A complex problem. We have a set $S:= \{e^{inr\pi} | n\in\Bbb N\}$. Where r is an irrational number. I wonder whether this set is dense in $\partial D(0,1)$. i.e. I want to see if $\overline S=\partial D(0,1).$ 
I think we can show that the set of arguments of $S$ in $[0,2\pi]$ is dense. And as $r$ is irrational, we can show that $\forall \epsilon \exists n\in \Bbb R$ s.t. $e^{inr\pi}$ has argument in $(0,\epsilon).$ Or equivalently $nr\pi \equiv \epsilon(mod 2\pi)$ 
Also I want to know if $r\in \Bbb Q$ then What can we say about above statement!!
 A: I will prove the following statement from which you will easily deduce your result :
The set $ A =\{ n + kr \ | \ (n,k) \in \mathbb{Z}^2  \}$ is a dense subgroup of $\mathbb{R}$ whenever $r$ is irrational.
1) It is obviously a subgroup of $(\mathbb{R},+)$ since it contains $0$ and is stable for the law $+$.
2) Suppose $\inf (A \cap \mathbb{R}^*_+ ) > 0 $. Then there exists $a\in A$ such that $ a = \inf (A \cap \mathbb{R}^*_+ )$, otherwise there would exist an uncreasing sequence of distincts elements of $A \ $ $a_n > a$ converging to $a$. But then $a_n - a_{n-1} > 0$ would to $0$ as $n$ goes to infinity (because $(a_n)$ is a Cauchy or any other argument using the quantification of converging sequences). But since $A$ is a subgroup $a_n - a_{n+1}$ belongs to $A \cap \mathbb{R}^*_+$ for all $n$. And then $\inf (A \cap \mathbb{R}^*_+ ) = 0 $.
So $\inf (A \cap \mathbb{R}^*_+ ) = a $ belongs to $A$. Then $a \cdot \mathbb{Z} \subset A$. Suppose $A \neq a \cdot \mathbb{Z}$. Then there existe $k \in \mathbb{Z}$ and $b \in A$ such that $ka < b < (k+1)a$. But then $0 < b - ka < a$, $b - ka \in A$ and $\inf (A \cap \mathbb{R}^*_+ ) = a $. This is absurd so $A = a \cdot \mathbb{Z}$. 
Now $r = ka$ and $1 = k'a$ with $k$ and $k'$ non-zero integers. Then $r = \frac{k}{k'}$ which is absurd since we've supposed $r$ irrational. 
So $\inf (A \cap \mathbb{R}^*_+ )$ must be $0$. Now we prove that $A$ is dense in $\mathbb{R}$. Let $x \in \mathbb{R}$ and $\epsilon > 0$. We just have to prove there exists $c \in A$ belonging to $]x-\epsilon, x + \epsilon[$. Since $\inf (A \cap \mathbb{R}^*_+ )= 0$ there exists $\alpha \in A$ such that $0 < \alpha < \frac{\epsilon}{3}$. Suppose the set $\{k\alpha \ | \ k\in \mathbb{Z}$ doesn't meet $]x-\epsilon, x + \epsilon[$. Then there exists $n \in \mathbb{Z}$ such that $n\alpha < x- \epsilon < x + \epsilon < (n+1)\alpha$. Then $\alpha > 2 \epsilon$ which is absurd. Soit $A$ meets $]x-\epsilon, x + \epsilon[$. Since $x$ and $\epsilon$ are unspecified, we have proved that $A$ is a dense subset of $\mathbb{R}$.
