TITLE: Are solutions $z$ of $Im(z^z)=0$ dense in $\mathbb{C}$?
QUESTION [8 upvotes]: Inspired by this question...
More specifically, say $z$ is a solution of $Im(z^z)=0$ if there exist real numbers $\theta$, $r>0$, and integer $n$ such that $z=r e^{i\theta}$ and
$$
r (\ln r \sin \theta + \theta \cos \theta) = n \pi
$$
The left hand side is $Im(z \ln z)$ for some branch of $\ln z$ (not necessarily the principal one).
Is the set of such $z$ dense in $\mathbb{C}$?

REPLY [5 votes]: Yes, it is dense. Fix $r > 0$ and let $f(\theta) = r (\ln(r) \sin(\theta) + \theta \cos(\theta))$.  Given an interval $\theta_0 < \theta < \theta_0 + \epsilon$, 
the same $z = r e^{i\theta}$ are obtained when replacing $\theta$ by $\theta + 2 \pi m$ for positive integer $m$.  But $f(\theta + 2 \pi m) = f(\theta) + 2 r \pi m \cos(\theta)$
As $\theta$ runs over the interval $\theta_0 < \theta < \theta_0 + \epsilon$, 
$f(\theta)$ runs over an interval of some length $\delta_1>0$, while $2 r \pi \cos(\theta)$ runs over an interval of some length $\delta_2 > 0$.  Then 
$f(\theta) + 2 r \pi m \cos(\theta)$ runs over an interval of length at least
$m \delta_2 - 2 \delta_1$.  If $m$ is sufficiently large, this length will be greater than $\pi$, so that it includes some integer multiple of $\pi$.
We conclude that every  arc $r e^{i\theta}$, $\theta_0 < \theta < \theta_0 + \epsilon$ contains solutions of $\text{Im}(z^z) = 0$.