Can anything be proven about this complex variant of the Collatz problem, or is it just as intractable? Given a Gaussian integer $z = a + bi$, where $a, b \in \mathbb{Z}$, $i = \sqrt{-1}$, iterate the function $$f(z) = \frac{z}{1 + i}$$ if $z$ has even Gaussian norm (that is, both $a$ and $b$ are odd, or they're both even), otherwise $f(z) = 3z + i$.
I conjecture that iterating this function eventually leads, if not to $1$, to one of the other Gaussian units ($-1, i, -i$).
For example, starting with $z = 14$, we get $$7 - 7i, -7i, -22i, -11 - 11i, -11, -33 + i, -16 + 17i, \ldots$$
(this is wrong, see edit below)
I have tried a few different values of $z$ with small norm, some purely real, with pencil and paper, haven't gotten far. Also I have tried it in Mathematica, but either I've made some mistakes in my programming that crash the program, or there really are a lot of values that escape to some infinity.
Surely someone else has studied this variant? If so, have they been able to determine anything (like finding a periodic orbit that doesn't include any units)?
EDIT: I made a mistake. Mathematica did save my notebook file at some point before it crashed, and I could have gotten the correct sequence from there instead of having to recalculate it anew. It should go like this: $$7 - 7i, -7i, -20i, 10 + 10i, 10, 5 - 5i, -5i, -14i, -7 - 7i, -7, -20, \ldots$$
Thanks to Mr. Cortek for pointing this out.
 A: The hope here is apparently to find that all Gaussian integers eventually lead to a number of the form $(1 + i)^n u$, where $n$ is a positive real integer and $u$ is a Gaussian unit.
Looking at a list of powers of $1 + i$, it seems to me like they take one of these forms: $\pm 2^n$, $\pm 2^n i$, $\pm 2^n \pm 2^n i$.
Then it's definitely possible for, say, $-3i$ to be followed by $-8i$ and form there it's a spiral drop to 1.
But when the real and imaginary parts are both nonzero, we run into the problem that $3z + i$ affects the real and imaginary parts unevenly.
For example, $5 + 5i$ multiplied by 3 is $15 + 15i$, but then you're only adding $i$ and so $15 + 16i$ falls short of $16 + 16i$ by 1.
Since $\mathbb Z[i]$ is a unique factorization domain and 3 is prime, a number with real and imaginary parts both nonzero is divisible by 3 only if each part is itself also divisible by 3.
Hence $3z$ fails to lead us to a number of the form $2^n + (2^n - 1)i$. A similar problem occurs with $3z + 1$ instead of $3z + i$.
A: Not a full answer, just some remarks and considerations.
In the norm language we can see that division of $z$ with $1+i$ produces some complex number $z^*$  that has the norm smaller by a factor of $\sqrt{2}$, and $3z+1$ approximately triples the norm.
Now since ${(\sqrt{2}})^3 \approx 2.8<3$ we see that for every $z \in \mathbb Z$ his associated Collatzian sequence must be generated in such a way that, on average, on every appliance of a function $f(z)=3z+1$ there are more than three appliances of a function $f(z)=\frac {z}{1+i}$ if we want that all subsequences of a Collatzian sequence associated to $z$ do not escape to infinity.
But this seems very unlikely to ever happen because there seems to be no reason that division $\frac {z}{1+i}$ will favour numbers of the form $O+Oi$ and $E+Ei$ quite more often than numbers $E+Oi$ and $O+Ei$, where $E$ stands for even and $O$ for odd integers.
A: The Collatz conjecture works heuristically because the map is contracting on average (to prove this it is easier to use $x \mapsto (3x+1)/2$ when $x$ is odd)
For your procedure this is no longer the case because dividing by $1+i$ divides the modulus by $\sqrt 2$. 
If $z$ is "odd" then you multiply it by approximately $3/ \sqrt 2$ in two steps, and if it is "even" you divide it by $\sqrt 2$ in one step, so on average, you multiply a number by $3/2$ in $3$ steps, which means a multiplication by $(3/2)^{1/3}$ per step. This is greater than $1$ so heuristically, most of the time the sequence should diverge to infinity.
For example if you start at $1+2i$, after 200 steps you are at $-673097903812-335968886130i$, whose norm is approximately $7.523 \times 10^{11}$ and it doesn't look like it wants to get back to small numbers.
For comparison with the heuristic, $\sqrt 5 \times (3/2)^{200/3} \approx 1.227 \times 10^ {12}$, which is decently close.
