Complex numbers: Represent complex numbers in a form appropriate to the context

# Unit 1: Introduction to imaginary and complex numbers

Dylan Busa

### Unit 1 outcomes

By the end of this unit you will be able to:

• Define imaginary numbers.
• Perform basic operations on imaginary numbers.
• Define complex numbers.
• Represent complex numbers in standard rectangular coordinate form.

## What you should know

Before you start this unit, make sure you can:

## Introduction

The history of numbers is long and interesting. At several points in Mathematics history, we have encountered problems that forced us to discover or invent new types of types of numbers to fill in the gaps.

First there were the counting numbers such as $\scriptsize 1$, $\scriptsize 20$, $\scriptsize 1\ 140$. These are the numbers we first learn about as young children. We call these natural numbers ($\scriptsize \mathbb{N}$) because they exist in the natural world around us – three chairs, five cows, etc. If we add the number zero to this set, we get whole numbers ($\scriptsize {{\mathbb{N}}_{0}}$).

But there were some gaps. Some of these gaps became evident when people tried to work out what $\scriptsize 4-7$ was equalled. To solve this problem, negative natural numbers such as $\scriptsize -1$, $\scriptsize -12$ and $\scriptsize -2\ 321$ were added to the set. We called this new expanded set of numbers the integers ($\scriptsize \mathbb{Z}$).

But there were still gaps. For example, dividing $\scriptsize 6$ by $\scriptsize 3$ is not a problem. The answer is $\scriptsize 2$. Even $\scriptsize \displaystyle \frac{6}{{-3}}=-2$ is fine. But what about $\scriptsize \displaystyle \frac{3}{6}$? To solve these kinds of problems, we needed to add the fractions to the set. We called this new expanded set of numbers rational numbers ($\scriptsize \mathbb{Q}$). Every number that can be written as one integer divided by another integer is a rational number.

But there were still more gaps which we discovered when we tried to figure out the answer to $\scriptsize \sqrt{2}$. This cannot be written as one integer divided by another, so we needed to expand the set of numbers yet again to include non-rational numbers ($\scriptsize \mathbb{Q}'$).

All these numbers together were called real numbers ($\scriptsize \mathbb{R}$) (see Figure 1).

At this point we seemed to be done. Phew! But…

Then someone tried to figure out what the answer to $\scriptsize \sqrt{{-1}}$ is. At first, this seemed to be impossible. $\scriptsize 1\times 1=1$ and $\scriptsize -1\times -1=1$. There is no real number that when we multiply it by itself gives us a negative answer. What to do?

The solution turned out to be to invent or discover (depending on your view) a new kind of number. This number is called $\scriptsize i$, the imaginary number and is defined as $\scriptsize {{i}^{2}}=-1$.

The imaginary number:

$\scriptsize {{i}^{2}}=-1$ or $\scriptsize \sqrt{{-1}}=i$

But what is this imaginary number good for? Well, now we can solve equations like this:
\scriptsize \begin{align*}{{x}^{2}}+9 & =0\\\therefore {{x}^{2}} & =-9\\\therefore x & =\pm \sqrt{{-9}}\\\therefore x & =\pm \sqrt{{(9)(-1)}}\\\therefore x & =\pm \sqrt{9}\times \sqrt{{-1}}\\\therefore x & =\pm 3\times i\\\therefore x & =\pm 3i\end{align*}

The name chosen for this number is a bit unfortunate because it turns up in all sorts of very non-imaginary places like three-phase electricity, sound analysis, all sorts of engineering problems and quantum physics.

If you have an internet connection, you should watch both these excellent videos that help explain why the imaginary numbers are as ‘real’ as any of the real numbers.

## Imaginary numbers

We now know how to find the square root of negative numbers. We simply write the square root of the corresponding positive number as a multiple of $\scriptsize i$.

$\scriptsize \sqrt{{-49}}=7i$
$\scriptsize \sqrt{{-25}}=5i$
$\scriptsize \sqrt{{-2}}=\sqrt{2}i$

Multiplying the imaginary number $\scriptsize i$ by every possible real number would yield the set of all imaginary numbers and, in general, $\scriptsize i$ follows the rules of real number arithmetic.

