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 Author Topic: Multi-dimensional complex plane equivalent ... finally?  (Read 1590 times) Description: I have a 3D complex plane equivalent, can someone make a Mandelbrot with it? 0 Members and 1 Guest are viewing this topic.
truth14ful
Guest
 « on: May 07, 2011, 05:27:30 PM »

An important property of the complex plane is the power property:
$i^1=i,\;i^2=-1,\;i^3=-i,\;i^4=1,\;\cdots$
If the real unit is specified as $j$, then the power property would look like this:
$i^1=i,\;i^2=-j,\;i^3=-i,\;i^4=j,\;\cdots$
And I see no reason why the pattern could not be continued into higher dimensions:
3D: $i^1=i,\;i^2=-j,\;i^3=-k,\;i^4=-i,\;i^5=j,\;i^6=k,\;\cdots$
4D: $i^1=i,\;i^2=-j,\;i^3=-k,\;i^4=-l,\;i^5=-i,\;i^6=j,\;i^7=k,\;i^8=l,\;\cdots$
5D: $i^1=i,\;i^2=-j,\;i^3=-k,\;i^4=-l,\;i^5=-m,\;i^6=-i,\;i^7=j,\;i^8=k,\;i^9=l,\;i^{10}=m,\;\cdots$
etc. So in 3D, with the formula $\left\{ z_n=(z_{n-1})^2+k\;|\;z_0=2i+3j \right\}$,
$z=\{2i+3j,-9i-4j+13k,-250i-185j+242k,\cdots\}$
Just to be sure, for the last step, I calculated $-9i-4j+13k$ as $-9i-4i^5+13i^6$ and $9i^4+4i^2+13i^6$ and got the same answer.