<s>[INST] Motivation behind the definition of multiplication in $ \mathbb{R}^2 $ 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.
 [/INST] 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} $.
</s>