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Fix a polynomial $g(x)\in\mathbb R[x]$ and a complex number $u\in\mathbb C\setminus\mathbb R$. My main question is

How can we construct a polynomial $s(x)\in\mathbb R[x]$ such that $s(x)$ is a sum of squares and $s(u)=g(u)$ and $s(\overline{u})=g(\overline{u})$?

I know that if $s(u)=g(u)$ then $s(\overline{u})=g(\overline{u})$, so we really only need the condition $s(u)=g(u)$. I am asking about this because I am trying to understand a proof of a theorem (as Theorem 7.3 in "Solving systems of polynomial equations" by Sturmfels) in which the author claims that we can write a polynomial $g(x)\in\mathbb R[x_1,\ldots,x_n]$ is congruent to a sum of squares $s(x)\in\mathbb R[x_1,\ldots x_n]$ modulo $\langle x-u\rangle\cap\langle x-\overline{u}\rangle$ for fixed $u\in\mathbb C^n$ strictly complex. The rest of the proof I can understand (and I could copy it into this question if requested), but I am having trouble seeing just why exactly this can be done. So my main question above is just asking about the univariate case of this, from which I believe I could generalize the argument.

So far I have only been able to think of the following. If we set $s(x):=\frac{g(x)^2+|g(u)|^2}{2\Re(g(u))}$ then we have $$g(u)^2+g(u)g(\overline{u})=g(u)\left(g(u)+\overline{g(u)}\right)=2\Re (g(u))g(u)$$ so $s(u)=g(u)$. But this only works if $\Re (g(u))>0$ since we need $s(x)$ to be a sum of squares in $\mathbb R[x]$.

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Note that since $u$ is not real, $u$ and $1$ are linearly independent over $\mathbb{R}$, so every complex number can be written as $au+b$ for some $a,b\in\mathbb{R}$. In particular, we can pick $a$ and $b$ such that $au+b$ is a square root of $g(u)$, and then take $s(x)=(ax+b)^2$.

Alternatively, pick $a\in\mathbb{R}$ large enough such that $v=u+a$ has positive real part. Now note that the consecutive powers of $v^2$ have an angle of less than $\pi$ between them. We can pick $n\in\mathbb{N}$ such that $g(u)$ has argument between the arguments of $v^{2n}$ and $v^{2n+2}$, and then $g(u)$ can be written as a linear combination of $v^{2n}$ and $v^{2n+2}$ with nonnegative real coefficients. This gives a polynomial $s(x)=b(x+a)^{2n}+c(x+a)^{2n+2}$ which is a sum of squares and has $s(u)=g(u)$.

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  • $\begingroup$ Yes I want $u\notin\mathbb R$, I will edit to say $u\in\mathbb C\setminus\mathbb R$. This solution is very interesting, thank you! I might leave the answer un-accepted for a little while just in case people have other approaches and want to share. $\endgroup$ – Dave Feb 19 at 22:02
  • $\begingroup$ On second thought I realiized my first approach was way more complicated than necessary; I've added a much simpler argument to my answer. $\endgroup$ – Eric Wofsey Feb 19 at 22:12
  • $\begingroup$ Yes I quite like this other approach, thanks for your help. $\endgroup$ – Dave Feb 20 at 1:23

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