Bulletin of the
Korean Mathematical Society BKMS

It is known that no two of the roots of the polynomial equation \begin{equation}\begin{split} \prod_{l=1}^n (x-r_l) + \prod_{l=1}^n (x+r_l) =0, \label{one-1} \end{split}\end{equation} where $0 < r_1 \leq r_2 \leq \cdots \leq r_n$, can be equal and all of its roots lie on the imaginary axis. In this paper we show that for $0 < h< r_k$, the roots of $$ (x-r_k+h)\prod_{\substack{l=1\\ l\neq k}}^n(x-r_l) + (x+r_k-h)\prod_{\substack{l=1\\l\neq k} }^n (x+r_l) = 0 $$ and the roots of (\ref{one-1}) in the upper half-plane lie alternatively on the imaginary axis.

Keywords: sums of polynomials, roots, root squeezing

MSC numbers: Primary 26C10; Secondary 11B83