Powers of complex numbers property I would like to prove the following statement.
Let $\lambda_1,\dots,\lambda_s \in \mathbb{C}$ be such that $|\lambda_1| = \dots = |\lambda_s|=1$. Then $\forall \varepsilon \gt 0$ there exist arbitrarily large integers $n$ such that $|\lambda_1^n+\dots+\lambda_s^n| \geq s-\varepsilon$.
I don't really know how to start. I've tried the following approach.
Write $\lambda_i = e^{i\alpha_i}$, then we want to find arbitrarily large integers $n$ such that $n\alpha_i \approx 2k\pi$ for some integer $k$ for all $i$. The problem is that I don't know how to make "$\approx$" rigorous and how to find those integers $n$.
 A: Your approach is good; combined with a simple linear transformation it takes us from the realm of complex powers to the realm of multiplication modulo 1.
Start by proving a lemma: for every $\epsilon>0$ and 
$x \in [0,1)$ there is $n \in \mathbb{N}, n<\epsilon^{-1}$ such that $|nx|\le\epsilon \mod 1$.
Then, let $\epsilon>0$ and $x_1,\ldots, x_s \in [0,1)$. From now on I'll denote $f(m)=\epsilon^{2^m}$ simply because otherwise the display will be unintelligible.
Prove by induction on $i$ that there are natural numbers $n_1,\ldots, n_s$, so that for all $i$ you have:


*

*$n_i < 1/f(s-i)$

*For all $j\le i$, $|n_1\cdots n_i x_j|\le f(s-i) \mod 1$.


The base case for $i=1$ follows from the lemma above, for threshold $f(s-1)$.
For the induction step: Use the lemma on $x=n_1\cdots n_{i-1}x_i$ and threshold $f(s-i)$ to find $n_i<1/f(s-i)$ such that $n_1\cdots n_ix_i \le f(s-i)$. For $j<i$ you use the hypothesis. $|n_1\cdots n_{i-1} x_j|\le f(s-i+1)$, and $n_i < 1/f(s-i)$, so $|n_1\cdots n_{i} x_j|< f(s-i+1)/f(s-i) = \epsilon^{2^{s-i+1}-2^{s-i}} = \epsilon^{2^{s-i}}=f(s-i)$.
It follows that there is $n=n_1\cdots n_s < 1/f(2^s-1)$ such that $nx_i \le \epsilon$ for all $i$.
All that's left is to show you have arbitrarily large $n$ with this property, but that's trivial - simply pick a smaller $\epsilon$ and larger $n$.
Note: The intuition behind the proof is really very simple, and I hope it came across in this writeup. It took me 1 minute to come up with the idea, 15 minutes to work out the details and an hour to write it down...
A: just a thought that wanted to share : for every $\alpha_i$ we can find an integer $k_i$ such that $k_i \alpha_i \mbox{mod} (2\pi) \in I_\delta$ where $I_\delta = (-\delta, \delta)$  for an arbitrary small $\delta$ (Equidistribution modulo $2\pi$). 
Now if $n \ge m \; lcm(k_1, k_2, \ldots, k_s)$ for an integer $m \ge 1$, then all $\beta_i = n \alpha_i \; \mbox{mod} ( 2\pi)$ are in $I_\delta$. 
So $\left|\sum{\lambda_k^n}\right| \ge s cos(\delta) = s - \epsilon$ where $\epsilon = s(1-cos(\delta))$.
