Suppose $p(x)$ and $q(x)$ are two quadratic polynomials in real coefficients such that: $$\lvert p(x) \rvert \leq \lvert q(x) \rvert ~ ~ ~ \text{for all} ~ x \in \mathbb{R} \tag{1}$$ Is the above a sufficient condition for the following extension to hold? $$\lvert p(z) \rvert \leq \lvert q(z) \rvert ~ ~ ~ \text{for all} ~ z \in \mathbb{C} \tag{2}$$ I have tried to find a counterexample but failed. In an attempt to study this conjecture, I found that if $p(x) = ax^2 + bx + c$ and $q(x) = a'x^2 + b'x + c'$ then $\lvert a \rvert < \lvert a' \rvert$ and $\lvert c \rvert < \lvert c' \rvert$. I also noticed that $p$ must have the same real roots as $q$, so that we can assume $q$ has no real roots. At that point I tried writing $z = u + vi$ and expanding $(2)$ but it got me nowhere, and I am at a loss as to what to try next?
Additionally, if the above is true, can it be generalized to all polynomials? Since if it is true for quadratics, and as it is true for constant and linear polynomials, there would seem to be the beginnings of a pattern...
$x^2 \le x^2 + 1$ for all $x \in \mathbb R$, but $|i^2| > |i^2 + 1|$.