TITLE: Can anything be proven about this complex variant of the Collatz problem, or is it just as intractable?
QUESTION [12 upvotes]: 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.

REPLY [4 votes]: 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.