It's true that if an $f:\mathbb{C} \to \mathbb{C}$ is an odd polynomial and $g:\mathbb{C} \to \mathbb{C}$ is an even polynomial are polynomials with reals coeficients and they share a root, then that zero must have a positive real part ?
I was wondering about that and have a feeling that I have done a mistake
first: for a very higher degree polynomials they must share a root
if $z = re^i\theta$ is root of $g$, then let's calculate $g(3z)$.
So $$g(3e^{i\theta}) = (3r)^n( \frac{a_n}{(3r)^n}e^{ni\theta} + \frac{a_{n-1}}{(3r)^{n-1}}e^{(n-1)i\theta} + \ldots )$$
$\Rightarrow$
$$(3r)^n\|( \frac{a_n}{(3r)^n}e^{ni\theta} + \frac{a_{n-1}}{(3r)^{n-1}}e^{(n-1)i\theta} + \ldots)\| \leq (3r)^n \|\frac{a_n}{(3r)^n}e^{ni\theta}\| + \|\frac{a_{n-1}}{(3r)^{n-1}}e^{(n-1)i\theta}\| + \ldots $$$$\leq\|a_n\|+\|a_{n-1}\| + \|a_{n-2}\|\ldots $$
$\Rightarrow$
$$\|g(3e^{i\theta})\| \leq (3r)^n\|g(e^{i\theta})\| $$
We can exchange $3$ for any real number $l$, so looks like, for very big polynomial , the root lies somewhere near another root if we take $0<l<1/r$ . Since $\|e^{i\theta}\|=\|e^{-i\theta}\| = \|- e^{i\theta}\|$, appears to have a high chance that, for very large degree, an odd polynomial share a root with an even polynomial and the whole thing looks like being converge to real line
Looks too good to be true