Hard polynomial question. I stuck with the following question.

Let $n\geq 1$, $A, B\in\mathbb{R}[x]$ be polynomials of degree $n+1$ and $n$, respectively. Let $a_i$ and $b_i$ be its roots, suppose they are all real and satisfy:
  $$a_1<b_1<a_2<b_2<\cdots<a_n<b_n<a_{n+1}.$$
  Show that one of the functions $E(z)=A(z)+iB(z)$ or $E(z)=A(z)-iB(z)$ satisfy the inequality
  $$|E(x-yi)|<|E(x+yi)|,$$
  for all $x,y\in\mathbb{R}$ with $y>0$.

I tried the particular case with $A(z)=z^2-6z+8$ and $B(z)=z-3$ and it worked but I couldn't find a pattern. Can anyone help?
 A: Let's define $E_+(z) := A(z) + iB(z)$, and $E_-(z) := A(z) - iB(z)$. Note that since all $a_i,b_j$ are real, $A(\bar{z}) = \overline{A(z)}$, and similarly for $B$.
Notice the following:
$$E_+({z}) = A(z) + iB(z) = \overline{A(\bar{z}) - i B(\bar{z})} = \overline{E_-(\bar{z})}$$
Similarly, one may prove that 
$$ E_+(\bar{z}) = \overline{E_-(z)}\\ \overline{E_+(z)} = E_-(\bar{z}) \\\overline{E_+(\bar{z})} = E_-(z).$$
Consequently,
$$\frac{|E_+(\bar{z})|}{|E_+(z)|} \cdot \frac{|E_-(\bar{z})|}{|E_-(z)|} = \frac{|E_+(\bar{z})|}{|E_+(z)|} \cdot \frac{ \left| \overline{E_+({z})}\right|}{\left| \overline{E_+(\bar{z})}\right|} = 1$$

To get to the strict inequality, we prove that $\frac{|E_+(\bar{z})|}{|E_+(z)|} \neq 1$ over the given set. To begin with,
\begin{align}|E_+(z)|^2 &= \left(A(z) + iB(z)\right)(A(\bar{z}) - iB(\bar{z}) )\\
&= \underbrace{|A(z)|^2 + |B(z)|^2}_{R(z)} +  \underbrace{i\left(A(\bar{z})B(z) - A(z) B(\bar{z}\right)}_{S(z)}) \\
&= R(z) + S(z)\end{align}
Similarly,
$$|E_+(\bar{z})|^2 = R(z) - S(z)$$
So all we have to show is that $S(z) \neq 0$ for all $z: \Im(z)>0$
$$S(z) = 0 \iff A(z)B(\bar{z}) = A(\bar{z}) B(z) \iff \Im\left(\frac{A(z)}{B(z)}\right) = 0$$
The above, however, is impossible for $y > 0$. This is most easily shown geometrically - on the argand plane, draw lines joining each of the $a_j$ and $b_j$ to $z$. Lastly, mark a point $X$ at $(p,0)$ with $p>a_{n+1}$. Then the argument of $\frac{z-a_i}{z-b_i}$ is equal to $\angle ZA_iX - \angle Z B_i X = -\angle A_iZB_i$, which is negative since $a_i < b_i$. Thus, $\arg(A(z)/B(z)) = \angle ZA_{n+1}X -\sum_i \angle A_i Z B_i$. However, $\angle ZA_{n+1}X = \angle ZA_1X +\angle A_1ZA_{n+1}$ by the exterior angle relation and $\sum_i \angle A_iZB_i< \angle A_1ZA_{n+1}$. Lastly, note that these angles are all contained in the triangle $A_1ZA_{n+1}$. This implies that $\angle ZA_1X \le \arg(A/B) \le \pi$. Lastly, since $y >0$, $\angle ZA_1X >0$, and $\arg(A/B) < \pi$, and the conclusion follows.
