# Root of a special polynomial

Given a polynomial $P(x)=\sum_{n=0}^{d}a_nx^n\in\mathbb{R}[x]$ with all roots on the unit circle.

Question: Is it true that all the roots of $Q(x)=\sum_{n=0}^{d}a_n{{x+d-n}\choose{d}}$ lie on a straight line? If it is not true,is it possible to give a counterexample?

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Very interesting. It appears they always lie on the line $x=-1/2$. Here's a plot. They do not seem to be bounded vertically. – Antonio Vargas May 1 '12 at 15:57
What makes you think the roots lie on a straight line? – Did May 1 '12 at 18:47
How did this question originate? – Olivier Bégassat May 1 '12 at 21:38
@Didier @ Olivier Bégassat: In fact,the original question is: Given a polynomial $P(x)$ that all the roots lie on the unit circle. Consider $P(x)/(1-x)^{d+1}=\sum_{n=0}^{\infty}c_nx^n$(d is the degree of $P(x)$). It is quite easy to find an interpolating polynomial $Q(x)$ that $c_n=Q(n)$. Then it is natural to ask about the location of the zeros. I find this question on a math forum, and this question was entitled with "Easy Riemann Hypothesis". This question seems very interesting, but I don't know how to work it out. (It seems that $P(x)\in\mathbb{C}[x]$ is OK) – y zh May 2 '12 at 1:41