TITLE: j-invariant fixed point?
QUESTION [8 upvotes]: If we view the j-invariant of a lattice as a map from the upper-half plane to the complexes by $\tau\mapsto j([1,\tau])$, then it is surjective, holomorphic, and has quite a number of other wonderful properties (see the third part of Cox's Primes of the Form for a great introductory reference).  
My question is: does $j$ have any fixed points?  If so, do we know what any/all of them are?  
I'm in particular curious what goes into the proof.  Specifically, whether the answer is immediate from some complex analysis, or whether you need to have a good handle on $j$ itself (or both!). A professor I asked suggested thinking about $j(\tau)-\tau$ on the compactification of the fundamental domain of $SL(2,\mathbb{Z})$, but we weren't able to clean it up.

REPLY [3 votes]: As already pointed out by Alexandre Eremenko, the $j$ function is a finite type map  in the sense of my thesis. It follows that any open set intersecting the domain boundary (here the extended real line) contains infinitely many fixed points which are repelling in the sense that the derivative there has modulus greater than 1. 
The normalization of $j$ is irrelevant: for any Mobius transformation $M$, the composition $M\circ j$ is also a finite type map.