Given $z_1,z_2\in\mathbb{C}\setminus\mathbb{R}$ with $\arg(z_1)\neq\arg(z_2)$ and $|z_1|=|z_2|$ and given $x\in\mathbb{R}\setminus\{0\}$.
Is it necessarily true that $(x+z_1)^n+(x-z_2)^n\notin\mathbb{R}$ for all integer $n\gt0$?
I encountered this hypothesis while trying to formulate a cleaner solution to an old question.
On
WLOG $z_1=r(\cos p+i\sin p) ;z_2=r(\cos q+i\sin q)$
$$(x+z_1)^n=(x^2+r^2+2rx\cos p)^{n/2}e^{n i\arctan\dfrac{r\sin p}{x+r\cos p}}$$
So, if $f(n)=(x+z_1)^n+(x-z_2)^n$ is real
$$\cos p=-\cos q$$
$$q=2m\pi+\pi\pm p$$
$$f(n)=(x^2+r^2+2rx\cos p)^{n/2}\left(e^{i\arctan\dfrac{r\sin p}{x+r\cos p}}+e^{i\arctan\dfrac{-r\sin(\pi\pm p)}{x-r\cos(\pi\pm p)}}\right)$$
So, we need $$\arctan\dfrac{r\sin p}{x+r\cos p}=-\arctan\dfrac{-r\sin(\pi\pm p)}{x+r\cos p}$$
which is true if $q=2m\pi+\pi-p$
$\implies z_2=r(-\cos p+i\sin p)$
No. Take, for instance, $z_1=1+i$, $z_2=-1+i$, $x$ is any real number, and $n$ is any natural number, then$$(x+z_1)^n+(x-z_2)^n\in\Bbb R.$$