Poincaré's functional equation

Remarks

Example: $f(w) = w^2 + 2w$ has $f(0) = 0$ and $\lambda = f'(0) = 2$. The function $\Psi(z) = e^z - 1$ satisfies $\Psi(0) = 0$, $\Psi'(0) = 1$ and $f(\Psi(z)) = (e^z - 1)^2 + 2(e^z - 1) = e^{2z} - 1 = \Psi(2z)$.

Non-example: for the same $f$, the function $\Psi(z) = z$ fails, since $f(z) = z^2 + 2z \ne 2z$ for $z \ne 0$.

If $\Phi$ is a solution of Schröder's equation $\Phi \circ f = \lambda \Phi$ with $\Phi(0) = 0$ and $\Phi'(0) = 1$, then its local inverse $\Psi = \Phi^{-1}$ solves Poincaré's equation. So for $\abs{\lambda} > 1$ a unique solution exists, by the Koenigs linearization theorem applied to $f$. Assume moreover that $f$ is entire. Then the formula $\Psi(z) = f^n(\Psi(z / \lambda^n))$, with $n$ so large that $z / \lambda^n$ lies in the domain of $\Psi$, extends $\Psi$ to an entire function. This function is the Poincaré function of $f$ at $0$.