Order of growth of $\frac{1}{f}$ An entire function is of finite order $\rho$ if $$\rho = \inf \{\lambda \geq 0  \ | \ \exists A, B > 0 \ s.t. \  |f(z)|\leq Ae^{B|z|^{\lambda}} \forall z \in \mathbb{C} \}$$
My question : if $f$ is of finite order $\rho$, $f(z) \neq 0 \ \forall z \in \mathbb{C}$, is $\frac{1}{f}$ of finite order $\rho$ ?

If $f$ is an entire zero-free function of finite order $\rho$, then by Hadamard's theorem
$$f(z) = e^{P(z)}$$
with a polynomial $P(z)$, and $\rho = \deg P$. Then
$$\frac{1}{f(z)} = e^{-P(z)}$$
is also of finite order $\deg (-P) = \deg P$.
