Having trouble understanding generalized complex numbers

Question

I'm reading a paper on the generalized complex numbers, but I have trouble in some of its fundamental properties. I have searched wiki but it left me none the wiser. Please see the image below, in which I have underlined the "dual numbers" with red because that's where I feel especially confused.

My problems are:
1).The definition of the imaginary unit $i$ as $i^2=iq+p$ doesn't make me comfortable. Because I think it is a recursive one. Is there any other way to define $i$ equivalently but without recursion?
2).Why are all "dual numbers" located on the parabola: $p+\frac14q^2=0$? How to show it is equivalent to the definition "$i^2=0$"? I have tried to derive it but I come to nothing. I don't think it should be very hard but I just fail do it.

Could anyone give me some explanation or hint that is not beyond my limit? (I'm a freshman and so far I have learned a little elementary calculus and linear algebra) Thanks in advance!

Answer 1

1) The equation $i^2=iq+p$ is not a recursive definition. It is defined as a solution of a quadratic equation. It is just like our normal definition $i^2=-1$. In this definition you can solve $i$ to get the two solutions, just like when you "solve" $i^2=-1$, you get our ordinary definition of $i$.

2) This is to say, when you solve $i^2=iq+p$, i.e., $i^2-qi-p=0$, you get a repeated root when the discriminant is $0$. It is like the case $i^2=0$, where you have a repeated root $0$.

Answer 2

I think the generalized complex number systems can simply be defined as rings of the form $\mathbb R[X]/(f(X))$ with $f(X)$ a real quadratic polynomial and $i$ is a notation of $\overline X$. For the meaning of this symbols maybe you can refer to wiki.

Answer 3

I think the article you quote is using slightly facetious language in order to seem more interesting. In particular, when it seems to frame its question as "what are some other definitions of $i$?", it would be more honest to say

What are the ways to define a multiplication operation on $\mathbb R^2$ which is compatible with normal vector addition, in a way such that there's a subspace of $\mathbb R^2$ that behaves like $\mathbb R$ with its usual operations?

That's not as immediately attention-catching, of course. And it would still be implicit that they want the result, in particular to be a commutative ring. (Or better yet a commutative unital $\mathbb R$-algebra, but you probably don't know what that is, so just ignore that).

Once you have decided to answer this, a natural first step is to choose a coordinate system such that the subspace that behaves like $\mathbb R$ is the $x$-axis and $x\in\mathbb R$ corresponds to $(x,0)\in\mathbb R^2$. This fixes how our multiplication is going to behave on pairs of this form, and if we also want multiplication with the "real" subset to correspond to scalar multiplication of vectors, we're forced to define $$ (x,0)\cdot(a,b)=(a,b)\cdot(x,0)=(ax,bx) $$

It now turns out that the only thing we still need to choose is what $(0,1)\cdot(0,1)$ will be. Once we've chosen $(0,1)\cdot(0,1)=(p,q)$, the associative and distributive laws forces the product of two arbitrary pairs to be $$ (a,b)\cdot(c,d) = (ac+bdp, bdq+ad+bc) $$

The complex numbers result when we choose $(p,q)=(-1,0)$, and in that case we call the pair $(0,1)$ $i$. The authors of your article extends that naming to systems where $(p,q)\ne (-1,0)$, which is confusing and nonstandard if you ask me, but there you go.

It is still the case, as for the complex numbers that $$ (a,b) = (a,0)+(0,b) = (a,0)+(b,0)\cdot(0,1) = a+bi $$ for all $a$ and $b$, but the rule for multiplying things differs from that of the complex numbers.

When the authors say $i^2=p+qi$, that's merely their notation for specifying $(0,1)\cdot(0,1)=(p,q)$.

The point of the claim in Figure 1 is that there are only three fundamentally different choices for $(p,q)$, namely $(-1,0)$ (the complex numbers), $(0,0)$ (dual numbers), or $(1,0)$ (double numbers). For any other choice of $(p,q)$ the resulting number system turns out to be isomorphic to one of these three cases -- or in other words, it is the same as one of these three cases, just viewed through a different coordinate system (which by the authors' convention means that it's a different member of the ring that gets to be called $i$).

(It's a particularly confusing feature that Figure 1 shows the $p$ axis going up, when $p$ is the real part of "$i^2$", which ought to go right in the usual diagrams of the complex plane. The figure really ought to be flipped around the diagonal $p=q$).