# Show basin of attraction has only one connected component

L.S.,

This is a homework question I find hard to answer, any help/hints would be greatly appreciated!

I have to prove that every complex polynomial of degree 2 with an attracting fixed point has a basin of attraction in the complex plane that has only one connected component.

My thoughts:

Only one connected component means the whole basin is connected, so maybe I can find an contradiction by first stating there are open $U$ and $V$ in $A$ that are disjunct and together make up $A$?

$|f'(z)| < 1$, because $f$ holomorf.

$A_0$ = $A$.

There is a critical point in $A_0$ (where $f'(z_{cr}) = 0)$

But I don't know how to put these facts into good use!

Thank you very much

• You didn't use the polynomial of degree 2 yet... If you look at $f^\prime(z)$, how many critical points are there? Can a critical point lie both in $U$ and $V$? Sep 13, 2013 at 14:43
• Thanks! Polynomial of degree $2 => 1$ critical point. This point cannot lie in $U$ and $V$, since $U$ and $V$ have to be disjoint! But still, how do we know there isn't another component of $A$, that doesn't have the fixed point in it (and doesn't have to have a critical point in it)? Sep 13, 2013 at 16:35

Let $A$ be the basin of attraction of the fixed point, and let $A_0\subseteq A$ be the immediate basin. We know that $A = \bigcup_{n\geq 0} f^{-n}(A_0)$, so it suffices to show that $f^{-1}(A_0) = A_0$. To prove this, consider the following questions:
1. For any point $x\in A_0$, how many preimages does $x$ have under the polynomial $f$?
2. The map $f\colon A_0\to A_0$ is a ramified covering map. What is its degree? Here's where you might want to know the critical point is contained in $A_0$.