# 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.