Dirichlet's Theorem for Turan's Method of Analysis This question concerns a special case of Turan's power sum, which looks reasonably straightforward.
Let $S(N,\nu)=\sum_{n=1}^N z_n^\nu,$ where all the $z_n$ are on the complex unit circle. Then, according to Gonek's paper A Note on Turan's Method, using Dirichlet's theorem on uniform approximation, one can show that there is some $\nu \in \{1,2,\ldots,6^N\},$ such that 
$$
\left \Vert \nu \arg \left(\frac{z_n}{2\pi}\right) \right\Vert \leq \frac{1}{6},\quad 1\leq n\leq N,
$$
where $||x||$ is the nearest distance from $x$ to an integer. This immediately implies that
$$
\max_{1\leq \nu \leq 6^N} |S(v,N)|\geq \frac{S(0,N)}{2}=\frac{N}{2}.
$$
How can one show this?
 A: The statement you want is this: 
Given $$ reals $_$ (here $_=\arg \frac{z_k}{2\pi}$), a positive integer $q$ (here 6), we can find $1\leq \nu \leq q^N,\nu$ integral, s.t $\Vert v a_k\Vert\leq \frac{1}{q}$ and the proof uses the standard $N$ dimensional unit cube, the $q^+1$ points $(_1,..._)$,$0\leq \leq ^$ modulo 1, so in the cube (by using the pigeonhole principle when you divide each edge into $$ parts) you have $^$ compartments. 
Armed with this result, you use the inequality $|\arg(w_k)| \leq \theta < \frac{\pi}{2}$, where $arg$ is taken in $[-\pi, \pi]$, implies $\Re{w_k} \geq |w_k|\cos \theta$, so if say $|w_k|=1$, $|\Sigma{w_k}| \geq |\Re\Sigma{w_k}| \geq N\cos \theta $, where $N$ is the number of terms in the sum
$\Vert \frac{1}{2\pi}\arg{z_n^\nu}\Vert = \Vert\nu \arg \left(\frac{z_n}{2\pi}\right)\Vert$, so we can take $|\arg{z_n^\nu}| \leq \frac{\pi}{3}, \cos ({\arg{z_n^\nu}}) \geq \frac{1}{2}$ and apply the above to deduce the inequality for precisely the $\nu$ given by Dirichlet which is bounded as noted
