Let $z=\operatorname{cis}\theta \in \mathbb{T}$ with $\theta \in \mathbb{Q}$. Show that the order of $z$ is infinite.

Question

I am stuck on the following question:

Let $z=\operatorname{cis}\theta \in \mathbb{T}$ with $\theta \in \mathbb{Q}$. Show that the order of $z$ is infinite.

I am trying to understand Daniel Robert-Nicoud's answer. Why is $\pi$ chosen and why not $0$? Are they saying that if $z$ has finite order, then there exists a positive integer $n$ such that $z^{n}=1$? If so, why is $n$ made rational and not a whole number, where they put $n=\frac{q}{p}$? (if that is what they are doing, I am not sure actually)...

Answer 1

For $u\in \Bbb{C}, e^u=1 \iff u=2k\pi i$ for some $k\in \Bbb{Z}$.

Given $z=e^{i\theta}$, with $\theta\in \Bbb{Q}$.
Suppose $|z|=n$.
Then $e^{in\theta}=1$.
This means that $in\theta=2k\pi i$ for some $k\in \Bbb{Z}-\{0\}$.
Hence $\pi=\frac{n\theta}{2k}\in \Bbb{Q}$, which is a contradiction.