Prove that if $\,\lvert\, f(\,f(z))\rvert>r,\,$ then $f$ is constant Let $r>0$. Prove that if $f$ is holomorphic on a whole complex plane and $|f(f(z))|>r$ for all $z\in\mathbb{C}$, then $f$ is constant. Can sb point me in the right direction?
 A: Let $g(z)=f\big(f(z)\big)$, then $g$ is also entire analytic, so is $h(z)=1/g(z)$, as $\lvert g(z)\rvert\ge r>0,\,$ for all $z\in\mathbb C$.
But $h$ is bounded by $1/r$, and hence constant due to Liouville's Theorem. Thus $g$ is also constant.
So $f\big(f(z)\big)=c$, for some $c\in\mathbb C$. If $f$ is not constant, then $f$ is open (Open Mapping Theorem), and hence $D(a,R)\subset f[\mathbb C]$, for some $a\in\mathbb C$ and $R>0$. But $f\big(f(z)\big)=c$ implies that $f$ is constant on $D(a,R)$, and hence it is constant everywhere.
Αvoiding the Open Mapping Theorem. If $f\big(f(z)\big)=c$, then $f\big[\,f[\mathbb C]\big]=\{c\}$.
But if $f$ is entire and not constant,then $f[\mathbb C]$ is dense in $\mathbb C$ - If not, then there exists disc $D(a,R)$, such that $D(a,R)\cap f[\mathbb C]=\varnothing$, and hence, 
$$
\left|\frac{1}{f(z)-a}\right|\le\frac{1}{R}.
$$
But $1/(f(z)-a)$ would be entire and bounded, and hence constant, and so would be $f$. Thus
$D=f[\mathbb C]$ is dense in $\mathbb C$. But then $f[D]=\{c\}$, which means that the roots of $g(z)=f(z)-c$ have a limit point, and thus (Identity Theorem) $g\equiv 0$ or $f\equiv c$.
A: Proof. According to Picard's theorem non-constant entire holomorphic function take all complex values, except maybe one. Then $f(f(z))$ have to take all values except maybe two complex numbers. But there are infinitely many complex numbers $z$ with property $|z| < r$. Contradiction.
Example. $e^z$ takes all values except $0$. $e^{e^z}$ takes all values except $0$ and $1$.
