Abstract

If q(z) is a polynomial of degree n with all zeros in the unit circle, then the self-reciprocal polynomial <TEX>$q(z)+x^nq(1/z)$</TEX> has all its zeros on the unit circle. One might naturally ask: where are the zeros of <TEX>$q(z)+x^nq(1/z)$</TEX> located if q(z) has different zero distribution from the unit circle? In this paper, we study this question when <TEX>$q(z)=(z-1)^{n-k}(z-1-c_1){\cdots}(z-1-c_k)+(z+1)^{n-k}(z+1+c_1){\cdots}(z+1+c_k)$</TEX>, where <TEX>$c_j$</TEX> > 0 for each j, and q(z) is a 'zeros dragged' polynomial from <TEX>$(z-1)^n+(z+1)^n$</TEX> whose all zeros lie on the imaginary axis.