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## Arithmetic

Before proceeding in this section let me first say that I’m assuming that you’ve seen arithmetic with complex numbers at some point before and most of what is in this section is going to be a review for you. I am also going to be introducing subtraction and division in a way that you probably haven’t seen prior to this point, but the results will be the same and aren’t important for the remaining sections of this document.

In the previous section we defined addition and multiplication of complex numbers and showed that $${i^2} = - 1$$ is a consequence of how we defined multiplication. However, in practice, we generally don’t multiply complex numbers using the definition. In practice we tend to just multiply two complex numbers much like they were polynomials and then make use of the fact that we now know that $${i^2} = - 1$$.

Just so we can say that we’ve worked an example let’s do a quick addition and multiplication of complex numbers.

Example 1 Compute each of the following.
1. $$\left( {58 - i} \right) + \left( {2 - 17i} \right)$$
2. $$\left( {6 + 3i} \right)\left( {10 + 8i} \right)$$
3. $$\left( {4 + 2i} \right)\left( {4 - 2i} \right)$$
Show Solution

As noted above, I’m assuming that this is a review for you and so won’t be going into great detail here.

a $$\left( {58 - i} \right) + \left( {2 - 17i} \right) = 58 - i + 2 - 17i = 60 - 18i$$

b $$\left( {6 + 3i} \right)\left( {10 + 8i} \right) = 60 + 48i + 30i + 24{i^2} = 60 + 78i + 24\left( { - 1} \right) = 36 + 78i$$

c $$\left( {4 + 2i} \right)\left( {4 - 2i} \right) = 16 - 8i + 8i - 4{i^2} = 16 + 4 = 20$$

It is important to recall that sometimes when adding or multiplying two complex numbers the result might be a real number as shown in the third part of the previous example!

The third part of the previous example also gives a nice property about complex numbers.

\begin{equation}\left( {a + bi} \right)\left( {a - bi} \right) = {a^2} + {b^2} \label{eq:mult} \end{equation}

We’ll be using this fact with division and looking at it in slightly more detail in the next section.

Let’s now take a look at the subtraction and division of two complex numbers. Hopefully, you recall that if we have two complex numbers, $${z_1} = a + bi$$ and $${z_2} = c + di$$ then you subtract them as,

\begin{equation}z_1 - z_2 = \left( a + bi \right) - \left( c + di \right) = \left( a - c \right) + \left( b - d \right)i \label{eq:sub} \end{equation}

And that division of two complex numbers,

\begin{equation}\frac{{{z_1}}}{{{z_2}}} = \frac{{a + bi}}{{c + di}} \label{eq:div} \end{equation}

can be thought of as simply a process for eliminating the $$i$$ from the denominator and writing the result as a new complex number $$u + vi$$.

Let’s take a quick look at an example of both to remind us how they work.

Example 2 Compute each of the following.
1. $$\left( {58 - i} \right) - \left( {2 - 17i} \right)$$
2. $$\frac{{6 + 3i}}{{10 + 8i}}$$
3. $$\frac{{5i}}{{1 - 7i}}$$
Show All Solutions Hide All Solutions
a Show Solution
There really isn’t too much to do here so here is the work, $\left( {58 - i} \right) - \left( {2 - 17i} \right) = 58 - i - 2 + 17i = 56 + 16i$

b Show Solution
Recall that with division we just need to eliminate the $$i$$ from the denominator and using $$\eqref{eq:mult}$$ we know how to do that. All we need to do is multiply the numerator and denominator by $$10 - 8i$$ and we will eliminate the $$i$$ from the denominator. \begin{align*}\frac{{6 + 3i}}{{10 + 8i}} & = \frac{{\left( {6 + 3i} \right)}}{{\left( {10 + 8i} \right)}}\frac{{\left( {10 - 8i} \right)}}{{\left( {10 - 8i} \right)}}\\ & = \frac{{60 - 48i + 30i - 24{i^2}}}{{100 + 64}}\\ & = \frac{{84 - 18i}}{{164}}\\ & = \frac{{84}}{{164}} - \frac{{18}}{{164}}i \\ & = \frac{{21}}{{41}} - \frac{9}{{82}}i\end{align*}