Just like if we add (or subtract) two real numbers, the answer is always another real number, when we add (or subtract) two imaginary numbers, the answer is always an imaginary number. For example, $\scriptsize 3i+5i=8i$ and $\scriptsize \sqrt{2}i-3i=\left( {\sqrt{2}-3} \right)i$.

### Example 1.1

Simplify the following, expressing your answers as imaginary numbers (where necessary):

1. $\scriptsize \sqrt{{-16}}$
2. $\scriptsize \sqrt{{-24}}$
3. $\scriptsize \sqrt{{-32}}-\sqrt{{-2}}$
4. $\scriptsize \sqrt{{80}}$

Solutions

1. .
\scriptsize \begin{align*} \sqrt{{-16}}&=\sqrt{{(16)(-1)}}\\ & =4\sqrt{{-1}}\\ & =4i\end{align*}
2. .
\scriptsize \begin{align*} \sqrt{{-24}}& =\sqrt{{(24)(-1)}}\\ & =\sqrt{{(4)(6)(-1)}}\\ & =2\sqrt{6}i\end{align*}
3. .
\scriptsize \begin{align*} \sqrt{{-32}}-\sqrt{{-2}} & =\sqrt{{16(-2)}}-\sqrt{{-2}}\\ & =4\sqrt{{-2}}-\sqrt{{-2}}\\ & =\sqrt{{-2}}\left( {4-1} \right)\\ & =3\sqrt{{-2}}\\ & =3\sqrt{2}i\end{align*}
4. .
\scriptsize \begin{align*} \sqrt{{80}}& =\sqrt{{(16)(5)}}\\ & =4\sqrt{5}\end{align*}

### Exercise 1.1

Simplify the following, expressing your answers as imaginary numbers (where necessary):

1. $\scriptsize \sqrt{{-400}}$
2. $\scriptsize \sqrt{{-49}}-\sqrt{{-100}}$
3. $\scriptsize \sqrt{{-4}}+\sqrt{{-25}}-\sqrt{3}$
4. $\scriptsize 5i-\sqrt{{-25}}+2\sqrt{2}i$

The full solutions are at the end of the unit.

## Multiply and divide imaginary numbers

Just as imaginary numbers obey the normal (real number) rules of addition and subtraction, they also obey the normal (real number) rules of multiplication and division. However, we always need to remain aware of the fact that $\scriptsize {{i}^{2}}=-1$. Whenever we see $\scriptsize {{i}^{2}}$, we can replace this with $\scriptsize -1$.

### Example 1.2

Simplify the following:

1. $\scriptsize 3i\times 5i$
2. $\scriptsize (-6i)(2i)$
3. $\scriptsize 7{{i}^{3}}\times 4{{i}^{2}}$

Solutions

1. We need to multiply the numbers and the $\scriptsize i$’s.
\scriptsize \begin{align*}3i\times 5i& =15{{i}^{2}}&& \text{But }{{i}^{2}}=-1\\ & =15(-1)\\ & =-15\end{align*}
2. .
\scriptsize \begin{align*} &(-6i)(2i) &&\text{Multiply the numbers and the }i \text{'s}\\& =-12{{i}^{2}} &&\text{But }{{i}^{2}}=-1\\& =-12(-1)\\ & =12\end{align*}
3. .
\scriptsize \begin{align*} 7i^3\times 4i^2 &=28i^5&&\text{The normal exponent rules still apply}\\ &=28\times i^2\times i^2\times i&&\text{But }i^2=-1\\ &=28\times(-1)\times(-1)\times i\\ &=28i \end{align*}

### Example 1.3

Simplify the following:

1. $\scriptsize \displaystyle \frac{{6i}}{{2i}}$
2. $\scriptsize \displaystyle \frac{{36{{i}^{2}}}}{{6i}}$
3. $\scriptsize \displaystyle \frac{{20{{i}^{2}}}}{4}$
4. $\scriptsize \displaystyle \frac{{18{{i}^{6}}\times 2{{i}^{4}}}}{{6{{i}^{3}}}}$

