Complex numbers

We encountered the set \(\mathbb{R}^2\) in the Matrices chapter, and defined an addition operation which made \((\mathbb{R}^2, +)\) an abelian group. In this section we will come across a multiplication operation on \(\mathbb{R}^2\) and we will see that with these two operations, \(\mathbb{R}^2\) can be made into a field, which is called the field of complex numbers. When we are making use of the field structure we will usually write this field as \(\mathbb{C}\) rather than \(\mathbb{R}^2\).

Complex numbers have a variety of applications, including in geometry, for e.g. representing figures in two dimensions, for modelling behaviour of electrical signals, and for analysing the behaviour of systems which can be modelled using differential equations, such as how populations of different species in a food web change over time, how heat flows through an object, or how mechanical systems like suspension in a car will behave. They also function as useful tools in many other areas of mathematics. For instance, they play a major role in the proof that quintic equations — that is, equations of the form \(ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0\) — cannot be solved in general, as well as offering some nifty tricks to perform otherwise difficult integrations of real-valued functions.

First, instead of writing elements of \(\mathbb{R}^2\) in the usual way, i.e. \((a, b)\), we will write them as \(a + bi\), where \(a\) and \(b\) are real numbers. For example, we write \((1,2)\) as \(1 + 2i\), we write \((1,0)\) as just \(1\), and we write \((0,1)\) as just \(i\). For a complex number \(a + bi\), we call \(a\) the “real part”, and \(b\) the “imaginary part”.

Therefore, to add two complex numbers together, we simply add the real and imaginary parts. That is, \((a + bi) + (c + di) = (a+c) + (b+d)i\).

For the multiplication operation, if we remember that \(i^2 = -1\), the rest sort of falls out. That is, to multiply two complex numbers together, we can write \((a + bi)(c + di) = ac + adi + bci + bdi^2\), making use of distributivity, and then using distributivity again (but in the reverse direction) and replacing \(i^2\) with \(-1\), we can write this as \((ac - bd) + (ad + bc)i\).


This approach is a bit sloppy because we are assuming that \(\mathbb{C}\) is a field before even defining its multiplication operation. I hope you can forgive me for this.

Notice that the subset of \(\mathbb{C}\) given by the complex numbers which have an imaginary part of \(0\) behaves in exactly the same way as \(\mathbb{R}\); because of this, we can consider \(\mathbb{R}\) as a subset of \(\mathbb{C}\). In the same vein we will describe elements of \(\mathbb{C}\) which have an imaginary part of \(0\) as “real”.

Another useful thing to notice is that to multiply a real number by a complex number, you simply multiply the real and imaginary parts by that number. That is, for \(a \in \mathbb{R}\),

\[(a+0i)(c+di) = (ac + 0d) + (ad + 0c)i = (ac) + (ad)i\]

Establishing that \(\mathbb{C}\) is a ring (in fact, a commutative ring) with respect to these addition and multiplication operations is a little tedious, so I won’t set it as an ‘official’ exercise, but you may find it worth doing anyway.

However, establishing that \(\mathbb{C}\) is a field — in particular, showing that all nonzero elements have multiplicative inverses — is more interesting.

To find the multiplicative inverse of an arbitrary complex number, we use an operation called conjugation. The conjugate of a complex number is obtained by negating the imaginary part, i.e. the conjugate of \(a + bi\) is \(a - bi\). The first thing to notice is that multiplying a complex number by its conjugate always yields a real number:

\[\begin{split}(a + bi)(a - bi) &= a^2 + abi - abi - b^2i^2 \\ &= a^2 + b^2\end{split}\]

Now, if we write the multiplicative inverse of \(a + bi\) as a fraction \(\frac{1}{a+bi}\), the answer becomes a little clearer. Just as with real numbers, we can multiply top and bottom by the same quantity:

\[\begin{split}\frac{1}{a+bi} &= \frac{a-bi}{(a+bi)(a-bi)} \\ &= \frac{a-bi}{a^2+b^2}\end{split}\]

Now we have the product of a complex number with a real number, i.e. we have \(a - bi\) multiplied by \(\frac{1}{a^2 + b^2}\). As we showed before we can simply multiply the real and imaginary parts by this real number, which gives us the inverse of \(a + bi\) as:

\[\frac{a-bi}{a^2+b^2} = \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i\]

Let’s check that this really is the multiplicative inverse of \(a + bi\), just to be safe:

\[\begin{split}(a+bi) \left( \frac{a}{a^2+b^2} - \frac{b}{a^2+b^2}i \right) &= \frac{a^2}{a^2+b^2} - \frac{ab}{a^2+b^2}i + \frac{ab}{a^2+b^2}i - \frac{b^2}{a^2+b^2}i^2 \\ &= \frac{a^2 + b^2}{a^2 + b^2} \\ &= 1\end{split}\]

So there you have it — we can indeed make \(\mathbb{R}^2\) into a field!

There is so, so much more I could say about complex numbers, but I think I will leave it here for now.