What can be said about $\lim_{n\to\infty}\frac{\alpha z^{n+1}+\overline{\alpha}\overline{z}^{n+1}}{\alpha z^n+\overline{\alpha}\overline{z}^n}$?

Question

Let $z\in\mathbb{C}$; then what can be said about the precise value of $$\lim_{n\rightarrow\infty}\frac{\alpha z^{n+1} + \overline{\alpha}\overline{z}^{n+1}}{\alpha z^n + \overline{\alpha}\overline{z}^n}$$ (In the special case I am dealing with, $z$ is a root of a polynomial $p(z)$ and $$\alpha = \prod_{\xi\neq z:p(\xi)=0}\left(\xi - z\right)^{-1}$$ Some observations:

• the numerator and denominator resemble the solutions to a linear recurrence relation
• the quotient is equal to $$\frac{\Re\left(\alpha z^{n+1}\right)}{\Re\left(\alpha z^{n}\right)}$$

Answer 1

Using the polar representation,

$$\dfrac{|\alpha||z|^{n+1}\cos(\angle\alpha+(n+1)\angle z)}{|\alpha||z|^{n}\cos(\angle\alpha+n\angle z)}=|z|\left(\cos(\angle z)-\tan(\angle\alpha+n\angle z)\sin(\angle z)\right) \\=\Re(z)-\tan(\angle\alpha+n\angle z)\,\Im(z).$$

This does not converge.