if $|z|>1$ then there is $q \in \mathbb{Q} \cap [0, 2\pi)$ such that $Re(e^{iq}z)>1$ I want to show that if $z \in \mathbb{C}$ and $|z|>1$, then there is $q \in \mathbb{Q} \cap [0, 2\pi)$ such that $Re(e^{iq}z)>1$.
Can you help me, please?. Thanks in advance.  
 A: One can write for every $z\in\mathbb{C}$: $z=|z|\exp(i\arg(z))$. Now take $p=-\arg(z)\in\mathbb{R}$. Then $ze^{ip}=|z|>1$. Moreover $f(t):=ze^{it}$ is continuous. So $g(t):=\Re(f(t))$ is continuous as well and we know that there is some $p$ such that $g(p)>1$. By continuity we can find an interval around $p$ with radius $\delta>0$, namely $I=(p-\delta,p+\delta)$, such that for all $t\in I$ one has $g(t)>1$.
Can you take it from here? Just a simple argument is need to finish it. 
A: I would like to add a sequential demonstration to this question.
There exists $x \in \mathbb{R}$ such that $Re(e^{ix}z)>1$, for instance $x=2\pi-arg(z)$. Moreover $\mathbb{Q}$ is dense in $\mathbb{R}$, which implies that there exists $(q_n)\in \mathbb{Q}^{\mathbb{N}}$ with limit $x$. Let's note that the sequence $(e^{iq_n}z)$ has limit $e^{ix)z$.
The subspace of $\mathbb{C}$: $S = \{y\in\mathbb{C} | Re(y)>1\}$ is open in $\mathbb{C}$. Therefore any sequence in $\mathbb{C}$ (for instance $(e^{iq_n}z)$) with limit $l \in S $ (for instance $e^{ix}z$ ) will be included at some point in $S$. 
Therefore: $\exists n \in \mathbb{N} | \forall k>n, Re(e^{iq_k}z)>1$.
This proves that there exists an infinity of such wanted $q$.
A: Write $z=re^{i c}$, $e^{iq}z=re^{i(q+c)}=r(cos(q+c)+isin(q+c))$, $rcos(q+c)>1$ implies that $cos(q+c)>{1\over r}$ $q+c$  since ${1\over r}<1$, there exists $u$ such that ${1\over r}<u<1$, and $\theta$ such that $cos(\theta)=u$, take $q=\theta-c$ mod $2\pi$.
