In a paper I came across, I have a complex number $z = \frac{-3+4i}{5}$. Hence $z$ lies on a unit complex circle $\mathbb{T}: = \{z \in \mathbb{C}: |z|=1\}$. It is stated that "$z$ is not a complex root of unity, so its orbit $\{z^{n}: n \in \mathbb{N}\}$ is dense in $\mathbb{T}$ ". This statement is not obvious to me.

(1). How do I show that $z$ is not a complex root of unity? (I assume in this context I need to show that $z^{m} \neq 1$ for any $m$). So this would amount to writing $z$ in polar form $e^{i\theta}$ - but how do I then show that $\theta$ is irrational?

(2). Assuming (1) is true, why does it follow the orbit $\{ z^{n}\}$ is dense?

  • 2
    $\begingroup$ This has been asked before $\endgroup$ – orole Feb 2 '18 at 22:57

A complex root of unity is a complex number $z$ for which for some $n\in\{1,2,3,\ldots\},$ you have $z^n=1.$ That implies $z= e^{i\theta}$ with $n\theta\equiv2\pi\bmod1.$ That means $\theta$ is a rational multiple of $2\pi$, or of $360^\circ$ if you like. According to Niven's theorem, that implies $\sin\theta$ is not rational. But you have sine${}=4/5.$

So $\alpha=\theta/(2\pi)$ is irrational if $e^{i\theta} = (-3/5) + i(4/5).$ That the sequence $(n\alpha\bmod 1)_{n=1}^\infty$ is dense in $[0,1)$ follows from Diophantine approximations.

| cite | improve this answer | |
  • $\begingroup$ Thank you - I was not aware of Niven's Thm. @Michael Hardy $\endgroup$ – Alex Feb 2 '18 at 23:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.