TITLE: Is a meromorphic function satisfying $f(2z)=\frac{f(z)}{1+f(z)^2}$ constant?
QUESTION [6 upvotes]: Let $f(z)$ be a holomorphic function on the unit disk satisfying $f(0)=0$ and $$f(2z)=\frac{f(z)}{1+f(z)^2}.$$
Extend it to a meromorphic function on the entire complex plane using this recursion. Must $f(z)$ be constant? I think so, but I can't prove it.
Substituting $g(z)=1/f(z)$ and rearranging yields
$$g(2z)=g(z)+\frac{1}{g(z)},$$
which seems useful, but I'm not sure how to proceed after that.

REPLY [7 votes]: Write $f(z)=z^kg(z)$ with $k>0$, $g$ holomorhic, $g(0)\ne0$.
Then $$2^kg(2z)=\frac{g(z)}{1+z^{2k}g(z)^2}$$
which gives a contradiction if we let $z=0$.

REPLY [4 votes]: Since $f(0)=0$, write $f(z)=a_nz^n+O(z^{n+1})$ for some $n>0$.
Then
$$
\frac{f(z)}{1+f(z)^2}=(a_nz^n+O(z^{n+1}))(1-a_n^2z^{2n}+O(z^{4n}))=a_nz^n+O(z^{n+1})
$$
and
$$
f(2z)=2^na_nz^n+O(z^{n+1})
$$
which after equating coefficients implies that $a_n=0$. Since $n$ was arbitrary, $f$ must be identically zero.