A sum of $k$-squares of complex numbers on the unit circle that converges to 0 This problem is suggested from a friend. Let me state it.

Let $a_1,\ldots,a_n$ be complex numbers and $c_1,\ldots,c_n$ be a complex numbers on the unit circle with $c_i \neq c_j$ for all $i,j \in \{1,\ldots,n\}$. Suppose that $a_1c_1^k+\cdots+a_nc_n^k$ converges to $0$ as $k \to \infty$. Then $a_1,\ldots,a_n$ must be $0$.

Actually, I have two solutions. One is from me, and the other is from my friend (I post them below). I wonder there is another approach. It seems to have various solutions, so I hope to get them.
Welcome any approach or vague idea. 
 A: *

*(From me)


For $\epsilon>0$, there exists $K>0$ such that $|a_1c_1^k+\ldots+a_nc_n^k|<\epsilon$ wherever $k>K$. Let $$A=\begin{bmatrix}a_1 \\ a_2 \\ \vdots \\ a_n 
\end{bmatrix}, \quad C=\begin{bmatrix}c_1^{K+1} & c_2^{K+1} & \cdots & c_n^{K+1} \\
c_1^{K+2} & c_2^{K+2} & \cdots & c_n^{K+2} \\
\vdots & \vdots & \ddots & \vdots \\
c_1^{K+n} & c_2^{K+n} & \cdots & c_n^{K+n} \end{bmatrix}.$$ 
Since $c_i \neq c_j$, by the well-known van der monde determinant, we know $\det C \neq 0$. Thus we have
$$
A=C^{-1}CA.
$$
Taking the operator norms, we have
$$
||A|| \leq ||C^{-1}||\times||CA||.
$$
The right-hand side is bounded by $c(n)\epsilon$ for some constant $c(n)$ that only depends on $n$. Since $\epsilon$ is arbitrary, this completes the proof.


*(From a friend)


We may assume that $c_1=1$. Let $b_k=a_1+a_2c_2^k+\cdots+a_nc_n^k$. The convergonce to $0$ implies that $\sum_{i=1}^k \frac{b_i}{k}=0$. The left-hand side is $a_1$, so $a_1=0$. By repeating this procedure, we get the result.
A: I have some other methods
Let $b_k = \sum_{s=1}^n a_s c_s^k$. I suppose $\lim_{k\rightarrow \infty} b_k = 0$.
Method 1  : 
Firstly, suppose the $c_s$ are of angle $\mathbb{Q}$-linearly independant.
Show the following lemma : 
Let $c$ in the unit circle of irrational angle. Le $\alpha$ in the unit circle. For $\epsilon > 0$ there exists infinitely many $n$ such that $|c^n - \alpha| \leq \epsilon$.
And its refinement : 
Let $c_1, .. c_n$ in the unit circle such that the angles are $\mathbb{Q}$-linearly independent.. Let $\alpha_1, .., \alpha_n$ in the unit circle. For $\epsilon > 0$ there exists infinitely many $n$ such that $|c_s^n - \alpha_s| \leq \epsilon$ for $1 \leq s \leq n$.
This lemma shows the result in the case where all the angles of the $c_s$ are $\mathbb{Q}$-linearly independant (all the $\sum a_s \alpha_s$ are zero, this implies $a_s = 0$).
Secondly, in the general case, it is possible to show that : 
Let $c_1, .. c_n$ on the unit circle. Let $\alpha_1, .. \alpha_n$ on the unit circle such that $(*)$ for all integers $n_1, .. n_s$, $\prod c_s^{n_s} = 1$ implies $\prod \alpha_s^{n_s} = 1$ (for instance $\alpha_i = c_i^k$ satisfies this property). Then for $\epsilon > 0$ there exists infinitely many $n$ such that $|c_s^n - \alpha_s| \leq \epsilon$ for $1 \leq s \leq n$. This property implies that the set of accumulation points of $b_k$ is the set constitued by the $\sum_{s=1}^n a_s \alpha_s$, where the $\alpha_s$ satisfying this property $(*)$. In particular, the hypothesis made on $b$ implies that the $\sum_{s=1}^n a_s \alpha_s$ are all zero, and so all the $b_k$ are $0$. Thus, for all polynom $P$, $\sum_{s=1}^n a_s P(c_s) = 0$, and you can conclude by interpolation. 
Method 2 : 
Consider the set $A$ of n-uplets $a = (a_1, .. a_n)$ such that $b_k(a) = \sum_{s=1}^n a_s c_s^k$ converges to $0$.


*

*$A$ is a subspace of $\mathbb{C}^n$ 

*$A$ is stable by multiplication by $c$ : if $(a_1, .., a_n)$ is in $A$, then $(a_1 c_1, .., a_n c_n)$ is in $A$.


Since the multiplication by $c$ is a cyclic endmorphism (vandermonde determinant ! ), $A$ must be $\{0\}$ or $\mathbb{C}^n$. The choice is trivial. 
