The distribution of roots of elliptic polynomial If $p(x)$ is an $n$ variables polynomial of even degree with complex coefficients which satisfies the strong elliptic condition, that is, Re$p(x) \ge C|x|^{2m}$ for any $x \in \mathbb R^n$ where $2m$ is the degree of $p$.  Then can we have the following conclusion:
Foy any $y \neq0 \in \mathbb R^{n-1}$, consider the roots of $p(y,z)=0$ in the complex plane. Then is the number of roots (counting with multiplicity) in the upper half plane equal to that of lower half?
 A: First of all, the notion of ellipticity refers to the principal part of the differential operator. In terms of the symbol, this means a property for the so-called principal symbol. Thus we may assume that $p$ is a homogeneous polynomial, of degree $2m$.
Then the answer to your question is Yes, when $y\in{\mathbb R}^{n-1}$ is non-zero, the univariate polynomial $z\mapsto p(y,z)$ has $m$ roots of positive (resp. negative) imaginary part. Here is the proof. First of all, there are $2m$ roots by assumptions, and none of the roots are real. If $n\ge3$ there imaginary parts keep a constant sign as $y$ varies, because the complement of the origin is connected. Thus let $m_\pm$ be (constant) number of roots of positive/negative imaginary parts. We have $m_-+m_+=2m$. Because the roots $z_j(-y)$ are nothing but the $-z_j(y)$ (by homogeneity), we have $m_-=m_+$. Therefore $m_\pm=m$.
If $n=2$ instead, one build the strongly elliptic polynomial $q(x,t)=p(x)+t^{2m}$ in $3$ variables. The previous analysis applies to $q$ and gives the result for $p$.
Now, the non-homogeneous case. We still know that $z\mapsto p(y,z)$ has $2m$ roots, with a constant number $m_\pm$ of positive/negative imaginary part. Fix $y\ne0$ real and d consider the polynomial $p_s(y,z)=s^{-2m}p(sy,sz)$. When $s\rightarrow+\infty$, $p_s$ tends to the homogeneous part $p_\infty$ of degree $2m$. The assumption implies that $P_\infty$ is strongly elliptic, to which the previous analysis applies. By continuity of the roots as functions of $\frac1s$, we find again that $m_\pm=m$.
