Motivation behind the definition of multiplication in $ \mathbb{R}^2 $

Question

I was reading Titu Andreescu’s book on complex numbers, Complex Numbers A to Z. In the book, the author defines multiplication in $ \mathbb{ R }^2 $ as,

$ (a,b)( c,d) = (ac - bd, ad + bc) $

I understand how this definition relates to multiplication in complex numbers, and I see how with the knowledge of complex numbers one would be motivated to define multiplication in $ \mathbb{R}^2 $. But what I am wondering is, how one would motivate this definition purely from the geometry of $ \mathbb{R}^2 $, from which we can then construct the field of complex numbers.

Can you recommend me any resources where I can find this approach ? Thank you very much in advance.

Answer 1

Write cis $\psi= $cos $\psi$+$i$ sin $\psi$. Then$$(a+bi)(\alpha +\beta i)=(r \space \text {cis} \theta)(\rho \space \text{cis} \phi)$$ $$=r \rho \space\text {cis}(\theta + \phi)$$That explains the geometric motivation for the definition of multiplication.

Answer 2

Here is a reasoning which shows that two rather simple requirements might lead us to this kind of multiplication.

We consider the $\mathbb{R}$-vector space $\mathbb{R}^2$ and the standard basis \begin{align*} \{(1,0),(0,1)\} \end{align*}

Let's assume we are looking for a multiplication which maps

• The vector $(1,0)$ to the unit element

• The square of $(0,1)$ to the negative unit element $(0,1)^2=-(1,0)$

Under these assumptions we derive \begin{align*} \color{blue}{(x_1,y_1)(x_2,y_2)} &=\left(x_1(1,0)+y_1(0,1)\right)\left(x_2(1,0)+y_2(0,1)\right)\\ &=x_1x_2(1,0)^2+\left(x_1y_2+y_1x_2\right)(0,1)+y_1y_2(0,1)^2\\ &=(x_1x_2-y_1y_2)(1,0)+(x_1y_2+y_1x_2)(0,1)\\ &\,\,\color{blue}{=(x_1x_2-y_1y_2,x_1y_2+y_1x_2)} \end{align*}

Answer 3

You have a problem in that the Euclidean plane $\mathbb{R}^2$ is isotropic and the Complex numbers are not. It can't arise purely from the geometry of $\mathbb{R}^2$, you have to introduce something else to break the symmetry.

However, there is a way to connect the two if you generate a Real Clifford algebra on $\mathbb{R}^2$. The Clifford algebra describes the geometry of the plane, and arises 'purely' from the geometry of $\mathbb{R}^2$ in this sense, and the Complex numbers are a sub-algebra of the resulting Clifford algebra.

Elements of the Clifford algebra may be written $\alpha+\beta \mathbf{e_x}+\gamma \mathbf{e_y}+\delta \mathbf{e_x}\mathbf{e_y}$, where $\mathbf{e_x}$ and $\mathbf{e_y}$ are orthonormal basis vectors of $\mathbb{R}^2$. The geometric interpretation of this picture is (roughly) that scalars ($\alpha$) represent scale transformations of the plane and zero-dimensional magnitudes, vectors ($\beta \mathbf{e_x}+\gamma \mathbf{e_y}$) represent reflections of the plane and 1D directed lengths, and bivectors ($\delta \mathbf{e_x}\mathbf{e_y}$) represent rotations of the plane and 2D directed areas. Reflect in the same line twice, and you get the identity - a scalar. Reflect in two different lines and you get a rotation. Rotate $90^\circ$ twice (i.e. square it) and you get $-1$. So the geometry of the plane tells us how to build the algebra, and the geometry naturally contains two elements ($\pm 90^\circ$ rotation) that correspond to square roots of $-1$ in the algebra ($\pm\mathbf{e_x}\mathbf{e_y}$).

