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Let $\theta\in (0,2\pi)$ be a real number such that $\displaystyle\frac{\theta}{\pi}\notin\mathbb{Q}$. We define $z:=\cos(\theta)+i\sin(\theta)\in S^1\subseteq\mathbb{C}$ and let $\{z_n\}_{n\in\mathbb{N}}\subseteq S^1$ be the sequence defined by $z_n:=z^n=\cos(n\theta)+i\sin(n\theta)$.

Prove that the set of the limit points of $\{z_n\}$ is $S^1$.

Frankenstein
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  • Going to polar coordinates, you are talking about the points with radius $1$ and argument $n\theta$, $n=1,2,\dots$, and you are asking for a proof that the points $n\theta$ are dense in $[0,2\pi]$. I'm sure this has been asked and answered on this site many times, but it may not be easy to find it here. – Gerry Myerson Mar 07 '13 at 12:18
  • You can see the idea at http://math.stackexchange.com/questions/39299/are-rotations-of-0-1-by-n-arccos-frac13-dense-in-the-unit-circle and at http://math.stackexchange.com/questions/73262/proving-that-mn-sqrt2-is-dense-in-r and http://math.stackexchange.com/questions/272545/multiples-of-an-irrational-number-forming-a-dense-subset? and http://math.stackexchange.com/questions/136665/how-to-show-this-is-a-dense-set and http://math.stackexchange.com/questions/189397/why-is-this-quotient-space-not-hausdorff/189402#189402 (for starters). – Gerry Myerson Mar 08 '13 at 01:14

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