TITLE: Can the argument of an algebraic number be an irrational number times pi?
QUESTION [8 upvotes]: This is mainly out of curiosity.  Let $\nu$ be an algebraic number.  Can Arg($\nu$) be of the form $\pi \times \mu$ for an irrational number $\mu$?

REPLY [2 votes]: To extend @ChrisEagle’s answer without using Niven, let me point out that any nonreal Gaussian integer off the lines $y=\pm x$ will give an example of $z$ with $\arg z=\lambda\pi$ and $\lambda$ irrational, because of unique factorization in the Gaussian integers. For, notice that the $\lambda$ above is rational if and only if $z^m$ is real for some positive integer $m$. And yet a Gaussian integer is real if and only if it is of the form
$$
2^r\prod_jq_j\prod_k(\rho_k\bar\rho_k)\,,
$$
where the $q_j$ are ordinary primes congruent to $3$ modulo $4$ and each $\rho_k\bar\rho_k$ is an ordinary prime congrent to $1$ mod $4$, just as $13=(3+2i)(3-2i)$. Note that $\rho_k$ and $\bar\rho_k$ are different Gaussian primes, i.e. not related by a unit factor. So take your $z$ satisfying the conditions I specified at the outset; then its factorization into primes must involve at least one $\rho$ that’s not matched by a $\bar\rho$, and every power of $z$ satisfies this condition as well, so is not real.