It is easy to check that every Gaussian integer can be written uniquely as a finite sum of the form $\sum_{n\geq 0}\epsilon_n(-1+i)^n$ for ‘digits’ $\epsilon_n$ in $\lbrace 0,1\rbrace$.
The sequence $\epsilon_0,\epsilon_1,\dotsc$ can be computed by the algorithm:
Input: z in $\mathbb Z[i]$.
iterate until z=0:
t=Re(z)+Im(z) modulo 2 (value taken in $\lbrace 0,1\rbrace$)
print(t) and replace z by (z-t)/(-1+i):
enditerate
Changing the above pseudocode slightly by stopping (for non-zero $z$ in $\mathbb Z[i]$) when $z$ is among the fourth-roots of unity $\lbrace \pm 1,\pm i\rbrace$ and coloring the initial Gaussian integer with yellow, green, red, blue if the algorithm stops with $1$, $-1$, $i$, $-i$ yields the following picture (for Gaussian integers with real and imaginary values in $\lbrace -100,100\rbrace$):
Striking features (experimentally on the picture):
The picture is fractal in nature (this is more or less obvious from the definition, the picture should be approximately invariant (perhaps up to a colour permutation) under multiplication by $-1+i$).
‘Connected components’ (defined in the obvious way) are bounded except for green (corresponding to $-1$ if I am not mistaken) where there seem to be three infinite components meeting at the origin.
There is a sort of curious symmetry of order $3$ (up to a dilatation).
Are there any nice explanations of these features?
Remark: Changing the picture algorithm slightly by dividing by $(1+i)$ instead of $(-1+i)$ leads to a sort of dual picture:
Colours have a unique ‘connected component’ except for red (corresponding to $i$) which has infinitely many.
Observe that powers of $(1+i)$ yield a slightly less satisfying numeral system: One has to accept leading coefficients in $\lbrace 1,i\rbrace$ in order to get finite sums for all Gaussian integers.
Added elementary explanation and analogue for Eisenstein integers:
$b=-1+i$ defines a nice numeral system, see for example the reference mentionned in the comment of Alison Miller below. Every Gaussian integer can be written uniquely as a finite sum of distinct natural powers of $b$. Colouring a non-zero Gaussian integer $z$ with the first unit-root hit during the computation of '$b$-digits' (in $\lbrace 0,1\rbrace\}$) of $z$ seems like a natural idea and the result depends of course only on the leading digits of the '$b$-expansion' of $z$. More precisely, writing $(\epsilon_n\ldots \epsilon_0)_b$ for the $b$-expansion of $\sum_{j=0}^n\epsilon_j(-1+i)^j$, we have \begin{align*} 1&=1_b,\\ i&=11_b,\\-i&=111_b,\\ -1&=11101_b. \end{align*} The colour of the non-zero Gaussian integer encoded by $w=(\epsilon_n\ldots \epsilon_0)_b$ is therefore the colour of $1$ if $w=1$ or $w=10*$ (where $*$ represents an arbitrary finite word with letters $0$ and $1$).
The colour of $w$ is the colour of $i$ if $w=11$ or $w=110*$.
The colour of $w$ is the colour of $-i$ if $w=111$ or $w=1111*$ or $w=11100*$.
The colour of $w$ is the colour of $-1$ in the remaining cases, i.e. if $w=11101*$.
Colours can therefore be computed by a finite state automaton defining a $2$-automatic sequence (after interpreting a $b$-expansion $w\in\{0,1\}^*$ as a binary digit).
There is a similar phenomenon for Eisenstein integers $\mathbb Z[(1+i\sqrt{3})/2]$: A numeral basis with digits in $\{0,1,2\}$ (giving rise to a bijection between Eisenstein integers and $\mathbb N$ by considering ternary expansions) is given by $b=(-3+i\sqrt{3})/2$. Unit-roots have the following $b$-expansions with digits in $\{0,1,2\}$: \begin{align*} 1&=1_b,\\ (1+i\sqrt{3})/2&=12_b,\\ (-1+i\sqrt{3})/2&=11_b,\\ (-1-i\sqrt{3})/2&=121_b,\\ (1-i\sqrt{3})/2&=122_b,\\ -1&=12102_b. \end{align*} One observes that all $b$-expansions of unit roots start with $1$. Colouring Eisenstein integers with the first unit root occuring during the computation of the $b$-digit-expansion fails therefore for Eisenstein integers with leading coefficient $2$ in their $b$-expansion. There are two natural ways to deal with this problem: One can either use an additional seventh colour for such Eisenstein integers or one can colour them with the same colour as $1$. The second solution gives rise to the following picture:
The colour code is:
yellow for $1$ or if no unit root occurs during the computation of the $b$-expansion,
green for $-1$,
red for $(1+i\sqrt{3})/2$,
blue for $-(1+i\sqrt{3})/2$,
white for $(-1+i\sqrt{3})/2$,
black for $(1-i\sqrt{3})/2$.
