On real roots of a polynomial in the form $ap(x)+bq(x)$

Question

Assume $m\in \mathbb{N}$, $a,b\in \mathbb{R}$, $a\neq 0$ and $(1+ix)^m=p(x)+iq(x)$.

Can we conclude that all roots of the polynomial $ap(x)+bq(x)$ are real numbers?

Answer 1

Set $x=\tan u$ then $p(x)+iq(x)=c(u)·(\cos (mu)+i\sin(mu))$ and $$ ap(x)+bq(x)=c(u)·r·\sin(mu+\phi) $$ where $a=r\sinϕ$ and $b=r\cosϕ$. Thus the roots are determined as $$ x_k=\tan u_k\text{ where } u_k=\frac{-\phi+\pi·k}m, \ k\in \Bbb Z,\text{ and } u_k\in(-\tfrac\pi2,\tfrac\pi2) $$ Now you have to argue that there are $m$ different roots in this set.