# Intro to the imaginary numbers

CCSS Math: HSN.CN.A.1
Learn about the imaginary unit i, about the imaginary numbers, and about square roots of negative numbers.
In your study of mathematics, you may have noticed that some quadratic equations do not have any real number solutions.
For example, try as you may, you will never be able to find a real number solution to the equation $x^2=-1$. This is because it is impossible to square a real number and get a negative number!
However, a solution to the equation $x^2=-1$ does exist in a new number system called the complex number system.

## The imaginary unit

The backbone of this new number system is the imaginary unit, or the number $i$.
The following is true of the number $i$:
• $i=\sqrt{-1}$
• $i^2=-1$
The second property shows us that the number $i$ is indeed a solution to the equation $x^2=-1$. The previously unsolvable equation is now solvable with the addition of the imaginary unit!

## Pure imaginary numbers

The number $i$ is by no means alone! By taking multiples of this imaginary unit, we can create infinitely many more pure imaginary numbers.
For example, $3i$, $i\sqrt{5}$, and $-12i$ are all examples of pure imaginary numbers, or numbers of the form $bi$, where $b$ is a nonzero real number.
Taking the squares of these numbers sheds some light on how they relate to the real numbers. Let's investigate this by squaring the number $3i$. The properties of integer exponents remain the same, so we can square $3i$ just as we'd imagine.
\begin{aligned}(3i)^2&=3^2i^2\\ \\ &=9{i^2}\\\\ \end{aligned}
Using the fact that $i^2=-1$, we can simplify this further as shown.
\begin{aligned}\phantom{(3i)^2} &=9\goldD{i^2}\\\\ &=9(\goldD{-1})\\\\ &=-9 \end{aligned}
The fact that $(3i)^2=-9$ means that $3i$ is a square root of $-9$.

What is $(4i)^2$?

\begin{aligned}(4i)^2&=4^2i^2\\ \\ &=16\goldD{i^2}\\\\ &=16(\goldD{-1})&\small{\gray{\text{(Because {i^2=-1})}}}\\\\ &=-16 \end{aligned}
Which of the following is a square root of $-16$?

The fact that $(4i)^2=-16$ means that $4i$ is a square root of $-16$.
The number $-4$ is a square root of positive $16$ since $(-4)^2=16$.
In this way, we can see that pure imaginary numbers are the square roots of negative numbers!

## Simplifying pure imaginary numbers

The table below shows examples of pure imaginary numbers in both unsimplified and in simplified form.
Unsimplified formSimplified form
$\sqrt{-9}$$3i$
$\sqrt{-5}$$i\sqrt{5}$
$-\sqrt{-144}$$-12i$
But just how do we simplify these pure imaginary numbers?
Let's take a closer look at the first example and see if we can think through the simplification.
Original equivalenceThought process
\begin{aligned}\sqrt{-9} = 3i \end{aligned}The square root of $-9$ is an imaginary number. The square root of $9$ is $3$, so the square root of negative $9$ is $\textit 3$ imaginary units, or $3i$.
The following property explains the above "thought process" in mathematical terms.
For $a>0$, $\Large\sqrt{-a}=i\sqrt{a}$
If we put this together with what we already know about simplifying radicals, we can simplify all pure imaginary numbers. Let's look at an example.

### Example

Simplify $\sqrt{-18}$.

### Solution

First, let's notice that $\sqrt{-18}$ is an imaginary number, since it is the square root of a negative number. So, we can start by rewriting $\sqrt{-18}$ as $i\sqrt{18}$.
Next we can simplify $\sqrt{18}$ using what we already know about simplifying radicals.
The work is shown below.
\begin{aligned}\sqrt{-18}&=i\sqrt{18}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\cdot\sqrt{9\cdot 2}&&\small{\gray{\text{9 is a perfect square factor of 18}}}\\\\ &=i\sqrt{9}\cdot\sqrt{2}&&\small{\gray{\sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \text{ when } a, b\geq0}} \\\\ &=i\cdot 3\cdot \sqrt2&&\small{\gray{\sqrt{9}=3}}\\\\ &=3i\sqrt{2}&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}
So it follows that $\sqrt{-18}=3i\sqrt{2}$.
Sure. Watch this video to see Sal simplify $\sqrt{-52}$.

## Let's practice some problems

### Problem 1

Simplify $\sqrt{-25}$.

\begin{aligned}\sqrt{-25}&=i\sqrt{25}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\cdot 5&&\small{\gray{\sqrt{25}=5}}\\\\ &=5i&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}
So it follows that $\sqrt{-25}=5i$.

### Problem 2

Simplify $\sqrt{-10}$.

\begin{aligned}\sqrt{-10}&=i\sqrt{10}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ \end{aligned}
Since there are no perfect square factors of $10$, this is as far as the simplification goes.
So it follows that $\sqrt{-10}=i\sqrt{10}$ or $\sqrt{10}i$.

### Problem 3

Simplify $\sqrt{-24}$.

\begin{aligned}\sqrt{-24}&=i\sqrt{24}&&\small{\gray{\text{For a>0, \sqrt{-a}=i\sqrt{a}}}}\\\\ &=i\sqrt{4\cdot 6}&&\small{\gray{\text{4 is a perfect square factor of 24}}}\\\\ &=i\sqrt{4}\cdot\sqrt{6}&&\small{\gray{\sqrt{ab}=\sqrt{a}\cdot\sqrt{b} \text{ when } a, b\geq0}} \\\\ &=i\cdot 2\cdot \sqrt6&&\small{\gray{\sqrt{4}=2}}\\\\ &=2i\sqrt{6}&&\small{\gray{\text{Multiplication is commutative}}} \end{aligned}
So it follows that $\sqrt{-24}=2i\sqrt{6}$ or $2\sqrt{6}i$.

## Why do we have imaginary numbers anyway?

The answer is simple. The imaginary unit $i$ allows us to find solutions to many equations that do not have real number solutions.
This may seem weird, but it is actually very common for equations to be unsolvable in one number system but solvable in another, more general number system.
Here are some examples with which you might be more familiar.
• With only the counting numbers, we can't solve $x+8=1$; we need the integers for this!
• With only the integers, we can't solve $3x-1=0$; we need the rational numbers for this!
• With only the rational numbers, we can't solve $x^2=2$. Enter the irrational numbers and the real number system!
And so, with only the real numbers, we can't solve $x^2=-1$. We need the imaginary numbers for this!
As you continue to study mathematics, you will begin to see the importance of these numbers.