Twitter

## 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 + 6*j*

**Polar form: ***r*(cos θ + *j* sin θ)

Example: 8(cos24° + *j* sin 24°)

**Exponential form: ***r*e^{jθ}

Example: 6*e*^{2.5j}

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.

### Search IntMath, blog and Forum

### Online Algebra Solver

This algebra solver can solve a wide range of math problems.

Go to: Online algebra solver

### Algebra Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand algebra lessons on DVD. See samples before you commit.

More info: Algebra videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!