c Show Solution
We’ll do this one a little quicker. $\frac{{5i}}{{1 - 7i}} = \frac{{5i}}{{\left( {1 - 7i} \right)}}\frac{{\left( {1 + 7i} \right)}}{{\left( {1 + 7i} \right)}} = \frac{{ - 35 + 5i}}{{1 + 49}} = - \frac{7}{{10}} + \frac{1}{{10}}i$

Now, for the most part this is all that you need to know about subtraction and division of complex numbers for this rest of this document. However, let’s take a look at a more precise and mathematical definition of both of these. If you aren’t interested in this then you can skip this and still be able to understand the remainder of this document.

The remainder of this document involves topics that are typically first taught in a Abstract/Modern Algebra class. Since we are going to be applying them to the field of complex variables we won’t be going into great detail about the concepts. Also note that we’re going to be skipping some of the ideas and glossing over some of the details that don’t really come into play in complex numbers. This will especially be true with the “definitions” of inverses. The definitions I’ll be giving below are correct for complex numbers, but in a more general setting are not quite correct. You don’t need to worry about this in general to understand what were going to be doing below. I just wanted to make it clear that I’m skipping some of the more general definitions for easier to work with definitions that are valid in complex numbers.

Okay, now that I’ve got the warnings/notes out of the way let’s get started on the actual topic…

Technically, the only arithmetic operations that are defined on complex numbers are addition and multiplication. This means that both subtraction and division will, in some way, need to be defined in terms of these two operations. We’ll start with subtraction since it is (hopefully) a little easier to see.

We first need to define something called an additive inverse. An additive inverse is some element typically denoted by $$- z$$ so that

\begin{equation}z + \left( { - z} \right) = 0\end{equation}

Now, in the general field of abstract algebra, $$- z$$ is just the notation for the additive inverse and in many cases is NOT given by $$- z = \left( { - 1} \right)z$$ ! Luckily for us however, with complex variables that is exactly how the additive inverse is defined and so for a given complex number $$z = a + bi$$ the additive inverse, $$- z$$, is given by,

$- z = \left( { - 1} \right)z = - a - bi$

It is easy to see that this does meet the definition of the additive inverse and so that won’t be shown.

With this definition we can now officially define the subtraction of two complex numbers. Given two complex numbers $${z_1} = a + bi$$ and $${z_2} = c + di$$ we define the subtraction of them as,

\begin{equation}{z_1} - {z_2} = {z_1} + \left( { - {z_2}} \right)\end{equation}

Or, in other words, when subtracting $${z_2}$$ from $${z_1}$$ we are really just adding the additive inverse of $${z_2}$$ (which is denoted by $$- {z_2}$$) to $${z_1}$$. If we further use the definition of the additive inverses for complex numbers we can arrive at the formula given above for subtraction.

${z_1} - {z_2} = {z_1} + \left( { - {z_2}} \right) = \left( {a + bi} \right) + \left( { - c - di} \right) = \left( {a - c} \right) + \left( {b - d} \right)i$

So, that wasn’t too bad I hope. Most of the problems that students have with these kinds of topics is that they need to forget some notation and ideas that they are very used to working with. Or, to put it another way, you’ve always been taught that $$- z$$is just a shorthand notation for $$\left( { - 1} \right)z$$, but in the general topic of abstract algebra this does not necessarily have to be the case. It’s just that in all of the examples where you are liable to run into the notation $$- z$$ in “real life”, whatever that means, we really do mean $$- z = \left( { - 1} \right)z$$.

