I'm going over a passed exam paper and the part a is to define Hadamard's Factorisation theorem:
Let $f$ be entire of order $ρ$ such that $f(0) \neq 0$. Then
$$f(z) = P(z)e^{Q(z)}
$$
where $P(z)$ is the canonical product formed with the zeros of $f$ and $Q(z)$ is a polynomial
of degree at most $ρ$.
So I assume I will need to use it for part b:
Let $p(z)$ be a polynomial for which $p(0) \neq 1$. Show that the equation
$\cosh(z) = p(z)$ has infinitely many solutions.
But $cosh(z)=0$ doesn't exist. How far I've gotten:
$$\cosh(z)=P(z)e^{Q(z)}$$
as $\cosh$ is order 1, $Q(z)$ is a polynomial of order 1
$$\cosh(z)=P(z)e^{bz}$$
I don't know where to go next.
Best Answer
In order to show that $\cosh(z) = p(z)$ has infinitely many solutions you need to consider the factorization of $\cosh(z) - p(z)$, not that of $\cosh(z)$.
Assume that there are only finitely many solutions. Then $$ \cosh(z) - p(z) = \frac 12 (e^z + e^{-z}) - p(z) = P(z) e^{bz} $$ with a polynomial $P$ and a constant $b \in \Bbb C$.
If $\operatorname{Re}b \le 0$ this implies that $e^x$ has at most polynomial growth for $x \in \Bbb R$, $x \to + \infty$, which is not possible.
If $\operatorname{Re}b \ge 0$ then you get a similar contradiction for $x \in \Bbb R$, $x \to - \infty$.