Solutions

1. We need to divide the numbers and the $\scriptsize i$‘s. If $\scriptsize i=\sqrt{{-1}}$ then $\scriptsize \displaystyle \frac{i}{i}=\displaystyle \frac{{\sqrt{{-1}}}}{{\sqrt{{-1}}}}=1$.
$\scriptsize \displaystyle \frac{6i}{2i}=3$
2. .
$\scriptsize \displaystyle \frac{36i^2}{6i}=6i$
3. .
\scriptsize \begin{align*} \displaystyle \frac{{20{{i}^{2}}}}{4}& =5{{i}^{2}}&& \text{But }{{i}^{2}}\\&=-1\\ & =-5\end{align*}
4. .
\scriptsize \begin{align*} \displaystyle \frac{{18{{i}^{6}}\times 2{{i}^{4}}}}{{6{{i}^{3}}}} & =\displaystyle \frac{{36{{i}^{{10}}}}}{{6{{i}^{3}}}}\\ & =6{{i}^{7}}\\ & =6\times {{i}^{2}}\times {{i}^{2}}\times {{i}^{2}}\times i\\ & =6(-1)(-1)(-1)i\\ & =-6i\end{align*}

The powers of $\scriptsize i$ are cyclic. Look at what happens when we raise $\scriptsize i$ to increasing powers.

\scriptsize \displaystyle \begin{align*} & {{i}^{1}}=i\\ & {{i}^{2}}=-1\quad \text{By definition}\\ & {{i}^{3}}={{i}^{2}}.i=-1.i=-i\\ & {{i}^{4}}={{i}^{3}}.i=-i.i=-{{i}^{2}}=-(-1)=1\\ & {{i}^{5}}={{i}^{4}}.i=1.i=i\quad \text{This is the same as }{{i}^{1}}\\ & {{i}^{6}}={{i}^{5}}.i=i.i={{i}^{2}}=-1\quad \text{This is the same as }{{i}^{2}}\\ & {{i}^{7}}={{i}^{6}}.i=-1.i=-i\quad \text{This is the same as }{{i}^{3}}\\ & {{i}^{8}}={{i}^{7}}.i=-i.i=-{{i}^{2}}=-(-1)=1\quad \text{This is the same as }{{i}^{4}}\end{align*}

We can see that the cycle repeats itself every fourth power of $\scriptsize i$. This means that we can simplify large powers of $\scriptsize i$ by factoring out as many factors of $\scriptsize {{i}^{4}}$ as possible. For example:

\scriptsize \begin{align*} & {{i}^{{35}}}\\ & ={{i}^{{(4\times 8)+3}}}\\ & ={{\left( {{{i}^{4}}} \right)}^{8}}.{{i}^{3}}\\ & ={{1}^{8}}.{{i}^{3}}\quad ({{i}^{4}}=1)\\ & ={{i}^{3}}\\ & ={{i}^{2}}.i\\ & =-i\end{align*}

### Note

Watch the video called “Ex: Raising the imaginary unit i to powers” if you would like to see more examples.

### Exercise 1.2

Simplify the following:

1. $\scriptsize \displaystyle \frac{{7{{i}^{5}}}}{{49{{i}^{7}}}}$
2. $\scriptsize \displaystyle \frac{{\sqrt{6}i}}{{\sqrt{{-24}}}}$
3. $\scriptsize \displaystyle \frac{{7{{i}^{3}}.3{{i}^{8}}.2{{i}^{4}}}}{{21{{i}^{5}}.3{{i}^{5}}}}$
4. $\scriptsize \displaystyle \frac{{\sqrt{{12}}i.(-4i).3{{i}^{5}}}}{{\sqrt{{-24}}.6{{i}^{4}}}}$

The full solutions are at the end of the unit.

## Complex numbers

We have seen that when we add or subtract imaginary numbers, the answer is always imaginary and that when we multiply or divide imaginary numbers, the answer is either real or imaginary depending on the powers of $\scriptsize i$.

But what happens when we add a real number to an imaginary number? The answer is neither a pure real number nor a pure imaginary number. Instead, we get a complex number. A complex number is simply the sum of a real number and an imaginary number. A complex number is expressed in standard form when written as $\scriptsize a+bi$ where $\scriptsize a$ is the real part and $\scriptsize b$ is the imaginary part (see Figure 2).

