Can we always pick an integer power of $e^{i \theta}$ such that its difference with $1$ has rational argument?

Question

Let $z = e^{i\theta}, \theta \in \mathbb{R}$. Then, does there exist $n \in \mathbb{N}$ such that:

$$1 - z^n = re^{2 \pi i \tau}$$

for some $\tau \in \mathbb{Q}$?

Naturally, this exists if $\theta$ is a rational multiple of $\pi$. However, does this hold for any $\theta$?

Although this question appears quite simple, I have no idea how I would approach it, and I suspect that its proof or disproof would be very difficult.

Link to motivation (it may appear entirely unrelated (it almost is); I wish to show that $\mathbb{C}$ has a certain property that I defined on fields with the addition of some analysis.)

Answer 1

No, on basis of set cardinalities. The possible pairings of $n \in \mathbb{N}$ and $\tau \in \mathbb{Q}$ are countable. Each of those equations has at most $n$ solutions for $z$, so a finite count. So the set of $\theta$ where the problem has any solution is countable. $[0, 2\pi] \subset \mathbb{R}$ is uncountable, so it contains many values of $\theta$ with no solution.

Answer 2

$\arg (1-z^n)=n\theta/2-\pi/2=2\pi \tau+2k\pi$ for some $k \in \mathbb Z$ so $\theta=\frac{4\pi (\tau+k+1/4)}{n}$ is a rational multiple of $\pi$