However, in this interpretation Complex numbers are not made up from vectors of the plane $\mathbb{R}^2$ itself, but those vectors multiplied by a constant vector. ($\mathbf{e_xe_x}=1$, $\mathbf{e_xe_y}=i$, so $\mathbf{e_x}(\alpha \mathbf{e_x}+\beta \mathbf{e_y})=\alpha+\beta i$.) Instead of using $\mathbf{e_x}$ and $\mathbf{e_y}$ which represent reflections, we reflect both to end up with linear combinations of the identity and $90^\circ$ rotations. This choice breaks the symmetry, selecting a vector to represent the Real axis. If you take just those even elements made up from scalars and bivectors, they are closed under addition and multiplication, and so form a sub-algebra. For the Clifford algebra built on $\mathbb{R}^2$, this even sub-algebra is isomorphic to the Complex numbers. For the Clifford algebra built on $\mathbb{R}^3$, this even sub-algebra is isomorphic to the Quaternions. And the construction can easily be extended to higher-dimensional geometries - just take the elements made up from products of an even number of vectors.

See p8-12 of Primer on Geometric Algebra for introductory mathematics and physics by David Hestenes for a very quick 'lecture notes' sketch, or search for 'Clifford algebra' and 'Geometric Algebra'.

Answer 4

Lets approach the problem from a purely geometric perspective and draw analogies with $\mathbb{R}$ to motivate our choices. Towards this end since addition and multiplication are binary operations it's going to be useful to fix one argument and let the other act on it by that operation. This distinction will merely be notational but it can be useful to keep the two ideas separated, they'll just be points on a line or in the plane. I'll use Latin letters for the fixed element and Greek for the actions so that $x+\alpha$ or $\alpha x$ can be interpreted as $\alpha$ acting on $x$ by addition and multiplication respectively.

So geometrically, what do the actions do? Addition in $\mathbb{R}$ will translate each point on the line by $\alpha$ which also encodes directional information. This generalizes readily to $\mathbb{R}^2$ by adding componentwise which also encodes translations.

However we start to run into problems if we multiply componentwise because we get zero-divisors. For example $(1,0)(0,1)=(0,0)$ even though neither of $(1,0)$ or $(0,1)$ are zero which causes problems since zero-divisors don't have multiplicative inverses. You can see pretty easily that finding an number $(a,b)$ so that $(1,0)(a,b)=(1,1)$.

So instead lets look for geometric intuition. In $\mathbb{R}$ if we consider $\alpha x$ then $x$ will be scaled by $\vert \alpha \vert$ but will also change direction if $\alpha < 0$. We can either consider that change of direction a reflection or a rotation by a half turn. Reflections come with a problem however since they reflect about line so we have to specialize some axis rather than just the origin which rotation can be done about the origin. This means rotations requires us to choose less stuff.

So again we have to define $\vert \alpha \vert$ in a way so that the scaling component is $1$ which in $\mathbb{R}$ was just $\pm 1$ which also have the property that the Euclidean distance from the origin to them is $1$. Using the same principle we now recognize that the unit circle should all scale the distance by $1$. In particular $(0,1)$ will be on the unit circle and is conveniently a quarter turn anticlockwise from $(1,0)$. Since we can write $(a,b)=a(1,0) + b(0,1)$ we can now write any point in the plane using these two points. For simplicity lets call $a$ the real part and $b$ the imaginary part so that we can write this as $a+bi$.

Now also want the encode the rotation data which should be a quarter turn anticlockwise for $i$. Since we're interested in the action $\alpha x$ we want $i$ to rotate itself by a quarter turn as well, which would mean $i^2=-1$.

But now we're done. We're at the complex numbers and we have all the algebraic machinery necessary to prove that it's a field. Note the we could have examined how rotation matrices act on vectors in $\mathbb{R}^2$ for another way to arrive at the same structure.

And if I could add one thing, I don't call them the complex numbers anymore. I call them the circle numbers. They encode the radius and the angle of points on a circle. You can rotate things with them. It's really a geometric playground hidden behind some algebra.

