On a circle (or a line) $\Omega$ in the complex plane that is not symmetric w.r.t. the real axis, choose $n\ge5$ 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 $k\ne k_0$?

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 $k\ne k_0$?