Complex Numbers $\stackrel{?}{=} \mathbb{R}^ 2$ Suppose we have a vector field over real numbers $\mathbb R^2$. In additon to vector field proporties define inner product $(x,y) = x_1\cdot y_1 + x_2\cdot y_2$, where $x_1,x_2,y_1,y_2$ are real numbers.This structure is called $2$-dimensional Euclidean Space. ( call it $\text{E-2}$ ).
I want to define complex number field from $\text{E-2}$. I have to satify $5$ addition axiom, $5$ multiplication axiom and a distribution axiom. The problem is although addition axioms are trivial, for multiplication I have to define; $xy =   (ac-bd,ad+bc)$ where $x = ( a,b )$, $y = (c,d)$ as an extra axiom.
My questions are:  


*

*Is there any natural way to define complex numbers without such
additional axioms, by using inner product ?

*By defining complex numbers from $\text{E-2}$, there is no need to use inner
product between two numbers, this bothers me as well, am I missing
something ?

 A: I will denote the (standard) inner product by $\langle\cdot,\cdot\rangle$ and consider the elements of $\Bbb R^2$ as column vectors. Put $$A=\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$$ and $$B=\begin{bmatrix}0 & -1\\1 & 0\end{bmatrix}.$$
We define an operation $\otimes$ on $\Bbb R^2$ by $$\vec w\otimes\vec z:=\langle\vec w,A\vec z\rangle\vec e_1+ \langle\vec w,BA\vec z\rangle\vec e_2,$$ where $$\vec e_1=\begin{bmatrix}1\\0\end{bmatrix}\text{  and  }\vec e_2=\begin{bmatrix}0\\1\end{bmatrix}.$$
It's rather contrived, but it does the job. Note that left-multiplication by $A$ is the operation of complex conjugation, that $\vec e_1$ is the $\otimes$-identity, and that left-multiplication by $B$ is the same as $\otimes$-multiplication by $\vec e_2,$ which serves as our "$i$".

Let me see if I can shed some light on the connection between inner product and complex multiplication (and how I developed the peculiar operation above).
First, I observed that, if there was some function $f:\Bbb C^2\to\Bbb R$ satisfying $$f(a+bi,c+di)=ac+bd$$ for all $a,b,c,d\in\Bbb R$ (that is, if $f$ acts like our inner product), then $$w\overline z=(w_1+iw_2)(z_1-iz_2)=w_1z_1+w_2z_2+i(w_2z_1-w_1z_2)=f(w,z)+i(w_2z_1-w_1z_2)$$ and similarly $$\overline wz=f(w,z)+i(w_1z_2-w_2z_1),$$ whence we find that $$f(w,z)=\frac12\left(w\overline z+\overline wz\right).$$ Readily, then, we have that $$f(w,\overline z)=\frac12\left(wz+\overline w\overline z\right),$$ and (a bit less obviously) that $$f(w,i\overline z)=\frac12\left(-iwz+i\overline w\overline z\right)=\frac1{2i}\left(wz-\overline w\overline z\right).$$ Therefore we have that $$f(w,\overline z)\cdot 1+f(w,i\overline z)\cdot i=wz.\tag{$\star$}$$ All that remains, then, is to decide what the equivalents to $1$ and $i$ would be in $\Bbb R^2$ (which was fairly natural), and how to express conjugation and multiplication by $i$ as linear transformations on $\Bbb R^2$ (not too tricky).

Added: We have some freedom in the representatives we can choose, but we can't just pick any two vectors. Let's call our vector representatives of $1$ and $i$ by the names $\vec 1$ and $\vec i,$ respectively. Let's first find some necessary conditions for them.

First of all, note that we obviously need $\vec 1$ and $\vec i$ to be linearly independent. Assume for the moment that there exist $2\times 2$ real matrices $C$ and $R$ corresponding to complex conjugation and multiplication by $i$ (the latter of which rotates the plane). In particular, this means that the following hold:

*

*$C\vec 1=\vec 1$

*$R\vec 1=\vec i$
By our work in the above section--in particular, by translating $(\star)$ into the desired terms--we find that we need $$\langle\vec w,C\vec z\rangle\vec 1+\langle\vec w,RC\vec z\rangle\vec i=\vec w\otimes\vec z\tag{$\heartsuit$}$$ for all $\vec w,\vec z\in\Bbb R^2.$
Now, since $\vec 1$ needs to be our $\otimes$-identity, then in particular, $$\vec 1=\vec1\otimes\vec1=\langle\vec 1,C\vec 1\rangle\vec1+\langle\vec1,RC\vec1\rangle\vec i=\langle\vec1,\vec1\rangle\vec1+\langle\vec1,R\vec1\rangle\vec i=\langle\vec1,\vec1\rangle\vec1+\langle\vec1,\vec i\rangle\vec i,$$ whence linear independence shows us that $\langle\vec 1,\vec1\rangle=1$ and $\langle\vec1,\vec i\rangle=0.$ Another property we need is that $\vec1=-\vec i\otimes\vec i,$ from which we can further deduce that $\langle\vec i,\vec i\rangle=1.$ Therefore, we require that $\left\{\vec1,\vec i\right\}$ be an orthonormal basis for $\Bbb R^2.$
From this, we can conclude that $$\vec1=\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$$ for some $\theta\in\Bbb R.$ Now, having chosen such a vector, we have two choices for $\vec i.$ (Can you see why, and what they are?) This might seem a bit alarming, but many constructions of $\Bbb C$ run into precisely this same problem, as discussed for example here and here. Regardless, it turns out that we must have either $R=B$ or $R=-B$ (with $B$ as given earlier), which depends on $\vec1$ and on which of the two options we choose for $\vec i.$
One final observation I will make (omitting the proof) is that we require $$C=\begin{bmatrix}\cos2\theta & \sin2\theta\\\sin2\theta & -\cos2\theta\end{bmatrix}.$$ In the case that $\theta$ is an integer multiple of $2\pi$--that is, that $\vec1=\vec e_1$--we have $C=A.$

I leave it to you to verify that the necessary conditions derived above are also sufficient to make $\otimes$ behave on $\Bbb R^2$ exactly the way we want complex multiplication to work. Observe, though, that there are uncountably-many different ways that we could define $\otimes,$ depending on our choice of $\theta$ and subsequent choice of $\vec i.$ The definition I gave above is about as close to canonical as you can get.
