The imaginary unit i

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.

$\begin{align}(3i)^2&=3^2i^2\
\
&=9{i^2}\\
\end{align}$

Using the fact that $i^2=-1$, we can simplify this further as shown.

$\begin{align}\phantom{(3i)^2}
&=9\goldD{i^2}\\
&=9(\goldD{-1})\\
&=-9
\end{align}$

The fact that $(3i)^2=-9$ means that $3i$ is a square root of $-9$.

Check your understanding

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[[☃ explanation 1]]

[[☃ graded-group 1]]

[[☃ explanation 1]]

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.

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 equivalence | Thought process

• | -
  $\begin{align}\sqrt{-9} = 3i \end{align}$| 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{align}\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{align}$

So it follows that $\sqrt{-18}=3i\sqrt{2}$.

[[☃ explanation 1]]

##Let's practice some problems

###Problem 1

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###Problem 2

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###Problem 3

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[[☃ explanation 1]]

##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.

We know that $i=\sqrt{-1}$ and that $i^2=-1$.

But what about $i^3$? $i^4$? Other integer powers of $i$? How can we evaluate these?

##Finding $i^3$ and $i^4$

The properties of exponents can help us here! In fact, when calculating powers of $i$, we can apply the properties of exponents that we know to be true in the real number system, so long as the exponents are integers.

With this in mind, let's find $i^3$ and $i^4$.

We know that $i^3=i^2\cdot i$. But since ${i^2=-1}$, we see that:

$\begin{align} i^3 &= {{i^2}}\cdot i\\ \\ &={ (-1)}\cdot i\\ \\ &= \purpleD{-i} \end{align}$

Similarly $i^4=i^2\cdot i^2$. Again, using the fact that ${i^2=-1}$, we have the following:

$\begin{align} i^4 &= {{i^2\cdot i^2}}\\ \\ &=({ -1})\cdot ({-1})\\ \\ &= \goldD{1} \end{align}$

##More powers of $i$

Let's keep this going! Let's find the next $4$ powers of $i$ using a similar method.

$\begin{align} i^5 &= {i^4\cdot i}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot i&&{\gray{\text{Since }i^4=1}} \\\\ &= \blueD i \\\\ \\\\ i^6 &= {i^4\cdot i^2}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot (-1)&&{\gray{\text{Since }i^4=1\text{ and }i^2=-1}} \\\\ &=\greenD{-1} \\\\ \\\\ i^7 &= {i^4\cdot i^3}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot (-i)&&{\gray{\text{Since }i^4=1\text{ and }i^3=-i}} \\\\ &=\purpleD{-i} \\\\ \\\\ i^8 &= {i^4\cdot i^4}&&{\gray{\text{Properties of exponents}}} \\\\ &=1\cdot 1&&{\gray{\text{Since }i^4=1}} \\\\ &=\goldD 1 \end{align}$

The results are summarized in the table.

$i^1$|$i^2$|$i^3$|$i^4$|$i^5$|$i^6$|$i^7$|$i^8$ :-:|-:|-:|-:|-:|-:|-:|-: $\blueD i$|$\greenD{-1}$|$\purpleD{-i}$|$\goldD 1$|$\blueD i$|$\greenD{-1}$|$\purpleD{-i}$|$\goldD 1$

##An emerging pattern

From the table, it appears that the powers of $i$ cycle through the sequence of $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$ and $\goldD1$.

Using this pattern, can we find $i^{20}$? Let's try it!

The following list shows the first $20$ numbers in the repeating sequence.

$\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$, $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$

According to this logic, $i^{20}$ should be equal to $\goldD 1$. Let's see if we can support this by using exponents. Remember, we can use the properties of exponents here just like we do with real numbers!

$\begin{align} i^{20} &= (i^4)^5&&{\gray{\text{Properties of exponents}}} \\\\ &= (1)^5 &&{\gray{i^4=1}} \\\\ &= \goldD 1 &&{\gray{\text{Simplify}}} \end{align}$

Either way, we see that $i^{20}=1$.

##Larger powers of $i$

Suppose we now wanted to find $i^{138}$. We could list the sequence $\blueD i$, $\greenD{-1}$, $\purpleD{-i}$, $\goldD 1$,... out to the $138^\text{th}$ term, but this would take too much time!

Notice, however, that $i^4=1$, $i^8=1$, $i^{12}=1$, etc., or, in other words, that $i$ raised to a ***multiple of $4$*** is $1$.

We can use this fact along with the properties of exponents to help us simplify $i^{138}$.

###Example

Simplify $i^{138}$.

###Solution

While $138$ is not a multiple of $4$, the number $136$ is! Let's use this to help us simplify $i^{138}$.

$\begin{align} i^{138}&=i^{136}\cdot i^2&&{\gray{\text{Properties of exponents}}} \\\\ &=(i^{4\cdot 34})\cdot i^2&&{\gray{136=4\cdot 34}} \\\\ &=(i^{4})^{34}\cdot i^2&&{\gray{\text{Properties of exponents}}} \\\\ &=(1)^{34}\cdot i^2 &&{\gray{i^4=1}} \\\\ &=1\cdot -1&&{\gray{i^2=-1}} \\\\ &=-1 \end{align}$

So $i^{138}=-1$.

Now you might ask why we chose to write $i^{138}$ as $i^{136}\cdot i^2$.

Well, if the original exponent is not a multiple of $4$, then finding the closest multiple of $4$ less than it allows us to simplify the power down to $i$, $i^2$, or $i^3$ just by using the fact that $i^4=1$.

This number is easy to find if you divide the original exponent by $4$. It's just the quotient (without the remainder) times $4$. [[☃ explanation 1]]

[[☃ explanation 2]]

##Let's practice some problems

###Problem 1

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[[☃ explanation 1]]

###Problem 2

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###Problem 3

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##Challenge Problem

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