When is $x^{2^n}$ dense in $\mathbb{S}^1$, for $|x|=1$?

Question

Motivation: So I just saw this question: Limit when $n\rightarrow\infty$ of $\text{sgn}(\sin(2^n \pi x))$ with $x\in(0,1)$ fixed., and the answer involves diadic numbers and things of the kind. Most answers involve some things like the the fractional part of $2^n x$ assuming sufficiently many values in $(0,1)$ when $x$ is non-diadic. A natural question that arises from this is "is the set of remainders of $2^n x$ is dense in $(0,1)$ when $x$ is non-diadic?"

This question admits some (trivially) equivalent forms:

1. If $x$ is non-diadic, is the set of remainders of $2^n x$ dense in $(0,1)$?
2. If $x$ is non-diadic, is $\left\{2^nx+\mathbb{Z}:n=1,2,\ldots\right\}$ dense in $\mathbb{R}/\mathbb{Z}$ (with quotient topology)?
3. If $x$ is non-diadic, is $\left\{e^{i\pi2^n x}:n=1,2,\ldots\right\}$ dense in $\mathbb{S}^1$?
4. If $z\in\mathbb{S}^1$ is not a $2^k$-th root of unity for any $k$, is $\left\{z^{2^n}:n=1,2,\ldots\right\}$ dense in $\mathbb{R}^1$?

(Of course I'm asking this for diadic numbers, but the same question is valid changing $2$ by another natural number, perhaps prime.)

There is a simpler result that states that if $\theta$ is an irrational, then $\left\{e^{i\pi\theta n}:n=1,2,\ldots\right\}$ is dense in $\mathbb{S}^1$, but the proof I know does not translate to this case.

Answer 1

It is not hard to check that the set of fractional parts of the numbers $2^n x$ for $n\ge 0$ is dense in $[0,1]$ if and only if any finite sequence of $0$ and $1$'s appears in the binary expansion of $x$ ( equivalently, every pattern appears infinitely often).

Here is an example of an irrational number $x$ for which this does not happen since its binary expansion does not contain two consecutive $1$'s $$x = \sum_{n=1}^{\infty} 2^{-n^2}$$

Here is an $x$ for which the set is dense in $[0,1]$. Take all sequences of length $1$, then all the sequences of length $2$ and so on and concatenate to get

$$x =0.0100011011000001010011...$$

If $x$ is a normal number ( see http://en.wikipedia.org/wiki/Normal_number) then the set of fractional parts of $2^n x$ is dense since all patterns of a given length are equiprobable ( the definition of normal) and so they all appear. In fact $x$ normal is equivalent to : the sequence of fractional parts is uniformly distributed in $[0,1]$

Almost all the numbers in $[0,1]$ are normal. It is widely believed that all the algebraic irrationals (eg: $1/\sqrt{7}$, $\sqrt[3]{2}-1$) and a large class of transcendentals ( eg:$\ \pi$, $e$, $\log 2$ ) are normal.