3. Find the smallest real number $c$ such that:

For any real-coefficient polynomial of degree 2012 with leading coefficient 1,
$$
P(x)=x^{2012}+a_{2011} x^{2011}+a_{2010} x^{2010}+\cdots+a_{0} \text {, }
$$

some of its coefficients can be multiplied by -1, while the others remain unchanged, so that each root $z$ of the new polynomial satisfies $|\operatorname{Im} z| \leqslant c|\operatorname{Re} z|$, where $\operatorname{Re} z$ and $\operatorname{Im} z$ denote the real and imaginary parts of the complex number $z$, respectively.
（Supplied by Hua-Wei Zhu）
3. First, prove: $c \geqslant \cot \frac{\pi}{4022}$.

Consider the polynomial $P(x)=x^{2012}-x$.
By changing the signs of the coefficients of $P(x)$, we obtain four polynomials $P(x), -P(x), Q(x)=x^{2012}+x$, and $-Q(x)$.

Notice that, $P(x)$ and $-P(x)$ have the same roots, one of which is
$$
z_{1}=\cos \frac{1006}{2011} \pi + \mathrm{i} \sin \frac{1006}{2011} \pi,
$$

Moreover, $Q(x)$ and $-Q(x)$ have the same roots, which are the negatives of the roots of $P(x)$, so $Q(x)$ has a root $z_{2}=-z_{1}$. Therefore,
$$
c \geqslant \min \left(\frac{\left|\operatorname{Im} z_{1}\right|}{\left|\operatorname{Re} z_{1}\right|}, \frac{\left|\operatorname{Im} z_{2}\right|}{\left|\operatorname{Re} z_{2}\right|}\right)=\cot \frac{\pi}{4022} \text {. }
$$

Next, prove: $c=\cot \frac{\pi}{4022}$ satisfies the problem's requirements.
For any
$$
\begin{aligned}
P(x)= & x^{2012} + a_{2011} x^{2011} + a_{2010} x^{2010} + \\
& \cdots + a_{0},
\end{aligned}
$$

by appropriately changing the signs of its coefficients, we can obtain the polynomial
$$
R(x)=b_{2012} x^{2012} + b_{2011} x^{2011} + \cdots + b_{0} \text {, }
$$

where $b_{2012}=1$; for $j=0,1, \cdots, 2011$,
$$
b_{j}=\left\{\begin{array}{ll}
\left|a_{j}\right|, & j \equiv 0,1(\bmod 4), \\
-\left|a_{j}\right|, & j \equiv 2,3(\bmod 4) .
\end{array}\right.
$$

We will use proof by contradiction to show that every root $z$ of $R(x)$ satisfies
$|\operatorname{Im} z| \leqslant c|\operatorname{Re} z|$.
Assume $R(x)$ has a root $z_{0}$ such that
$\left|\operatorname{Im} z_{0}\right| > c\left|\operatorname{Re} z_{0}\right|$,

then $z_{0} \neq 0$, and either the angle between $z_{0}$ and $\mathrm{i}$ is less than $\theta = \frac{\pi}{4022}$, or the angle between $z_{0}$ and $-\mathrm{i}$ is less than $\theta$.

Assume the angle between $z_{0}$ and $\mathrm{i}$ is less than $\theta$, the other case can be considered by the conjugate imaginary root of $z_{0}$.
Consider two cases.
(1) $z_{0}$ is in the first quadrant (or on the imaginary axis).
Let $\angle\left(z_{0}, \mathrm{i}\right)=\alpha<\theta$, where $\angle\left(z_{0}, \mathrm{i}\right)$ is the smallest angle to rotate $z_{0}$ counterclockwise to align with $\mathrm{i}$.
For $0 \leqslant j \leqslant 2012$, if $j \equiv 0,2(\bmod 4)$, then $\measuredangle\left(b_{j} j_{0}, 1\right)=j \alpha \leqslant 2012 \alpha < 2012 \theta$;
if $j \equiv 1,3(\bmod 4)$, then
$\measuredangle\left(b_{j} z_{0}, \mathrm{i}\right)=j \alpha < 2011 \theta$.
Thus, the principal argument of each $b_{j} z_{0}$ is in
$$
[2 \pi - 2012 \alpha, 2 \pi) \cup \left[0, \frac{\pi}{2} - \alpha\right] \text {. }
$$

This angular region has a vertex angle of
$2012 \alpha + \frac{\pi}{2} - \alpha = \frac{\pi}{2} + 2011 \alpha < \pi$.
Since $b_{j} z_{0} (0 \leqslant j \leqslant 2012)$ are not all zero, their sum cannot be zero.
(2) $z_{0}$ is in the second quadrant.
Let $\angle\left(\mathrm{i}, z_{0}\right)=\alpha<\theta$.
If $j \equiv 0,2(\bmod 4)$, then
$$
\measuredangle\left(1, b_{j} j_{0}\right)=j \alpha < 2012 \theta ;
$$

if $j \equiv 1,3(\bmod 4)$, then
$\measuredangle\left(\mathrm{i}, b_{j} j_{0}\right)=j \alpha \leqslant 2011 \alpha < \frac{\pi}{2}$.
Thus, the principal argument of each $b_{j} z_{0}$ is in $\left[0, \frac{\pi}{2} + 2011 \alpha\right]$.
Since $\frac{\pi}{2} + 2011 \alpha < \pi$, and $b_{j} 0_{0} (0 \leqslant j \leqslant 2012)$

are not all zero, their sum cannot be zero.
In conclusion, the smallest real number $c$ is $\cot \frac{\pi}{4022}$.