How to find the initial point whose sequence of iterates converges {$z_n$}? Consider the iterration of rational function $R$($z$),
Suppose the sequence of iterates {$z_n$} of initial point $z_0$ converges to $w$. Then (because $R$ is continuous at $w$),
$w$ = $\displaystyle \lim_{n \to \infty}$$z_{n+1}$ = $\displaystyle\lim_{n \to \infty}$$R$($z_{n}$) = $R$($\displaystyle\lim_{n \to \infty}$$z_n$) = $R$($w$)
From the above equation, it is clear that the sequence of iterations of any initial points always converges to the fixed points of that function.
e.g. Consider, $R$($z$) = $z^2$$ -$ $4$$z$ $ + $$6$ , It has two fixed points $2$ and $3$ (here, we are not considering $\infty$ as fixed point)
Here, I want to know for which initial point $z_0$ its sequence of iterates will converge to $2$? How to calculate these initial points mathematically?
 A: You're continuity argument shows that if the sequence of iterates of $R$ converges, then the limit must be a fixed point. It's a stretch, though, to say that "sequence of iterations of any initial points always converges to the fixed points of that function." Indeed, the sequence of iterates many not converge at all and the exact behavior may be quite complicated. The example involving $z^3-1$ in this answer indicate that the problem of determining the basin of attraction of a fixed point may be quite difficult.
For your particular function $R(z)=z^2-4z+6$ though, a simple computer experiment indicates that the basin of attraction might be circle. You can generate the Julia set of most polynomials using this webpage, for example. Once you suspect that, it's not too hard to prove. To do so, I suggest you rewrite your function as
$$
R(z) = (z-2)^2+2
$$
and consider the behavior of points for $|z-2|<1$ vs $|z-2|>1$. You've probably studied the dynamics of $f(z)=z^2$ and should find some similarities.
A: Consider the recursive sequence $z_{n+1} = \cos z_n$, converging to some $w \in (0,1)$. You can prove (using the fixed point theorem) that the sequence will converge to $w$, regardless of the choice of $z_0$. So you see that your question must be clarified.
Another example, the picture below represents the convergence pattern of Newton's method applied to the numerical solution of $z ^3 -1=0$. Each point is coloured according to which root newton's method (starting there) is converging to.
