Prove the existence of a specific conformal mapping

Question

Let $U$ be an open set containing $0$ and $f:U \rightarrow C$ a holomorphic function such that $f(0)=0$ and $f^{'}(0)=2$.Prove that there exists an open neighbourhood $0 \in V \subset U $ and a holomorphic injective function $h:V \rightarrow V$ such that $h(f(z))=2h(z)$. Since I don't have any idea where to start, I'd appreciate a small hint rather then a full solution. Thank you for all your answers.

Answer 1

Define a sequence $z_0=z$, $z_{n+1}=f^{-1}(z_n)$. We need $$ h(z) = 2h(z_1) = 4h(z_2) = \ldots $$ We know that $h(w)\approx c\cdot w$ for small $w$; try $$ h(z) = \lim 2^n z_n. $$

Answer 2

One way to do this is to use power series. Assume that $h$ has a power series, and solve for its coefficients in terms of the coefficients of the power series for $f$ using the equation $h(f(z))=2h(z)$. You then just have to check that the coefficients have a nonzero radius of convergence and so actually do define a function.