# Complex Numbers - Basic Definitions

by M. Bourne

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

### On this page:

Roots of cubic equations

Imaginary Numbers

Powers of j

Complex Numbers

Equivalent complex numbers

Forms of a complex number

**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.

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:

General form | Example | |
---|---|---|

Rectangular form: | `x + yj` | `5 + 6j` |

Polar form: | `r(cos\ θ + j\ sin\ θ)` | `8(cos\ 24^"o" + j\ sin\ 24^"o")` |

Exponential form: | `r\ e^(\ j\ θ)` | `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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