Complex Multiplication with Euclidean Geometry

2023-08-08

Introduction

Throughout this proof, I will not use the $r e^{i θ}$ formulation of complex coordinates, to keep things straightforward. The goal here is to develop a construction that could have been developed before such a formulation was known.

I prefer to make this proof as "primitive" as possible, for a variety of reasons:

It develops an intuition regarding complex multiplication,
It shows that if complex numbers were discovered earlier, that multiplication of complex numbers could still have been defined the way we use them today.
It will allow for an earlier introduction to complex multiplication in education, which may prove to be useful.

For these reasons, I prefer to refrain from using concepts that are too advanced, such as trigonometric functions or matrix multiplication - or at least I will not rely on those concepts directly. I will mention matrices briefly however, only to make the connection between the Euclidean construction of the product of complex numbers, and the usual algebraic product that we use. The use of linear transformation severely simplifies the proof.

Notation

Lines in complex plane will be denotes via a pair of complex numbers $(x, y)$ , where both $x$ and $y$ are complex. This denotes the line from $x$ to $y$ .

The length of a line $(x, y)$ , or it's distance from the origin of the complex plane $O (0, 0)$ , is denoted $| (x, y) |$ .

Triangles will be denoted with 3 complex numbers $△ (x, y, z)$ . Again, the 3 variables are the vertices defining the triangle.

Similarity of triangles is stated via " $\sim$ ": $△ (u, v, w) \sim △ (x, y, z)$ .

Congruency is declared via " $≅$ ": $△ (u, v, w) ≅ △ (x, y, z)$ .

Functions are declared via the usual notation $F (x) = y$ . Specific mappings of a function will use the notation $F : a \mapsto b$ .

Underlying Intuition

The intuition remains much the same as the computing the product of real numbers via classical Euclidean Geometry: To compute $a \cdot b$ , create a linear mapping $T$ such that $T : 1 \mapsto a$ . Then apply the same linear mapping to $b$ . The main difference: our attention is towards complex numbers rather than real numbers.

Recall that any linear mapping $T$ conserves scaling (ie. $T (x)$ such that $T (c x) = c T (x)$ ). If $T$ maps $T : 1 \mapsto a$ , then it also maps $T : b \mapsto a b$ since $T (b) = T (1 \cdot b) = T (1) \cdot b = a b$ . Euclidean constructions that multiply lengths of line segments focus on producing such transformations. I am applying this logic to complex numbers, but working with Euclidean constructions to make things straightforward and as elementary as possible.

It is this geometric definition that I will provide that I believe is more fitting to cover our modern notion of multiplication, one that includes complex numbers.

Euclidean Construction

Let $a$ and $b$ be 2 complex numbers.

Mark points 0, 1, $a$ , and $b$ , on the complex plane.
Draw triangles $△ (0, 1, a)$ and $△ (0, 1, b)$ .
Draw circle centred at $0$ , using the line $(0, 1)$ as your radius. Mark the intersection with the line $(0, a)$ as $1^{'}$ . Extend $(0, a)$ if necessary.
Measure lines $r = | (0, b) |$ and $s = | (1, b) |$ with a compass. Then, draw a circle of radius $r$ with centre $0$ , and another circle about radius $s$ with centre $1^{'}$ . Mark this intersection as $1^{'} b$ .

Note that by side-side-side, $△ (0, 1, b) ≅ △ (0, 1^{'}, 1^{'} b)$ , as $| (0, 1) | = | (0, 1^{'}) |$ , $| (1, b) | = | (1^{'}, 1^{'} b) |$ , and $| (0, b) | = | (0, 1^{'} b) |$ .
Draw a line parallel to $(1^{'}, 1^{'} b)$ that intersects with line $(0, 1^{'} b)$ . Extend $(0, 1^{'} b)$ if necessary. Mark the point of intersection of these lines as $z$ .

Algebraic Proof of Construction

Claim: $z = a \cdot b$ . That is, $z$ is the complex product of the numbers $a = a_{0} + i a_{1}$ and $b = b_{0} + i b_{1}$ in the complex plane. More specifically,

z = a b = (a_{0} + i a_{1}) (b_{0} + i b_{1}) = (a_{0} b_{0} - a_{1} b_{1}) + i (a_{1} b_{0} + a_{0} b_{1})

Although the Euclidean construction in the previous section can be treated as a definition, I'd like to ensure that it corresponds to our usual notion of a product of two complex numbers.

This proof will first determine a linear transformation $T$ such as $T : 1 \mapsto a$ (which will capture the rotation and scaling simultaneously - as opposed to in individual steps as in the construction).

The transformation will take the following form:

T (x, y) = (A x - B y, B x + A y)

I will ignore proving that this preserves scaling and rotations as that can be covered in linear algebra courses, given that this corresponds to the transformation matrix corresponding to a scaling and rotation combined:

(\begin{matrix} A & - B \\ B & A \end{matrix}) = (A^{2} + B^{2}) (\begin{matrix} \cos θ & - \sin θ \\ \sin θ & \cos θ \end{matrix})

Using a function $T : R^{2} \to R^{2}$ as defined by $T (x, y)$ rather than a scaled rotation matrix allows us to skip the need for introducing trigonometry or linear algebra, as the same linear transformation involving rotations and scaling was applied on all 3 points of $△ (0, 1, b)$ .

Returning to the proof, we first map $1 \mapsto a = (a_{0}, a_{1})$ (as done in the construction):

T (1, 0) = (A - 0, B + 0) = (A, B) = (a_{0}, a_{1})

This gives us our coefficients. Thus:

T (x, y) = (a_{0} x - a_{1} y, a_{1} x + a_{0} y)

Applying it to $b$ :

T (b_{0}, b_{1}) = (a_{0} b_{0} - a_{1} b_{1}, a_{1} b_{0} + a_{0} b_{1})

In the complex plane, this corresponds to $z = (a_{0} b_{0} - a_{1} b_{1}) + i (a_{1} b_{0} + a_{0} b_{1})$ , which is what we desired.