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 x2=1x^2=-1. This is because it is impossible to square a real number and get a negative number!
However, a solution to the equation x2=1x^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 ii.
The following is true of the number ii:
  • i=1i=\sqrt{-1}
  • i2=1 i^2=-1
The second property shows us that the number ii is indeed a solution to the equation x2=1x^2=-1. The previously unsolvable equation is now solvable with the addition of the imaginary unit!

Pure imaginary numbers

The number ii is by no means alone! By taking multiples of this imaginary unit, we can create infinitely many more pure imaginary numbers.
For example, 3i3i, i5i\sqrt{5}, and 12i-12i are all examples of pure imaginary numbers, or numbers of the form bibi, where bb 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 3i3i. The properties of integer exponents remain the same, so we can square 3i3i just as we'd imagine.
(3i)2=32i2=9i2\begin{aligned}(3i)^2&=3^2i^2\\ \\ &=9{i^2}\\\\ \end{aligned}
Using the fact that i2=1i^2=-1, we can simplify this further as shown.
(3i)2=9i2=9(1)=9\begin{aligned}\phantom{(3i)^2} &=9\goldD{i^2}\\\\ &=9(\goldD{-1})\\\\ &=-9 \end{aligned}
The fact that (3i)2=9(3i)^2=-9 means that 3i3i is a square root of 9-9.

Check your understanding

What is (4i)2(4i)^2?
  • Your answer should be
  • an integer, like 66
  • a simplified proper fraction, like 3/53/5
  • a simplified improper fraction, like 7/47/4
  • a mixed number, like 1 3/41\ 3/4
  • an exact decimal, like 0.750.75
  • a multiple of pi, like 12 pi12\ \text{pi} or 2/3 pi2/3\ \text{pi}

(4i)2=42i2=16i2=16(1)(Because )i2=1=16\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 16-16?
Choose 1 answer:
Choose 1 answer:

The fact that (4i)2=16(4i)^2=-16 means that 4i4i is a square root of 16-16.
The number 4-4 is a square root of positive 1616 since (4)2=16(-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
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
9=3i\begin{aligned}\sqrt{-9} = 3i \end{aligned}The square root of 9-9 is an imaginary number. The square root of 99 is 33, so the square root of negative 99 is 3\textit 3 imaginary units, or 3i3i.
The following property explains the above "thought process" in mathematical terms.
For a>0a>0, a=ia\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.


Simplify 18\sqrt{-18}.


First, let's notice that 18\sqrt{-18} is an imaginary number, since it is the square root of a negative number. So, we can start by rewriting 18\sqrt{-18} as i18i\sqrt{18}.
Next we can simplify 18\sqrt{18} using what we already know about simplifying radicals.
The work is shown below.
18=i18For , a>0a=ia=i929 is a perfect square factor of 18=i92ab=ab when a,b0=i329=3=3i2Multiplication is commutative\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 18=3i2\sqrt{-18}=3i\sqrt{2}.
Sure. Watch this video to see Sal simplify 52\sqrt{-52}.

Let's practice some problems

Problem 1

Simplify 25\sqrt{-25}.

25=i25For , a>0a=ia=i525=5=5iMultiplication is commutative\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 25=5i\sqrt{-25}=5i.

Problem 2

Simplify 10\sqrt{-10}.

10=i10For , a>0a=ia\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 1010, this is as far as the simplification goes.
So it follows that 10=i10\sqrt{-10}=i\sqrt{10} or 10i\sqrt{10}i.

Problem 3

Simplify 24\sqrt{-24}.

24=i24For , a>0a=ia=i464 is a perfect square factor of 24=i46ab=ab when a,b0=i264=2=2i6Multiplication is commutative\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 24=2i6\sqrt{-24}=2i\sqrt{6} or 26i2\sqrt{6}i.

Why do we have imaginary numbers anyway?

The answer is simple. The imaginary unit ii 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=1x+8=1; we need the integers for this!
  • With only the integers, we can't solve 3x1=03x-1=0; we need the rational numbers for this!
  • With only the rational numbers, we can't solve x2=2x^2=2. Enter the irrational numbers and the real number system!
And so, with only the real numbers, we can't solve x2=1x^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.