$\def\Im{\mathrm{Im}}$Let $a,z,a_1,a_2,\cdots,a_n\in \mathbb {C}$ such that $\Im(a_1),\Im(a_2),\cdots,\Im(a_n) $ have the same sign and $$a(z+a_1)(z+a_2)\cdots(z+a_n)+(z+\overline a_1)(z+\overline a_2)\cdots(z+\overline a_n)=0.$$ Prove that $z\in \mathbb{R}$ only and only if $|a|=1$.
Any ideas please?
One direction is easy:
We have $$a\prod_{i=1}^n(z+a_i)=-\prod_{i=1}^n(z+\bar{a_i}). $$ Taking norm: $$|a|\prod_{i=1}^n|z+a_i|=\prod_{i=1}^n|z+\bar{a_i}|. $$ Now if $z$ is real, then $|z+a_i|=|\overline{z+a_i}|=|z+\bar{a_i}|$ thus $|a|=1$.
For other direction, I have no idea!