Complex numbersΒΆ

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, or representing projections of three-dimensional objects onto a plane, 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.

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 \((ac - bd) + (ad + bc)i\).

Note

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 with 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 \(a \in R\),

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

Establishing that \(\mathbb{C}\) is 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 given that it is a commutative ring β€” that is, showing that all nonzero elements have multiplicative inverses β€” 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 β€” 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.