[Math] the difference between complex numbers and 2D vectors

complex numbersgeometryvectors

This is a follow-up to a previous question regarding complex numbers. Many people there compared complex numbers to vectors, and there was disagreement about what the difference was. Some options people have:

  1. Complex numbers and vectors are exactly identical.

  2. The $i$ in complex numbers changes the way it describes the space, if so, how and why?

  3. Complex numbers are 2D but can be used cleverly to describe larger dimensional systems. If so how and why?

What are the differences?

Best Answer

Complex numbers are not identical to two dimensional vectors, but in some ways complex numbers and vectors have similar behaviours. There are also similarities between complex numbers and two dimensional matrices. Mathematicians call these similarities isomorphisms.

There are two things you can do with a pair of complex numbers. You can add (or subtract) them and you can multiply (or divide) them.

Let's think about addition/subtraction first.

Suppose we map each complex number to a two dimensional vector as follows:

$a+ib \mapsto (a,b)$

Then in the world of complex numbers

$(a+ib) + (c+id) = (a+c) +i(b+d)$

and in the world of vectors

$(a,b) +(c,d) = (a+c,b+d)$

So we can see that

$(a+ib) + (c+id) \mapsto (a,b) + (c,d)$

where the addition on the left hand side is complex number addition, and the addition on the left hand side is vector addition. And we could reverse the mapping and go from vector addition to complex number addition instead.

We say that addition of complex numbers is isomorphic to addition of two dimensional vectors. Note that we are not saying that complex numbers and vectors are the same - we are just saying that for the purposes of addition/subtraction they behave in the same way.

Now let's think about multiplication/division.

When we multiply a complex number by $i$ then we get:

$i(c+id) = -d + ic$

In the vector world we start with a vector $(c,d)$ and end up with a vector $(-d,c)$. We have rotated the vector anticlockwise by $90^o$. We can represent this rotation by the matrix

$\begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}$

With a little more thought, we can come up with a mapping from complex numbers to two dimensional matrices:

$a+ib \mapsto \begin{pmatrix}a & -b \\ b & a \end{pmatrix}$

and then we have

$(a+ib)(c+id) = (ac-bd) +i(ad+bc)$

and

$\begin{pmatrix}a & -b \\ b & a \end{pmatrix}\begin{pmatrix}c & -d \\ d & c \end{pmatrix} = \begin{pmatrix}ac-bd & -ad-bc \\ ad+bc & ac-bd \end{pmatrix}$

so

$(a+ib)(c+id) \mapsto \begin{pmatrix}a & -b \\ b & a \end{pmatrix}\begin{pmatrix}c & -d \\ d & c \end{pmatrix}$

So we have found another isomorphism. We say that multiplication of complex numbers is isomorphic to multiplication of two dimensional matrices.

Related Question