Answer 5

You can define geometric rotation in $ \mathbb{R}^{2} $ like this:

$$ \begin{align*} \operatorname{cis}(\theta) &= \cos(\theta) + \sin(\theta) \cdot \mathrm{i}\\ (a_{1} + b_{1} \cdot \mathrm{i}) (a_{2} + b_{2} \cdot \mathrm{i}) &= (\sqrt{a_{1}^{2} + b_{1}^{2}} \cdot \operatorname{cis}(\arg(a_{1} + b_{1} \cdot \mathrm{i}))) \cdot (\sqrt{a_{2}^{2} + b_{2}^{2}} \cdot \operatorname{cis}(\arg(a_{1} + b_{1} \cdot \mathrm{i}))\\ (a_{1} + b_{1} \cdot \mathrm{i}) (a_{2} + b_{2} \cdot \mathrm{i}) &= \sqrt{(a_{1}^{2} + b_{1}^{2}) \cdot (a_{2}^{2} + b_{2}^{2})} \cdot \operatorname{cis}(\arg((a_{1} + a_{2}) + (b_{1} + b_{2}) \cdot \mathrm{i}))\\ &\Rightarrow (r_{1} \cdot \operatorname{cis}(\theta_{1})) \cdot (r_{2} \cdot \operatorname{cis}(\theta_{2})\\ &\Rightarrow (r_{1} \cdot r_{2}) \cdot \operatorname{cis}(\theta_{1} + \theta_{2})\\ \end{align*} $$

You can also represent complex numbers as versors (similar to vectors): $$ \begin{align} x + y \cdot \mathrm{i} &\widehat{=} \left[\begin{matrix} x\\ y \end{matrix}\right]\\ w + x \cdot \mathrm{i} + y \cdot \mathrm{j} + z \cdot \mathrm{k} &\widehat{=} \left[\begin{matrix} w\\ x\\ y\\ z \end{matrix}\right]\\ \\ \text{quaternion: }\mathrm{i}^{2} = \mathrm{j}^{2} = \mathrm{k}^{2} = \mathrm{i} \cdot \mathrm{j} \cdot\mathrm{k} &= -1 \end{align} $$

You also can show it with Brahmagupta–Fibonacci identity or Euler's four-square identity: $$ \begin{align} \left(a^2 + b^2\right)\left(c^2 + d^2\right) & {}= \left(ac-bd\right)^2 + \left(ad+bc\right)^2 & & (1) \\ & {}= \left(ac+bd\right)^2 + \left(ad-bc\right)^2. & & (2) \end{align} $$ $$ \begin{align} \left(a_1^2+a_2^2+a_3^2+a_4^2\right)\left(b_1^2+b_2^2+b_3^2+b_4^2\right) = {} &\left(a_1 b_1 - a_2 b_2 - a_3 b_3 - a_4 b_4\right)^2 \\ &+ \left(a_1 b_2 + a_2 b_1 + a_3 b_4 - a_4 b_3\right)^2 \\ &+ \left(a_1 b_3 - a_2 b_4 + a_3 b_1 + a_4 b_2\right)^2 \\ &+ \left(a_1 b_4 + a_2 b_3 - a_3 b_2 + a_4 b_1\right)^2. \end{align} $$ While the Brahmagupta–Fibonacci identity has a direct relationship to the multiplication of complex numbers, i.e. in $ \mathbb{R}^{2} $, the Euler's four-square identity has a direct relationship to the quaternions (an extension of the complex numbers), i.e. in $ \mathbb{R}^{4} $ and $ \mathbb{R}^{2} $.

Answer 6

Let $z=a+ib \neq 0$.

• there is one and only one direct similarity of Euclidean $\mathbb{R}^2$ which transforms $(1,0)$ into $z$: it is the similarity whose matrix in the canonical base of $\mathbb{R}^2$ is $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$.
• Let $z'=c+id$. $zz'$ is then defined as the image of $(1,0)$ by $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}.\begin{bmatrix} c & -d \\ d & c \end{bmatrix}$ which justifies the definition $(a+ib).(c+id):= (ac-bd)+i(bc+ad)$.

I hope not to betray his thought by stating that this is how Jean Dieudonné constructs the multiplication of complex numbers in $\mathbb{C}$, after a rigorous study of the similarities in $\mathbb{R}^2$. You can find this construction in his book Algèbre linéaire et géométrie élémentaire.