# 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:

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`.

### 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?

## 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`.

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

## 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.

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)`

**b. ** `sqrt(-100)`

**c. ** `sqrt(-7)`

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

**e. ** `sqrt(-2 × -18)`

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

## 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)`

**2.** Simplify each of the following:

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

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

**c.** `j^2 −
j^6`

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

## 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.

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