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?



2 Answers 2


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.

  • $\begingroup$ that's really what I wanted. $\endgroup$
    – Herband
    Nov 27, 2011 at 18:03

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.

  • $\begingroup$ @Leonardo, thank you very much. $\endgroup$
    – Herband
    Nov 27, 2011 at 17:59
  • 1
    $\begingroup$ You're welcome; but I think Harry answered your question more thoroughly. $\endgroup$
    – student
    Nov 27, 2011 at 18:02
  • 2
    $\begingroup$ Dear Leandro, your sense of modesty and fair-play is quite welcome, but your answer is very nice too: +1. $\endgroup$ Nov 27, 2011 at 21:05
  • $\begingroup$ Dear Leandro, Sorry, for unaccepting, but I have +'ed all your answers and your questions on MAth.SE to encourage you answering more... $\endgroup$
    – Herband
    Nov 28, 2011 at 14:46

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