Primes and the Unit circle. Consider the "prime spiral" $f(z) = \sqrt{z}\exp(2\pi i \sqrt{z})$, for integer $z$. It has been shown that the intersections of $f$ with some quadratic curves contain a significantly disproportionate number of primes. Let $P = \{\exp(2\pi i \sqrt{p}) \;:\ p \;prime \}$, the set of unitized prime spiral coordinates.
1) Is $P$ equal to the circle group or is it a proper subset? (note that $P$ is countable)
2) What is the complement of $P$ in the circle group?
3) Is the complement a group?
4) Is the circle group equal to the closure of $P$?
 A: *

*As you note, $P$ is countable. The circle group (by which I take it you mean the set of complex numbers of modulus one) is uncountable. So, $P$ is a proper subset. 

*I don't think there's much you can say about the complement (note spelling), except that it's the complement. 

*No, the complement is not a group. Take any transcendental $\alpha$; then $e^{2\pi i\alpha}$ and $e^{2\pi i(1-\alpha)}$ are in the complement, but their product isn't. 

*Yes. It's well-known, and not hard to prove, that the numbers, fractional part of $\sqrt n$, are dense in the unit interval. 
A: As a complement to Gerry's answer, another straightforward way of showing that $P$ can't be the circle group is to note that $P$ isn't even a group; there is no $n$ with $\sqrt{n}\equiv\sqrt{2}+\sqrt{3}\pmod 1$.  For if there were, then there would be some $m$ with $\sqrt{n}=\sqrt{2}+\sqrt{3}+m$; but then $n=(2+3+m^2)+2m\sqrt{2}+2m\sqrt{3}+2\sqrt{6}$ ; and $\sqrt{2}, \sqrt{3}$ and $\sqrt{6}$ can be easily shown to be linearly independent over the integers.