For example, $\scriptsize 5+2i$ is a complex number, as is $\scriptsize 3+4\sqrt{3}i$. The entire number system is represented in Figure 3.

A complex number is a number of the form $\scriptsize a+bi$ where:

• $\scriptsize a$ is the real part
• $\scriptsize b$ is the imaginary part.

If $\scriptsize b=0$, then $\scriptsize a+bi$ is a real number. If $\scriptsize a=0$ and $\scriptsize b\ne 0$, then the complex number is called a pure imaginary number.

### Take note!

Standard form is sometimes also referred to as rectangular form.

### Example 1.4

Express the following numbers in standard form:

1. $\scriptsize \sqrt{{-36}}$
2. $\scriptsize \sqrt{{-42}}$
3. $\scriptsize \sqrt{{-32}}-\sqrt{7}$
4. $\scriptsize \sqrt{{80}}-\sqrt{{-4}}$

Solutions

1. .
\scriptsize \begin{align*} \sqrt{{-36}}& =\sqrt{{(36)(-1)}}\\ & =6\sqrt{{-1}}\\ & =6i\end{align*}
In standard form this is $\scriptsize 0+6i$. There is no real part, just an imaginary part.
2. .
\scriptsize \begin{align*} \sqrt{{-42}}& =\sqrt{{(42)(-1)}}\\ & =\sqrt{{42}}i\end{align*}
In standard form this is $\scriptsize 0+\sqrt{{42}}i$. There is no real part, just an imaginary part.
3. .
\scriptsize \begin{align*} \sqrt{-32}-\sqrt{7}&=\sqrt{(32)(-1)}-\sqrt{(7)}\\ &=\sqrt{(16)(2)(-1)}-\sqrt{7}\\ &=4\sqrt{2}i-\sqrt{7}\\ &=-\sqrt{7}+4\sqrt{2}i \end{align*}
4. .
\scriptsize \begin{align*} \sqrt{{80}}-\sqrt{{-4}}& =\sqrt{{(16)(5)}}-\sqrt{{(4)(-1)}}\\ & =4\sqrt{5}-2i\end{align*}

### Did you know?

The German mathematician Leopold Kronecker once said “God made the integers; all else is the work of man”. Complex numbers represent an incremental journey of investigation, struggle and invention over a thousand years by mathematicians such as Pythagoras, Descartes, De Moivre, Euler, Gauss, and many others. Complex numbers answer questions that for centuries had puzzled the greatest minds in science.

### Exercise 1.3

Express the following in standard form:

1. $\scriptsize \sqrt{{-144}}$
2. $\scriptsize \sqrt{{-600}}$
3. $\scriptsize \sqrt{{12}}$

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• $\scriptsize {{i}^{2}}=-1$ by definition, and therefore, $\scriptsize \sqrt{{-1}}=i$.
• How to write the square roots of negative numbers as imaginary numbers.
• How to add, subtract, multiply and divide imaginary numbers.
• How to simplify powers of $\scriptsize i$ by using the fact that $\scriptsize {{i}^{4}}=1$.
• Complex numbers are written in standard form as $\scriptsize a+bi$ where $\scriptsize a$ is the real part and $\scriptsize bi$ is the imaginary part.

# Unit 1: Assessment

#### Suggested time to complete: 30 minutes

Simplify the following writing your answers in standard form:

1. $\scriptsize \displaystyle \frac{{5i+6i}}{{2{{i}^{2}}}}$
2. $\scriptsize \displaystyle 7{{i}^{5}}\times 3{{i}^{{14}}}\div 7{{i}^{{12}}}$
3. $\scriptsize \displaystyle \frac{{-7{{i}^{5}}+13{{i}^{5}}}}{{-3{{i}^{4}}-5{{i}^{4}}}}$
4. $\scriptsize \displaystyle \frac{{\left( {\sqrt{{-16}}-\sqrt{{-144}}} \right)}}{{2i}}$
5. $\scriptsize \displaystyle \frac{{\sqrt{{-2}}.\sqrt{{-7}}.\sqrt{{-9}}}}{{\sqrt{{-4}}.3i}}-\displaystyle \frac{{\sqrt{{24}}}}{{\sqrt{{12}}}}$

The full solutions are at the end of the unit.

