China National Team Selection Test

First, we point out that $c\ge \cot \frac{\pi }{4022}$. Consider the polynomial $P(x)=x^{2012}-x$. Changing the sign of coefficients of $P(x)$, we obtain four polynomials $P(x)$, $-P(x)$, $Q(x)=x^{2012}+x$ and $-Q(x)$. Note that $P(x)$ and $-P(x)$ have the same roots; one of the roots is $z_1=\cos \frac{1006}{2011}\pi +i\sin \frac{1006}{2011}\pi$. $Q(x)$ and $-Q(x)$ have the same roots, and are the opposite number of the roots of $P(x)$. Thus $Q(x)$ has a root $z_2=-z_1$. Then, $$c\ge \min \left(\frac{|\mathrm{Im}{z}_1|}{|\mathrm{Re}{z}_1|},\frac{|\mathrm{Im}{z}_2|}{|\mathrm{Re}{z}_2|}\right)=\cot \frac{\pi }{4022}.$$ Next, we show that the answer is $c=\cot \frac{\pi }{4022}$. For any $$P(x)=x^{2012}+a_{2011}{x}^{2011}+a_{2010}{x}^{2010}+\cdots +a_{1}x+a_0,$$ we obtain a polynomial $$R(x)=b_{2012}{x}^{2012}+b_{2011}{x}^{2011}+b_{2010}{x}^{2010}+\cdots +b_{1}x+b_0,$$ by changing sign of some coefficient of $P(x)$, where $b_{2012}=1$, and for $j=1,2,\dots ,2011$, $$b_j=\{\begin{array}{ll}|a_j|,& j\equiv 0,1(\mathrm{mod}4),\\ -|a_j|,& j\equiv 2,3(\mathrm{mod}4).\end{array}$$ We show that, for each root $z$ of $R(x)$, we have $|\mathrm{Im}z|\le c|\mathrm{Re}z|.$ We prove this result by contradiction. Suppose there is a root $z_0$ of $R(x)$, such that $|\mathrm{Im}{z}_0|>c|\mathrm{Re}{z}_0|$, then $z_0\ne 0$ and either the angle of $z_0$ and $i$ is less than $\theta =\frac{\pi }{4022}$, or the angle of $z_0$ and $-i$ is less than $\theta$. Suppose that the angle of $z_0$ and $i$ is less than $\theta$; for the other case, we need only consider the conjugate of $z_0$. There are two cases:

If $z_0$ is on the first quadrant (or imaginary axis), suppose that $\angle (z_0,i)=\alpha <\theta$, where $\angle (z_0,i)$ is the least angle that rotates $z_0$ to $i$ anticlockwise. For $0\le j\le 2012$, if $j\equiv 0,2(\mathrm{mod}4)$, then $\angle (b_j{z}_0^j,1)=j\alpha \le 2012\alpha <2012\theta$. If $j\equiv 1,3(\mathrm{mod}4)$, then $\angle (b_j{z}_0^j,i)=j\alpha <2011\theta$ and $\angle (b_1{z}_0,i)=\alpha$. Thus, the principal argument of $b_j{z}_0^j\in [2\pi -2012\alpha ,2\pi )\cup [0,\frac{1}{2}\pi -\alpha ]$. The vertex angle of this angle-domain is $2012\alpha +\frac{1}{2}\pi -\alpha =\frac{1}{2}\pi +2011\alpha <\pi$. And $b_j{z}_0^j,0\le j\le 2012$, are not all zero, so their sum cannot be zero.

If $z_0$ is at the second quadrant, suppose that $\angle (i,z_0)=\alpha <\theta$, if $j\equiv 0,2(\mathrm{mod}4)$, then $\angle (1,b_j{z}_0^j)=j\alpha <2012\theta$. If $j\equiv 1,3(\mathrm{mod}4)$, then $\angle (i,b_j{z}_0^j)=j\alpha \le 2011\alpha <\frac{\pi }{2}$. Thus, every principle argument of $b_j{z}_0^j\in [0,\frac{\pi }{2}+2011\alpha ]$. Since $\frac{\pi }{2}+2011\alpha <\pi$, and $b_j{z}_0^j,0\le j\le 2012$, are not all zero, so their sum cannot be zero.

Summing up, the least real number $c=\cot \frac{\pi }{4022}$. $\square$