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?