There is something missing about the coefficients $a_j$. Because if just $a_j \in \mathbb{C}$, then such real number $c$ doesn't exist. For example $P(x)=x^{2012}-i x^{2011}$.

dgrozev wrote: There is something missing about the coefficients $a_j$. Because if just $a_j \in \mathbb{C}$, then such real number $c$ doesn't exist. For example $P(x)=x^{2012}-i x^{2011}$. The translation forgot that $P(x)$ is a polynomial with real coefficients.

Let plug $n$ instead of $2012$. We will prove that when $4 | n$, $c=\cot(\frac{\pi}{2(n-1)})$. Denote $D(c)=\{z\,|\,z\in \mathbb{C}, \Im z > c |\Re z| \} $. The problem could be restated as to find the min possible $c$, such that we can change the signs of some polynomial's coefficients so that no roots are in $D(c)$. Because $P(x)$ has real coefficients, $-D(c)$ will also be free of any roots, which is equivalent to the problem's statement. Notice also that $P(z)$ being monic is of no importance except that the coefficient at $x^n$ is not zero, i.e. omitting the word "monic" from the problem statement will change nothing. 1. Let $4 | n,\, \varphi =\frac{\pi}{2(n-1)}$ and $ P(x)$ is with real coefficients with degree $n$. Then we can change the signs of some polynomial's coefficients, so that the new polynomial has no roots in $D(\cot \varphi)$. Let $z=i.z',\, Q(z')=P(i.z'),\, O_{\varphi} = \{z'\,|\, |arg (z')| < \varphi \}$. We have: $Q(z') = Q_2(z')+i.Q_1(z')$ where $Q_2, Q_1$ are with the same real coefficiens as $P(z)$ but with some sign changes, $Q_2$ has only even degrees of $z'$, $Q_1$ only odd degrees of $z'$. Let now change coefficients signs of $Q_1,\, Q_2$ so that all coefficients become positive. So, we get new polynomial $Q'(z')=Q'_2(z')+i.Q'_1(z')$ where $Q'_1,\,Q'_2$ have positive coefficients. Suppose that $z'\in O_{\varphi}$ and $\arg (z') \geq 0$. In this case $0 \leq \arg (z'^m) < \frac{\pi}{2}$ when $0 \leq m < n$ and $0 < \arg (z'^n) < \pi$. This means that the arguments of all terms of $Q'(z')$ are between $0$ and $\pi$ and so their sum $Q'(z')$ can not be $0$. When $z'\in O_{\varphi}$ and $\arg (z') < 0$, the same reasons give that the arguments of all terms of $Q'(z')$ are between $-\pi $ and $0$ so $Q'(z')=0$ is impossible. 2. If $4 | n$ then $c=\cot(\frac{\pi}{2(n-1)})$ is sharp. In order to get a contradiction suppose that there exists constant $c_1 < c$ which satisfies the problem's conditions. Let's consider $p_1(z)=z^{n-1} - 1$ and $p_2(z)=z^{n-1} + 1$. Notice that both $p_1$ and $p_2$ have a root in $D(c_1)\cap \{z\,|\, |z| < 2 \} $, e.g. $p_1$ has a root: $e^{i \frac{n\pi}{2(n-1)} } = i.e^{i \frac{\pi}{2(n-1)}} $. Now if $\varepsilon >0$ is sufficiently small real, consider $P(z)=\varepsilon z^n + z^{n-1}+1 $. The idea is to show that $\pm\varepsilon z^n + p_1(z)$ and $\pm\varepsilon z^n + p_2(z)$ also have a root in $D(c_1)\cap \{z\,|\, |z| < 2 \}$ if $\varepsilon $ is sufficiently small. This can be obtained by applying Rouche's theorem.