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I have a definition of $i$ (here) that has come from observing Newton-Raphson (NR) on $(x^2+c)=0$. The attractors when plotting NR were $\pm \sqrt{c}$, which happen to be the coefficients of $i$ in the solution. Since neither attractor was dominant, the interpreted solution required the simultaneous implementation of these two real numbers $\pm \sqrt{c}$. $i$ therefore permits us to use two real numbers simultaneously.

For example, the simultaneous real numbers associated with:

$i \to [-1,1]$
$(-i) \to [1,-1]$
$2i \to [-2,2]$
$(8+5i) \to [3,13]$
$(-2-3i) \to [1,-5]$

and have the form $[(real - imaginary) , (real + imaginary)]$ for $+i$. $-i$ simulatneous reals are in descending order.

Using this definition of $i$ we are able to perform all mathematical operations on complex numbers using just real numbers. Further supporting arguments of this definition are given by factorization properties and complex step differentiation examples.

I would appreciate any feedback, suggestions to further reading on definitions of $i$ (aside from its square equals -1 property), or on the notion of simultaneous implementation of numbers.

Fred Viole
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    Have you seen the ring-theoretic approach? i.e. take the ring $\mathbb{R}[x]$ and quotient by the maximal ideal $\langle x^2+1 \rangle$? – Sam Weatherhog Sep 23 '15 at 05:39

2 Answers2

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A simpler approach than even the ring theoretic one is to just take matrices of the form:

$$ \left( \begin{matrix} a&-b \\ b&a \end{matrix} \right)\mapsto a+bi,\text{ }a,b\in\mathbb{R} $$

These matrices add and multiply exactly the same way as the complex numbers and: $$ \left( \begin{matrix} 0&-1 \\ 1&0 \end{matrix} \right)\mapsto i $$ Note that: $$ \left( \begin{matrix} 0&-1 \\ 1&0 \end{matrix} \right)^2=\left( \begin{matrix} -1&0\\0&-1 \end{matrix} \right)=-I $$

where $I$ is the identity matrix (corresponding to the real number 1).

Sam Weatherhog
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Your way may work, but it's definitely not the normal way. The normal way is to introduce calculation rules for pairs of reals:

$(x,y) + (z,w) = (x+z, y+w)$

$c + (x, y) = (c+x, y)$

$(x,y)(z,w) = (xz-yw, xw+yz)$

$c(x,y) = (cx,cy)$

etc...

Then we just call the pair $(0,1)$ for $j$ and note that pairs of the form $(c,0)$ can be identified as real numbers (and then instead of the above notation we can start using the notation $x+iy$ instead).

skyking
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