roots of the discrete Fourier transform of a polynomial

Question

Let $p(x)= a_0+a_1x+a_2x^2+...+a_{n-1}x^{n-1}$, $q(x)= b_0+b_1x+b_2x^2+...+b_{n-1}x^{n-1}$ be two polynomials with complex coefficients. if $(b_0,b_1,b_2,...,b_{n-1})$ is the discrete Fourier transform of $(a_0,a_1,a_2,...,a_{n-1})$ can we say anything about the roots of $q(x)$ if we know the roots of $p(x)$. I'm mainly interested in relations between the roots of $p$ and $q$ where $a_j$'s are positive integers. In this case $b_j=$ conjugate$(b_{n-j})$ for all meaningful $j$.

Answer 1

The roots of a polynomial are related to its coefficients by Vieta's relations.

For the fourth degree,

$$a_0=r_0+r_1+r_2+r_3,$$ $$a_1=r_0r_1+r_1r_2+r_2r_3+r_3r_0,$$ $$a_2=r_0r_1r_2+r_0r_2r_3+r_1r_0r_2+r_1r_2r_3+r_2r_0r_1+r_2r_1r_3,$$ $$a_3=r_0r_1r_2r_3$$

where for convenience we used the opposite of the root values.

Then taking the DFT and expressing the new coefficients in terms of the roots of the DFT, we get four relations

$$(r_0+r_1+r_2+r_3)+(r_0r_1+r_1r_2+r_2r_3+r_3r_0)+(r_0r_1r_2+r_0r_2r_3+r_1r_0r_2+r_1r_2r_3+r_2r_0r_1+r_2r_1r_3)+r_0r_1r_2r_3=s_0+s_1+s_2+s_3$$

$$(r_0+r_1+r_2+r_3)+i(r_0r_1+r_1r_2+r_2r_3+r_3r_0)-(r_0r_1r_2+r_0r_2r_3+r_1r_0r_2+r_1r_2r_3+r_2r_0r_1+r_2r_1r_3)-ir_0r_1r_2r_3=s_0s_1+s_1s_2+s_2s_3+s_3s_0$$

$$(r_0+r_1+r_2+r_3)-(r_0r_1+r_1r_2+r_2r_3+r_3r_0)+(r_0r_1r_2+r_0r_2r_3+r_1r_0r_2+r_1r_2r_3+r_2r_0r_1+r_2r_1r_3)-r_0r_1r_2r_3=s_0s_1s_2+s_0s_2s_3+s_1s_0s_2+s_1s_2s_3+s_2s_0s_1+s_2s_1s_3$$

$$(r_0+r_1+r_2+r_3)-i(r_0r_1+r_1r_2+r_2r_3+r_3r_0)-(r_0r_1r_2+r_0r_2r_3+r_1r_0r_2+r_1r_2r_3+r_2r_0r_1+r_2r_1r_3)+ir_0r_1r_2r_3=s_0s_1s_2s_3$$

I can't see a way to simplify this.

Unless there is some simplification due to the particular expression of the coefficients, in general it is not even possible to express the $s_k$ in terms of the $r_k$ in general, as the resolution of algebraic equations can't be done with closed-form expressions.