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Alexandre Eremenko
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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.

Alexandre Eremenko
  • 91.8k
  • 9
  • 259
  • 429