Fixed points of a holomorphic map on a simply connected domain Given a holomorphic map $f: \Omega\to \Omega$, where $\Omega$ is a simply-connected domain in $\mathbb{C}$, is the number of fixed points at most $1$ if $f$ is not the identity map? How many could they be?
By the Riemann Mapping Theorem, I am able to reduce the problem to finding a fixed point of a holomorphic map from the unit disc to itself. How should I proceed?
Thanks.
 A: The number of fixed points for such a holomorphic map doesn't have to be finite.
Consider for instance the mapping $f : \mathbb{C} \rightarrow \mathbb{C}$ given by $f(z) = z\cos(z)$. Then a fixed point of $f$ is just a point $z$ where $\cos(z) = 1$, and we know there are infinitely many such $z$.
However, the set of fixed points cannot have an accumulation point, for otherwise the Identity Principle applies and $f$ must be the identity map. This at least implies that the number of fixed points must be countable, for every uncountable subset of $\mathbb{C}$ has an accumulation point.
A: If $\Omega=\mathbb{C}$, there can be arbitrarily many (consider $z\mapsto z^n$).  (Edit: See also Leandro's answer for more about this case.)
However if we disallow this and require $\Omega\ne \mathbb{C}$, then unless f is the identity there can be at most 1.  For if $\Omega \ne \mathbb{C}$ and f has a fixed point, then by the Riemann mapping theorem we can assume $\Omega=D$, the open unit disk, and applying fractional linear transformations we can assume the fixed point is 0.  Then the Schwarz lemma shows that if there are any other fixed points, f must be of the form $z\mapsto cz$ for some c, and the only way this can have additional fixed points is for c to be 1, making f the identity.