Okay, now that we have subtraction out of the way, let’s move on to division. As with subtraction we first need to define an inverse. This time we’ll need a multiplicative inverse. A multiplicative inverse for a non-zero complex number $$z$$ is an element denoted by $${z^{ - 1}}$$ such that

$z\,z{^{ - 1}} = 1$

Now, again, be careful not to make the assumption that the “exponent” of -1 on the notation is in fact an exponent. It isn’t! It is just a notation that is used to denote the multiplicative inverse. With real (non-zero) numbers this turns out to be a real exponent and we do have that

${4^{ - 1}} = \frac{1}{4}$

for instance. However, with complex numbers this will not be the case! In fact, let’s see just what the multiplicative inverse for a complex number is. Let’s start out with the complex number $$z = a + bi$$ and let’s call its multiplicative inverse $${z^{ - 1}} = u + vi$$. Now, we know that we must have

$z\,z{^{ - 1}} = 1$

so, let’s actual do the multiplication.

\begin{align*}z{z^{ - 1}} & = \left( {a + bi} \right)\left( {u + vi} \right)\\ & = \left( {au - bv} \right) + \left( {av + bu} \right)i\\ & = 1\end{align*}

This tells us that we have to have the following,

$au - bv = 1 \hspace{0.5in} av + bu = 0$

Solving this system of two equations for the two unknowns $$u$$ and $$v$$ (remember $$a$$ and $$b$$ are known quantities from the original complex number) gives,

$u = \frac{a}{{{a^2} + {b^2}}} \hspace{0.5in} v = - \frac{b}{{{a^2} + {b^2}}}$

Therefore, the multiplicative inverse of the complex number $$z$$ is,

\begin{equation}{z^{ - 1}} = \frac{a}{{{a^2} + {b^2}}} - \frac{b}{{{a^2} + {b^2}}}\,\,i \label{eq:inv}\end{equation}

As you can see, in this case, the “exponent” of -1 is not in fact an exponent! Again, you really need to forget some notation that you’ve become familiar with in other math courses.

So, now that we have the definition of the multiplicative inverse we can finally define division of two complex numbers. Suppose that we have two complex numbers $${z_1}$$and $${z_2}$$ then the division of these two is defined to be,

\begin{equation}\frac{{{z_1}}}{{{z_2}}} = {z_1}\,z_2^{ - 1}\end{equation}

In other words, division is defined to be the multiplication of the numerator and the multiplicative inverse of the denominator. Note as well that this actually does match with the process that we used above. Let’s take another look at one of the examples that we looked at earlier only this time let’s do it using multiplicative inverses. So, let’s start out with the following division.

$\frac{{6 + 3i}}{{10 + 8i}} = \left( {6 + 3i} \right){\left( {10 + 8i} \right)^{ - 1}}$

We now need the multiplicative inverse of the denominator and using $$\eqref{eq:inv}$$ this is,

${\left( {10 + 8i} \right)^{ - 1}} = \frac{{10}}{{{{10}^2} + {8^2}}} - \frac{8}{{{{10}^2} + {8^2}}}i = \frac{{10 - 8i}}{{164}}$

Now, we can do the multiplication,

$\frac{{6 + 3i}}{{10 + 8i}} = \left( {6 + 3i} \right){\left( {10 + 8i} \right)^{ - 1}} = \left( {6 + 3i} \right)\frac{{10 - 8i}}{{164}} = \frac{{60 - 48i + 30i - 24{i^2}}}{{164}} = \frac{{21}}{{41}} - \frac{9}{{82}}i$

Notice that the second to last step is identical to one of the steps we had in the original working of this problem and, of course, the answer is the same.

As a final topic let’s note that if we don’t want to remember the formula for the multiplicative inverse we can get it by using the process we used in the original multiplication. In other words, to get the multiplicative inverse we can do the following

${\left( {10 + 8i} \right)^{ - 1}} = \frac{1}{{\left( {10 + 8i} \right)}}\frac{{10 - 8i}}{{\left( {10 - 8i} \right)}} = \frac{{10 - 8i}}{{{{10}^2} + {8^2}}}$

As you can see this is essentially the process we used in doing the division initially.