# Unit 1: Solutions

### Exercise 1.1

1. .
\scriptsize \begin{align*}\sqrt{-400}&=\sqrt{(400)(-1)}\\ &=20i \end{align*}
2. .
\scriptsize \begin{align*} \sqrt{-49}-\sqrt{-100}&=\sqrt{(49)(-1)}-\sqrt{(100)(-1)}\\ &=7i-10i\\ &=-3i \end{align*}
3. .
\scriptsize \begin{align*} \sqrt{{-4}}+\sqrt{{-25}}-\sqrt{{-3}}&=\sqrt{{(4)(-1)}}+\sqrt{{(25)(-1)}}-\sqrt{{(3)(-1)}}\\ &=2i+5i-\sqrt{3}i\\ &=7i-\sqrt{3}i\\ &=\left( {7-\sqrt{3}} \right)i\end{align*}
4. .
\scriptsize \begin{align*} 5i-\sqrt{-25}+2\sqrt{2}i&=5i-\sqrt{(25)(-1)}+2\sqrt{2}i\\ &=5i-5i+2\sqrt{2}i\\ &=2\sqrt{2}i \end{align*}

Back to Exercise 1.1

### Exercise 1.2

1. .
\scriptsize \begin{align*} \displaystyle \frac{7i^5}{49i^7}&=\displaystyle \frac{1}{7i^2}\\ &=-\displaystyle \frac{1}{7} \end{align*}
2. .
\scriptsize \begin{align*} \displaystyle \frac{\sqrt{6}i}{\sqrt{-24}}&=\displaystyle \frac{\sqrt{6}i}{\sqrt{(6)}(4)(-1)}\\ &=-\displaystyle \frac{\sqrt{6}i}{2\sqrt{6}i}\\ &=\displaystyle \frac{1}{2} \end{align*}
3. .
\scriptsize \begin{align*} \displaystyle \frac{7i^3.3i^8.2i^4}{21i^5.3i^5}&=\displaystyle \frac{42i^15}{63i^10}\\ &=\displaystyle \frac{2i^5}{3}\\ &=\displaystyle \frac{2\times i^4\times i}{3}\\ &=\displaystyle \frac{2}{3}i \end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{{\sqrt{{12}}i.(-4i).3{{i}^{5}}}}{{\sqrt{{-24}}.6{{i}^{4}}}} & =\displaystyle \frac{{\sqrt{{(4)(3)}}i(-4i).3{{i}^{5}}}}{{\sqrt{{(4)(6)(-1)}}.6{{i}^{4}}}}\\ & =\displaystyle \frac{{2\sqrt{3}i.(-12{{i}^{6}})}}{{2\sqrt{6}i.6{{i}^{4}}}}\\ & =\displaystyle \frac{{-24\sqrt{3}{{i}^{7}}}}{{12\sqrt{6}{{i}^{5}}}}\\ & =\displaystyle \frac{{-2\sqrt{3}{{i}^{2}}}}{{\sqrt{6}}}\\ & =\displaystyle \frac{{-2\sqrt{3}{{i}^{2}}}}{{\sqrt{2}\sqrt{3}}}\\ & =\displaystyle \frac{{-2(-1)}}{{\sqrt{2}}}\times \displaystyle \frac{{\sqrt{2}}}{{\sqrt{2}}}\\ & =\displaystyle \frac{{2\sqrt{2}}}{2}\\ & =\sqrt{2}\end{align*}

Back to Exercise 1.2

### Exercise 1.3

1. .
\scriptsize \begin{align*} \sqrt{-144}&=\sqrt{(144)(-1)}\\ &=12\sqrt{-1}\\ &=12i \end{align*}
Standard form: $\scriptsize 0+12i$
2. .
\scriptsize \begin{align*} & \sqrt{{-600}}\\ & =\sqrt{{(6)(100)(-1)}}\\ & =10\sqrt{{(6)(-1)}}\\ & =10\sqrt{6}i\end{align*}
Standard form: $\scriptsize 0+10\sqrt{6}i$
3. .
\scriptsize \begin{align*} & \sqrt{{12}}\\ & =\sqrt{{(4)(3)}}\\ & =2\sqrt{3}\end{align*}
Standard form: $\scriptsize 2\sqrt{3}+0i$

