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Suppose that there is number $a+jb$ where $j^2=1$ and the whole number is split-complex number. We want to set this number to satisfy the following:

A) $(a+jb)(a+jb) = k(c+jd)$ where $k$ is fixed integer

B) for any natural number $z$, there exists $z$ split-complex numbers that satisfy A) with same $k$, but for two such different numbers, $(a+jb)(c+jd) \neq k(e+jf)$ where $e+jf$ is any number.

Is this possible?

Anixx
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user72658
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    What is a "split complex number" for you? What have you done, how far did you get trying to solve this problem? – DonAntonio Apr 15 '13 at 10:31
  • http://en.wikipedia.org/wiki/Split-complex_number so it is something formal. – user72658 Apr 15 '13 at 11:55
  • I tried to expand the whole multiplication use several number theory rules regarding multiples of $k$, modulo, etc. But didn't work well to decide this decisively. – user72658 Apr 15 '13 at 11:56
  • Isn't at A) you rather want $(a+jb)^2=k\cdot (a+jb)$ instead of $k\cdot (c+jd)$? Also, in B), are we talking about a set $S$ of $z$ pieces of these "split-complex numbers", and I guess, $e+jf\in S$ is meant, isn't it? – Berci Apr 15 '13 at 13:16
  • user72658: please spend some time trying to clarify these conditions. Item A) seems already to be pretty good, but it's a little hard to tell what you mean in Item B). – rschwieb Apr 15 '13 at 13:28
  • @rschwieb: Please share us how do you interpret item A)? Are $a,b,c,d$ integers? I guess not, and $c+jd$ wants to be $a+jb$ on the RHS. – Berci Apr 15 '13 at 13:43
  • @Berci I was interpreting $a,b,c,d$ as used in the link the OP gave: real numbers. I am probably wrong though since he did put number-theory in the tags :) – rschwieb Apr 15 '13 at 14:22
  • So.. Isn't always $(a+jb)^2=k\cdot(c+jd)$ for some real numbers $c,d$ and $k=1$? If it's not $a+jb$ on the right hand side, I can't interpret the question... – Berci Apr 15 '13 at 16:42

1 Answers1

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Let $s:=1+j$. Then $s^2=2+2j=2s$. So, for a $t\in\Bbb R$, we have $$(t\cdot s)^2=t^2\cdot 2s=2t\,(t\cdot s)\,.$$ This answers A): for a fixed $k$, let $t:=k/2$, and consider $t\cdot s$, i.e. $\displaystyle\frac{k+kj}2$.

For B), let's look for all eigenvectors of this "split-complex" squaring. We already know that for $u:=s/2=\displaystyle\frac{1+j}2$ we have $u^2=u$, and similarly, $v:=\displaystyle\frac{1-j}2$, that gives $v^2=v$. We also have $$u\cdot v=v\cdot u=1/4\cdot(1+j)(1-j)=1/4\cdot(1^2-j^2)=0\,,$$ and that every split-complex number $a+jb$ can be expressed as $c\,u+d\,v$ (with $c=a+b$ and $d=a-b$). Now we have $$(c\,u+d\,v)^2=c^2\,u+d^2\,v$$ so, in order that A) be satisfied ($(cu+dv)^2=k\,(cu+dv)$), we need $c^2=k\cdot c$ and $d^2=k\cdot d$. That is, either $c=0$ or $c=k$, and the same for $d$: either $d=0$ or $d=k$. It gives all in all $4$ such numbers ($0,k,ku,kv$), but $0$ should be excluded if $z>1$ because of the hypothesis. So, $z$ can be at most $3$, and with $z=3$, $\ S:=\{k,\,ku,\,kv\}$ satisfies the condition: $k\cdot ku=k^2u\notin S$, $\ ku\cdot kv=0\notin S$...

Berci
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