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Suppose, I have a function $f(z)=\xi z$ where $|\xi|=1$ but $\xi$ is not a root of unity. Then, from the fact that the $n$-th iteration $f^{\circ n}(z)$, $z \in \mathbb{C}$, is dense on the circle around $0$ and radius $|z|$, how can I can deduce that $\{f^{\circ n}\}$ is a normal family?

Thanks for any help!

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up vote 3 down vote accepted

According to the Arzelà–Ascoli theorem (see, e.g., Ahlfors, Complex analysis, chapter 5, section 4.3) a family of continuous functions $f_\alpha: \Omega\to{\mathbb C}$ is normal iff it is equicontinuous on every compact set $E\subset\Omega$. Now given a $\xi$ on the unit circle one has

$$|f^{\circ n}(z_1)-f^{\circ n}(z_2)|=|\xi|^n\ |z_1-z_2|=|z_1-z_2|$$

for all $z_1$, $z_2\in\Omega:={\mathbb C}$, which shows that the condition of the theorem is more than fulfilled whether $\xi$ is a root of unity or not.

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