Can we find positive reals $a_1, a_2, \dots, a_{2002}$ such that for any positive integer $k$, with $1 \leq k \leq 2002$, every complex root $z$ of the following polynomial $f(x)$ satisfies the condition $|\text{Im } z| \leq |\text{Re } z|$, \[f(x)=a_{k+2001}x^{2001}+a_{k+2000}x^{2000}+ \cdots + a_{k+1}x+a_k,\]where $a_{2002+i}=a_i$, for $i=1,2, \dots, 2001$.
Problem
Source: China TST 2003
Tags: algebra, polynomial, algebra unsolved
MathPanda1
11.04.2016 06:49
Does anyone have a solution to this problem? Thanks!
P-H-David-Clarence
22.10.2016 11:14
Notice that $|\text{Im} z| \leq |\text{Re} z|$ implys $$\text{Re}z\geq0$$So suming squares over all roots:$$a^{2}_{k+2000}\geq 2a_{k+2001}a_{k+1999}$$Then get them all over $k=1,2,3\ldots $we get a contradiction.
MathPanda1
21.02.2017 17:34
Sorry, but how did you get $a^{2}_{k+2000}\geq 2a_{k+2001}a_{k+1999}$? Thanks! Edit: $Rez$ doesn't have to be nonnegative since it's $|Rez|$, not just $Rez$. Edit: Never mind, P-H-David-Clarence's solution is really nice! I don't know, but apparently I got that the answer is such $a_i$'s exist? This might be wrong, but I think the polynomial $(x+1)[(x + (cos(\pi / 1001)+i sin(\pi / 1001))][x+ (cos(\pi / 1001)-i sin(\pi / 1001))]^{1000}$ works? I think my idea was to choose roots such that $r^{2002} = 1$. Edit: This doesn't work
EthanWYX2009
21.02.2023 18:28
From $|\Im z|\leqslant |\Re z|$ we can get $\Re z^2=|\Re z|^2-|\Im z|^2\geqslant 0$. Therefore $\sum\limits_{k=1}^{2001}z_k^2=\frac{a_{k+2000}^2-2a_{k+1999}a_{k+2001}}{a_{k+2001}^2}\geqslant 0$. Let $a_{k+2000}=\min_{1\leq k\leq 2002}a_k$ and we can get contradiction.