distribution of gaussian primes here is a naive question that so far I don't have already found somewhere else.
In the following, I consider the norm on gaussian integers with $N(a+ib)=a^2+b^2$.
Consider prime gaussian integers whose norm is $>2$. For any of them, say $p$, find the unique integer $k$ such that $1<N(p (1+i)^{-k})<2$. Set $p'$ as this element. Is it true that the set of all elements $p'$ is dense in the set of complex numbers such that $1<N(z)<2$?
(up to conjugacy and multiplication by $i$, one may consider only the eighth part of the plane such that $0<Im(z)<Re(z)$ for instance, this does not change anything, it just adds uniqueness of representation of primes and their image $p'$).
Thanks in advance for any comment!
 A: Yes, it is true - or rather, it's almost surely possible to prove that it's true.
Consider a small open set $U$ in your annulus $\{z\colon 1<|z|<2\}$, and define
$$
V = \bigcup_{k=1}^\infty (1+i)^k U,
$$
the set of all points in $\Bbb C$ that map into $U$ under the operation you define. We want to be assured that some Gaussian prime lies in $V$.
In fact, one should be able to prove that there are infinitely many Gaussian primes in $V$. More specifically, let $\alpha$ be the ratio of the area of $U$ to $3\pi^2$ (the area of the annulus); then the set $V$ consists of "a positive proportion $\alpha$ of $\Bbb C$", no matter how you might define the notion of the relative area of an unbounded open set relative to all of $\Bbb C$. One should be able to prove that the number of Gaussian primes in $V$ with modulus at most $x$, say, is asymptotically $\alpha$ times the number of Gaussian primes with modulus less than $x$.
One example of this can be found in Harman and Lewis, "Gaussian primes in narrow sectors". In fact they treat a gradually narrowing set ("$\alpha\to0$ slowly enough"), so the case of a fixed positive $\alpha$ should be even easier.
So the bottom line is: almost surely this can be done.
