*i.e.*you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

## 1. The Definition

As I’ve already stated, I am assuming that you have seen complex numbers to this point and that you’re aware that \(i = \sqrt { - 1} \) and so \({i^2} =- 1\). This is an idea that most people first see in an algebra class (or wherever they first saw complex numbers) and \(i = \sqrt { - 1} \) is defined so that we can deal with square roots of negative numbers as follows,

\[\sqrt { - 100} = \sqrt {\left( {100} \right)\left( { - 1} \right)} = \sqrt {100} \,\,\sqrt { - 1} = \sqrt {100} \,\,i = 10\,i\]What I’d like to do is give a more *mathematical* definition of a complex numbers and show that \({i^2} = - 1\) (and hence \(i = \sqrt { - 1} \) can be thought of as a consequence of this definition. We’ll also take a look at how we define arithmetic for complex numbers.

What we’re going to do here is going to seem a little backwards from what you’ve probably already seen but is in fact a more accurate and mathematical definition of complex numbers. Also note that this section is not really required to understand the remaining portions of this document. It is here solely to show you a different way to define complex numbers.

So, let’s give the definition of a complex number.

Given two real numbers \(a\) and \(b\) we will define the complex number \(z\) as,

\[z = a + bi\]Note that at this point we’ve not actually defined just what \(i\) is at this point. The number \(a\) is called the **real part** of \(z\) and the number \(b\) is called the **imaginary part** of \(z\) and are often denoted as,

There are a couple of special cases that we need to look at before proceeding. First, let’s take a look at a complex number that has a zero real part,

\[z = 0 + bi = bi\]In these cases, we call the complex number a **pure imaginary** number.

Next, let’s take a look at a complex number that has a zero imaginary part,

\[z = a + 0i = a\]In this case we can see that the complex number is in fact a real number. Because of this we can think of the real numbers as being a subset of the complex numbers.

We next need to define how we do addition and multiplication with complex numbers. Given two complex numbers \({z_1} = a + bi\) and \({z_2} = c + di\) we define addition and multiplication as follows,

\begin{align} {z_1} + {z_2} & = \left( {a + c} \right) + \left( {b + d} \right)i \\ {z_1}{z_2} & = \left( {ac - bd} \right) + \left( {ad + cb} \right)i \end{align}Now, if you’ve seen complex numbers prior to this point you will probably recall that these are the formulas that were given for addition and multiplication of complex numbers at that point. However, the multiplication formula that you were given at that point in time required the use of \({i^2} = - 1\) to completely derive and for this section we don’t yet know that is true. In fact, as noted previously \({i^2} = - 1\) will be a consequence of this definition as we’ll see shortly.

Above we noted that we can think of the real numbers as a subset of the complex numbers. Note that the formulas for addition and multiplication of complex numbers give the standard real number formulas as well. For instance, given the two complex numbers,

\[{z_1} = a + 0i \hspace{0.5in} {z_2} = c + 0i\]the formulas yield the correct formulas for real numbers as seen below.

\begin{align}{z_1} + {z_2} & = \left( {a + c} \right) + \left( {0 + 0} \right)i = a + c\\ {z_1}{z_2} & = \left( {ac - \left( 0 \right)\left( 0 \right)} \right) + \left( {a\left( 0 \right) + c\left( 0 \right)} \right)i = ac\end{align}The last thing to do in this section is to show that \({i^2} = - 1\) is a consequence of the definition of multiplication. However, before we do that we need to acknowledge that powers of complex numbers work just as they do for real numbers. In other words, if \(n\) is a positive integer we will define exponentiation as,

\[{z^n} = \underbrace {z \cdot z \cdot \cdots z}_{n{\rm{ times}}}\]So, let’s start by looking at \({i^2}\), use the definition of exponentiation and the use the definition of multiplication on that. Doing this gives,

\[\begin{align}{i^2} & = i \cdot i\\ & = \left( {0 + 1i} \right)\left( {0 + 1i} \right)\\ & = \left( {\left( 0 \right)\left( 0 \right) - \left( 1 \right)\left( 1 \right)} \right) + \left( {\left( 0 \right)\left( 1 \right) + \left( 0 \right)\left( 1 \right)} \right)i\\ & = - 1\end{align}\]So, by defining multiplication as we’ve done above we get that \({i^2} = - 1\) as a consequence of the definition instead of just stating that this is a true fact. If we now take \(\sqrt{\,\,\,\,\,} \) to be the standard square root, *i.e.* what did we square to get the quantity under the radical, we can see that \(i = \sqrt { - 1} \).