TITLE: Is there an analytic function with $f(z)=f(e^{iz})$?
QUESTION [12 upvotes]: Does there exist a non-constant analytic function $f$ satisfying
  $f(z)=f(e^{iz})$ ?

I don't know where to start.

REPLY [6 votes]: We are looking for $f$ with  $f=f\circ g$ where $g(z)= e^{iz}$. We notice that $g$ happens to have a fixed point $w\approx0.576+0.375 i$. As $|g'(w)|=|g(w)|=|w|<1$ it is an attracting fixpoint. Consequently, there exists a sequence $z_0,z_1,\ldots$ of points with $z_{n+1}=g(z_n)$ converging to $w$ (we just need to pick $z_0$ sufficiently close to but different from $w$). We conclude that $f$ is constant on the set $\{z_0,z_1,\ldots\}$ with accumulation point $w$ and so by the identity theorem $f$ is constant.
Note that this proof does not require $f$ to be entire, it suffices that its domain $D$ is connected and contains a neighbourhood of $w$ (also, we require $f(z)=f(g(z))$ only for $z\in D$ with $g(z)\in D$).