Let $P_1(z)$ and $P_2(z)$ be monic polynomials with real coefficients and roots $\{z_1^{(1)},z_1^{(2)},...\}$ and $\{z_2^{(1)},z_2^{(2)},...\}$, respectively. Are there any results relating the non-trivial roots $\{z^{(1)}, z^{(2)},...\}$ of \begin{equation} P(z) = \xi P_1(z) + P_2(z) \end{equation} to the roots of $P_1$ and $P_2$?

In particular, if $P_1, P_2$ and $P$ have then same number of complex valued roots, under what circumstances do we have (up to a relabelling) \begin{equation} \min \{\mathrm{Im} \, z_1^{(k)}, \mathrm{Im} \, z_2^{(k)} \} \leq \, \mathrm{Im} \, z^{(k)} \leq \max \{\mathrm{Im} \, z_1^{(k)}, \mathrm{Im} \, z_2^{(k)} \} \end{equation} for all $k$ and for all $\xi \in [0,\infty)$?

Example: Let $P_1(z) = z (z^2 + 1)$ and $P_2(z) = z^2 + r^2$. The polynomial $P(z) = \xi P_1(z) + P_2(z)$ satisfies the above property given $r\leq 1$.

  • $\begingroup$ Perhaps the notation for roots is misleading, but it would appear you assume $P_1(z)$ has (some number of) roots that are consecutive powers of a common base $z_1$, and similarly for $P_2$ and roots $z_2^k$. If no such relation is intended, perhaps double subscripts or (more simply) two letters $u_i,v_i$ with single subscripts would be appropriate. Also, is anything known about degrees of $P_1,P_2$ ? $\endgroup$ – hardmath May 16 '13 at 12:53
  • $\begingroup$ @hardmath Thanks for the input. I've fixed the notation to make it more clear. I'm posing the question for general degrees, but one example is $P(z) = \xi z(z^2 + 1) + (z^2 + r^2)$ which satisfies the property above given $r \leq 1$. $\endgroup$ – paursand May 16 '13 at 13:46
  • $\begingroup$ In your example $P_1(z)$ is degree 3 and $P_2(z)$ is degree 2, so it's unclear what "up to a relabelling" and "for all $k$" allows (or constrains) about the imaginary parts of roots. $\endgroup$ – hardmath May 16 '13 at 13:57
  • $\begingroup$ @hardmath Thank you for pointing this out. If we insist that the number of complex valued roots are equal then the statement should be more clear. $\endgroup$ – paursand May 16 '13 at 15:30

One way to approach this is to bound the roots of your polynomials in terms of the polynomial coefficients (which, in contrast, behave nicely under addition):

Let $(K, \vert \cdot \vert)$ be a valued field and let $$f(X) = X^d +a_1 X^{d-1}+ \ldots + a_d = (X-b_1)(X-b_2)\cdots(X-b_d)$$ be a polynomial, where $a_1,\ldots, a_d,b_1,\ldots,b_d \in K$. Let $C=2$ if the absolute value is archimedean and let $C=1$ if the absolute value is non-archimedean. Then $$C^{-d} \vert f \vert \leq \prod_{i=1}^d \max\{\vert b_i \vert, 1\} \leq C^d \vert f \vert.$$

(The proof is by induction on $d$.) In particular, this gives bounds on the absolute values of the roots (both above and below) in terms of the coefficients of $f$. By adding two polynomials $f$ and $g$, we quickly obtain bounds for the roots of their sum (simply by replacing $\vert f \vert$ by $\vert f+g \vert$).


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