Prove that for any $\lambda \neq 0$ and any polynomial $p(z) \not\equiv 0,$ the function $g(z)=e^{\lambda z}-p(z)$ has infinitely many zeros.
My approach: Suppose to the contrary that $g$ has finitely many zeros, at $a_j$'s (say)
$$g(z)=\prod_{j=1}^{n} (z-a_j).$$
By Hadamard's factorization
$$g(z)=e^{h(z)} \cdot \prod_{j=1}^{n} (z-a_j),$$
for some polynomial $h(z).$ I don't know how to get a contradiction. Any help is much appreicated.
Best Answer
The same approach as in Showing that $e^z=z$ has infinitely many solutions can be used:
$h(z) = p(z) e^{-\lambda z}$ has an essential singularity at $z = \infty$, and has only finitely many zeros. Using Great Picard's Theorem it follows that $h$ takes any non-zero value infinitely often. In particular, $h(z) = 1$ has infinitely many solutions.
This is the desired conclusion because $h(z) = 1 \Longleftrightarrow e^{\lambda z} = p(z)$.