Complex Numbers are Very Real

A brief while ago I had a discussion with some people who claimed that complex numbers may be very hard to wrap your head around. This is arguably quite true, as intuition can be different for everyone. In this post, I give an alternate representation (!) of complex numbers that only uses the reals. Before we begin though, let us recall some basic properties of 2 \times 2 real matrices.

A 2 \times 2 matrix is an entry of 4 numbers a, b, c, d, written as \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}. We can add and subtract them component-wise, multiply them with \begin{pmatrix} a & b \\ c & d  \\ \end{pmatrix} \cdot \begin{pmatrix} a' & b' \\ c' & d' \\ \end{pmatrix} = \begin{pmatrix} aa' + bc' & ab' + bd' \\ ca' + dc' & cb' + dd' \\ \end{pmatrix}, and compute inverses (read: divide) when the quantity called the determinant, ad - bc, is non-zero. There are plenty of resources to learn about matrices, so use those to familiarize yourself before moving on to the next section.

In order to represent the complex numbers as matrices, we must first obtain a representation of the reals. This is quite easy. We can have the mapping x \mapsto \begin{pmatrix} x & 0 \\ 0 & x \\ \end{pmatrix}. This mapping makes x + y, x - y, x * y, x / y behave as one would expect. Now comes the hard part. How do we come up with a matrix i such that i^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}? To start off, we can write i = \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} and solve the equations but I will just give you the answer to verify for now. One representation is i = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}. And now that we can represent reals and the imaginary unit, it becomes quite easy to express any arbitrary complex number. a + bi = \begin{pmatrix} a & -b \\ b & a \\ \end{pmatrix}. To get division, recall that the determinant must not be zero, or a^2 + b^2 \neq 0. This is precisely the condition that the complex number is non-zero! Sweet, right?

To obtain the second representation of i, recall that (-i)^2 is also equal to -1. The rest is left as an exercise to the reader.

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