Back to Exercise 1.3

### Unit 1: Assessment

1. .
\scriptsize \begin{align*}\displaystyle \frac{{5i+6i}}{{2{{i}^{2}}}} & =\displaystyle \frac{{11i}}{{2(-1)}}\\ & =\displaystyle \frac{{11i}}{{-2}}\\ & =-\displaystyle \frac{{11}}{2}i\end{align*}
Standard form: $\scriptsize 0-\displaystyle \frac{{11}}{2}i$
2. .
\scriptsize \displaystyle \begin{align*} 7{{i}^{5}}\times 3{{i}^{{14}}}\div 7{{i}^{{12}}}& =\displaystyle \frac{{7{{i}^{5}}\times 3{{i}^{{14}}}}}{{7{{i}^{{12}}}}}\\ & =\displaystyle \frac{{21{{i}^{{19}}}}}{{7{{i}^{{12}}}}}\\ & =3{{i}^{7}}\\ & =3.{{i}^{4}}.{{i}^{3}}\quad \left( {{{i}^{4}}=1} \right)\\ & =3.{{i}^{2}}.i\quad \left( {{{i}^{2}}=-1} \right)\\ & =-3i\end{align*}
Standard form: $\scriptsize 0-3i$
3. .
\scriptsize \begin{align*}\displaystyle \frac{{-7{{i}^{5}}+13{{i}^{5}}}}{{-3{{i}^{4}}-5{{i}^{4}}}} & =\displaystyle \frac{{6{{i}^{5}}}}{{-8{{i}^{4}}}}\\ & =-\displaystyle \frac{{3i}}{4}\\ & =-\displaystyle \frac{3}{4}i\end{align*}
Standard form: $\scriptsize 0-\displaystyle \frac{3}{4}i$
4. .
\scriptsize \begin{align*} \displaystyle \frac{{\left( {\sqrt{{-16}}-\sqrt{{-144}}} \right)}}{{2i}}& =\displaystyle \frac{{\left( {\sqrt{{(16)(-1)}}-\sqrt{{(144)(-1)}}} \right)}}{{2i}}\\ & =\displaystyle \frac{{\left( {4i-12i} \right)}}{{2i}}\\ & =\displaystyle \frac{{-8i}}{{2i}}\\ & =-4\end{align*}
Standard form: $\scriptsize -4+0i$
5. .
\scriptsize \begin{align*} \displaystyle \frac{{\sqrt{{-2}}.\sqrt{{-7}}.\sqrt{{-9}}}}{{\sqrt{{-4}}.3i}}-\displaystyle \frac{{\sqrt{{24}}}}{{\sqrt{{12}}}}& =\displaystyle \frac{{\sqrt{{(2)(-1)}}.\sqrt{{(7)(-1)}}.\sqrt{{(9)(-1)}}}}{{\sqrt{{(4)(-1)}}.3i}}-\displaystyle \frac{{\sqrt{{6.4}}}}{{\sqrt{{3.4}}}}\\ & =\displaystyle \frac{{\sqrt{2}i.\sqrt{7}i.3i}}{{2i.3i}}-\displaystyle \frac{{2\sqrt{6}}}{{2\sqrt{3}}}\\ & =\displaystyle \frac{{3\sqrt{{14}}{{i}^{3}}}}{{6{{i}^{2}}}}-\displaystyle \frac{{\sqrt{2}.\sqrt{3}}}{{\sqrt{3}}}\\ & =\displaystyle \frac{{\sqrt{{14}}i}}{2}-\sqrt{2}\\ & =-\sqrt{2}+\displaystyle \frac{{\sqrt{{14}}}}{2}i\end{align*}
Standard form: $\scriptsize -\sqrt{2}+\displaystyle \frac{{\sqrt{{14}}}}{2}i$

Back to Unit 1: Assessment