I'm studying the book Iterations of Rational Functions by Alan F. Beardon.
Here is the page that all the quotations come from.
It is said that:
If $z$ is close to the fixed point $\zeta$, then, approximately, $|R(z)-\zeta|=|R(z)-R(\zeta)|=|R'(\zeta)|\cdot |z-\zeta|$.
My question is about this affirmation:
If $z_0$ lies sufficently close to an attraction fixed point $\zeta$, then $z_n \rightarrow \zeta$ as $n \rightarrow \infty$.
It makes sense that $z_0$ really close to $\zeta$ implies that $|z_0-\zeta|$ is really small but I can't prove that so easy.
These are my thoughts about it:
This equalty $|R(z)-\zeta|=|R(z)-R(\zeta)|=|R'(\zeta)|\cdot |z-\zeta|$ comes from the derivative right?
So for an arbitrary $\epsilon > 0$ there is a $\delta>0$ that: $|z-\zeta|< \delta \implies |R(z)-R(\zeta)|<\epsilon$.
To find the $z_0$ we must consider the "distance" from $z_0$ to $\zeta$ to all $\delta$ ?
But I can't find a way to choose the $z_0$.