On a circle (or a line) $\Omega$ in the complex plane that is not symmetric w.r.t. the real axis, choose $n\ge5$ distinct points $z_1,...,z_n$ and consider the polynomial $p(z)=\prod_j(z-z_j)=z^n+a_{n-1}z^{n-1}+\cdots+a_0$. Question: How many of the $a_k$'s can be reals at most ?
Intuitively, imagining something like a Java applet, there are "more or less" $n-1$ degrees of freedom to move the points around on $\Omega$, so we might reasonably expect that it is possible to choose them (say for an appropriate $\Omega$) such that all but one $a_{k_0}$ are reals.

  • How could that be rigorously proven?
  • If it is true, can it be done for any $k_0\in\lbrace0,...,n-1\rbrace$?
  • How to find a concrete solution?
  • Can such a polynomial even be rational, i.e. such that $a_{k_0}\in\mathbb Q[i]$ and $a_k\in\mathbb Q$ for $k\ne k_0$?
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  • $\begingroup$ Why not $n$ degrees of freedom? You're choosing $n$ points (one degree of freedom each) and trying to satisfy $n$ constraints: $\Re(a_i)=0$ for $i=0,\ldots,n-1$. Shouldn't you expect to be able to get them all real? $\endgroup$ – Anthony Quas Feb 25 '13 at 15:35
  • $\begingroup$ how's that? If they were all real, any non-real zeros would come in conjugate pairs and so $\Omega$ would be symmetric to the real line, which is excluded! The constraint of the points being on a circle sort of removes 1 degree of freedom. But only "sort of", that's why I have said vaguely "more or less" :) $\endgroup$ – Wolfgang Feb 25 '13 at 15:53
  • $\begingroup$ You want all the roots to be simple right? Otherwise you can pick any suitable circle and take as many real roots as you like from the one or two points where that circle intersects the real line. $\endgroup$ – Aaron Golden Feb 25 '13 at 17:14
  • $\begingroup$ yes, of course. Otherwise we could as well start with a circle that has 0 on it. OK, I'll add "distinct". $\endgroup$ – Wolfgang Feb 25 '13 at 17:30
  • $\begingroup$ @Aaron Golden: Well, if the circle meets the real line. I think the question is interesting when it doesn't meet the real line, as well. $\endgroup$ – David E Speyer Feb 25 '13 at 18:06

Suppose it is a line, passing through the origin under the angle $\phi$. Then your polynomial must be

$$z^n+a_1z^{n-1}+...+a_0=\prod_{j=1}^n(z-t_je^{i\phi}),$$ where $t_j$ are real. Vjeta's formulas give $$a_k=\pm\sum t_{i_1}...t_{i_k} e^{ik\phi},$$ Now how can $a_k$ be real?

First way: $e^{ik\phi}$ is real. For how many $k=1...n$ this can happen, is easy to find out.

Second way: $$b_k:=\sum t_{i_1}...t_{i_k}=0.$$ Of course this can happen for all $k$ if all $t_k=0$. If you want to exclude $t_j=0$ than the question is reduced to "how many zero coefficients can have a polynomial with all roots real and non-zero ?". I mean the real polynomial $\prod(z-t_k)$, whose coefficients are $\pm b_k$.

For this real polynomial, you can use the following theorem of Descartes: The number of positive zeros of a real polynomial is at most the number of sign changes in the sequence of coefficients (which is at most the number of non-zero coefficients minus 1). Same applies to the number of negative zeros if you make the change of the variable $x\to-x$, which changes the sign switches but does not change the number of non-zero coefficients.

If you want all roots to be distinct, at most one of them is zero.

I leave the details to you.

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  • 3
    $\begingroup$ @ Alexandre: I don't get your point. I did modify my question (after Aaron's comment, already before you posted your answer) by making it clear that the roots should of course all be different, otherwise it becomes trivial. Moreover, if the polynomial has only (distinct) real roots, Ω is necessarily the real line, which is symmetric w.r.t. itself, thus excluded. $\endgroup$ – Wolfgang Feb 26 '13 at 9:04
  • $\begingroup$ @Wolfgang: You modified your question (sorry I did not see it in time) and now I modified my answer. $\endgroup$ – Alexandre Eremenko Feb 27 '13 at 2:11

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