Does $$\operatorname{Re}((a+bi)^{n})=\operatorname{Re}((a+bi)^{n+1})$$ have a solution $a,b\in\mathbb{Q},|a|\neq |b|\neq0$, for all $n\in\mathbb{N}$?
I was working on my previous question, $\operatorname{Re}(a+bi)^m=\operatorname{Re}(a+bi)^n$, then all of the solutions I could find on PC was the case of $n-m=1$ and trivial cases. Is there a reason why many solutions can be found for $n-m=1$, and are there solutions for all $n\in\mathbb{N}$?
Let $w = x + y \, \mathrm i$ for $x,y \in \mathbb Q$ such that $w^n+\overline w^n \neq 0$ and $w^{n+1}+\overline w^{n+1} \neq 0$. Take $$z = w \cdot \frac{w^n+ \overline{w}^n}{w^{n+1}+\overline w^{n+1}}.$$ Then $z$ has rational real and imaginary parts and $$\operatorname{Re}(z^{n+1}) = \operatorname{Re}(z^n).$$ Note that $z$ only depends on the direction of $w$: multiplying $w$ by a non-zero rational factor results in the same $z$.
