A question from complex analysis Let $n\geq2$. We assume $0<\alpha_n<\cdots<\alpha_2<\alpha_1<1$ and $0<\beta_n<\cdots<\beta_2<\beta_1<1$ 
, $\alpha_n=\beta_n$, and there exists $1\leq j_0\leq n$ such that $\alpha_{j_0}\neq \beta_{j_0}$.
My question: is there a complex number $s$ such that 
$\sum_{j=1}^n s^{\alpha_j}=0$ and $\sum_{j=1}^n s^{\beta_j} \neq 0$ or the other hand  $\sum_{j=1}^n s^{\beta_j} =0$ and $\sum_{j=1}^n s^{\alpha_j}\neq0$?
Thank you for @Peter Mueller's counter example. Since we know that the two 'polynomials'
$\sum_{j=1}^n s^{\alpha_j}$ and $\sum_{j=1}^n s^{\beta_j}$
 in my question maybe have the same set of roots. But I want to ask that is it possible that all the same root has the same multiplicity? Since for the ordinary polynomials this statement is not true. But for my case, what will happen?
 A: The answer is no. Take the polynomials
\begin{align}
f(x) &= x^6 + x^5 + x^4 + x^3 + x^2 + x\\\
g(x) &= x^8 + x^6 + x^5 + x^4 + x^3 + x.
\end{align}
From
\begin{align}
f(x) &= x(x+1)(x^2+x+1)(x^2-x+1)\\\
g(x) &= x(x+1)(x^2+x+1)(x^2-x+1)^2
\end{align}
wee see that $f(x)$ and $g(x)$ have the same complex roots. Upon setting $x=s^{1/10}$ we see that
\begin{align}
\alpha_6,\dots,\alpha_1 &= 1/10,\; 2/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10\\\
\beta_6,\dots,\beta_1 &= 1/10,\; 3/10,\; 4/10,\; 5/10,\; 6/10,\; 8/10
\end{align}
is a counterexample for $n=6$.
Added: More examples, with $n=4$, can be constructed using
\begin{align}
f(x) &= x(1+x+x^v+x^{v+1})\\\
g(x) &= x(1+x^u+x^v+x^{u+v})
\end{align}
if $v$ is odd and $1\lt u\lt v$ is a divisor of $u$. In this case, $f(x)$ and $g(x)$ again have the same complex roots.
A: The answer is no. For example if $n=1$, there is no such number.
Or take any $n$, any $\alpha_j$ and multiply your first sum on $s^\beta$ where
$\beta$ is very small positive. Then the second sum will be with $\beta_j=\alpha_j+\beta$,
and the sums will have common zeros. 
