# 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?

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mathoverflow.net/howtoask – Anthony Quas Apr 18 '13 at 15:53
This is a reasonable question, especially now, when it is corrected. Please don't close it. – Alexandre Eremenko Apr 19 '13 at 13:15
When $n=2$, it is right. I hope it is right for $n\geq 3$. – CooLee Apr 19 '13 at 14:05

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.

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I think $(x^4+x)(x^2-x+1)$ is not equal to $x^6+x^5+x^4+x^3+x^2+x$ and $(x^4+x)(x^2-x+1)^2$ is not equal to $x^8+x^6+x^5+x^4+x^3+x$. – CooLee Apr 19 '13 at 16:51
Sorry, I fixed the typo. – Peter Mueller Apr 19 '13 at 17:14
@Peter Mueller,How do you find the Counter-example？ – bo.gu Apr 20 '13 at 3:27
@bo.gu: I do not understand your question. Are you asking how I did find the counter-examples, or do you want to know why these polynomials yield counter-examples? – Peter Mueller Apr 22 '13 at 13:41

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.

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Of course one can easily modify the statement to eliminate these simple counterexamples but I leave this to the author. – Alexandre Eremenko Apr 19 '13 at 1:25
very sorry for the unclear statement. Please see the question again. Thank you very much. – CooLee Apr 19 '13 at 7:04