Existence of real roots of a polynomial.

Question

Suppose we have a $P(z) = \sum_{i = 0}^{m} a_{i}z^{i}$ and this polynomial has $m$ real roots. Is it true or not that $P(z+qi) + P(z-qi)$ also has $m$ real roots ?

I still doesn't find any counter-examples. Any hints?

Answer 1

Suppose $P(z)=(z-r_1)\ldots(z-r_m)$. Let $z$ be a root of $P(z+qi)+P(z-qi)$ that is not real and suppose $q>0$ (the proof for $q<0$ is similar). Write $z=x+iy$ with $y> 0$ (the other case follows by using $\bar z$ instead of $z$). One has $$(z-r_1+qi)\ldots (z-r_m+qi)=-(z-r_1-qi)\ldots (z-r_m-qi),$$ and so $$\|z-r_1+qi\| \ldots \|z-r_m+qi\|=\|z-r_1-qi\|\ldots \|z-r_m-qi\|.$$ We show $\|z-r_k+qi\|<\|z-r_k-qi\|$ to derive a contradiction. Note that $$\|z-r_k+qi\|^2=\|(x-r_k+(y+q)i\|^2=(x-r_k)^2+(y+q)^2>(x-r_k)^2+(y-q)^2 \geq \|z-r_k-qi\|^2.$$