Talk:Complex-base system

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There is a discussion about the use of this idea in image compression. [1]

I preserved the phrase "is one of periodic cases" – but what does it mean? —Tamfang (talk) 00:32, 5 December 2008 (UTC)Reply

Seeing no clarification, I'll remove the phrase but keep the footnote. —Tamfang (talk) 04:45, 16 December 2009 (UTC)Reply

Listed below are those of the system (as a special case shown above systems) and shows code numbers 2, -2, -1.

Whatever language this came from, it's not clear in English. Can someone improve it? —Tamfang (talk) 17:38, 3 May 2011 (UTC)Reply

General notation:
Instead of:

${\displaystyle <\rho =\pm i{\sqrt {2}},[0,1]>}$

I would like to propose the notation:

${\displaystyle {\bigg \langle }\rho =\pm i{\sqrt {2}},\{0,1\}{\bigg \rangle }}$.

The ${\displaystyle \{0,1\}}$ because it is a set, so that one can say: set of digits ${\displaystyle D:=\{0,1\}}$ and ${\displaystyle {\big \langle }\rho =\pm i{\sqrt {2}},D{\big \rangle }}$, and the ${\displaystyle {\big \langle }...{\big \rangle }}$ because of optics.

Text below:
Shouldn't one split up

• ${\displaystyle <\rho ={\sqrt {2}}e^{\pm i\pi /2},B_{2}>}$, example, ${\displaystyle <\rho =[-1\pm i],[0,1]>}$ (see also section "Base −1±i" below).

into

• ${\displaystyle {\bigg \langle }\rho ={\sqrt {2}}e^{\pm i\pi /2},B_{2}{\bigg \rangle }}$, example, ${\displaystyle {\big \langle }\rho =(\pm i{\sqrt {2}}),\{0,1\}{\big \rangle }}$ .

and

• ${\displaystyle {\bigg \langle }\rho ={\sqrt {2}}e^{\pm 3i\pi /4},B_{2}{\bigg \rangle }}$, example, ${\displaystyle {\big \langle }\rho =(-1\pm i),\{0,1\}{\big \rangle }}$ (see also section "Base −1±i" below). -- Nomen4Omen (talk) 11:42, 2 August 2011 (UTC)Reply

The inline reference to Penney's paper was wrong - it linked to some other paper. Unfortunately the system for inline references on Wikipedia is complicated enough that I don't have time to figure out how it works! So I added the bibliographical information on Penney's paper to the 'References' section in a very crude way. I hope someone can fix things up. John Baez (talk) 16:07, 8 November 2012 (UTC)Reply

I've experimented with mixing the radices -1+i and -1-i. If they alternate, the rounding domain is a parallelogram with vertices at 0, 1, i, -1+i (or the conjugates); other sequences give other pretty shapes. —Tamfang (talk) 03:46, 3 March 2019 (UTC)Reply

You mean radices alternating between -1+i and -1-i ?

Isn't a0*(-1+i)+a1*(-1+i)*(-1-i) = a0*(-1+i)+a1*2 very close to ${\displaystyle {\big \langle }2i,\{0,-1+i,2,1+i\}{\big \rangle }}$ ? Which looks almost like the Quater-imaginary base system ${\displaystyle {\big \langle }2i,\{0,1,2,3\}{\big \rangle }}$ ? --Nomen4Omen (talk) 11:04, 3 March 2019 (UTC)Reply

There may be a place for this – with a better caption – but not filling the top of the page. —Tamfang (talk) 06:14, 4 May 2024 (UTC)Reply

The rounding region of an integer – i.e., a set of complex (non-integer) numbers that share the integer part of their representation in this system – has in the complex plane a fractal shape: the twindragon.

I would call it the truncation region. Rounding, it seems to me, is an arithmetic operation independent of base. —Tamfang (talk) 18:38, 1 July 2024 (UTC)Reply