More precisely, this is Theorem 16.13 of Bak's and Newman's Complex Analysis (Third Edition):
Suppose $f$ is an entire function of finite order. Then either $f$ has infinitely many zeroes or $$f(z) = Q(z) e^{P(z)}$$ where $Q$ and $P$ are polynomials.
To recall, an entire function $f$ is said to be of finite order if for some $k$ and some $R > 0$, $$|f(z)| \leq \exp(|z|^k)$$ for all $z$ with $|z| \geq R.$
Here is the proof of the theorem that is given in the book:
Suppose $f$ has a finite number of zeroes $\alpha_1,\dots,\alpha_k$. Then we may write $$f(z) = Q(z) g(z)$$ where $$Q(z) = (z-\alpha_1)\cdots(z-\alpha_k)$$ and $g$ is an entire function that is never zero. Thus we can define an entire function $$P(z) = \log{g(z)},$$ which by our hypothesis must satisfy $$|\operatorname{Re}{P(z)}| \leq |z|^k \quad \text{for }|z| \geq R $$ for some $k$ and $R$. Hence $P$ is a polynomial and $f(z) = Q(z) e^{P(z)}$ , as desired.
I have the following questions regarding this proof:
Why exactly is $g(z)$ an entire function (i.e. holomorphic on the whole complex plane)? I know that $g(z) = f(z)/Q(z)$, so why does the denominator $Q(z)$ not cause problems?
Why exactly is $P(z) = \log{g(z)}$ an entire function? It seems strange to me since the logarithm is not an entire function, so I cannot argue with the fact that the composition of entire functions is, again, an entire function.
I know that $|\operatorname{Re}{P(z)}| = \log{|g(z)|}$. So we obtain the inequality by using the fact that $g$ (and not $f$, as I thought at first) is an entire function. Is that correct?
Thank you in advance!
$Q$ has the same zeros as $f$(and with same multiplicities), so all the zeros are cancelled and the zeros of $Q$ are removable singularities(hence we just re-define $g$ as the limit at the singularity), and $g$ is analytic.
There must be a Theorem(or at least exercise) stating that if a function $g$ is holomorphic and nonzero on a simply-connected region $\Omega$, then there is a holomorphic branch of $\log g$(i.e. a holomorphic function $h$ on $\Omega$ s.t $g = e^h$ on $\Omega$). This can be proved in the same spirit as argument principle.
The inequality is from the fact that $f$ has finite order. Note that $1/Q$ vanishes(or is nonzero constant) near $\infty$, so it is bounded for some $|z|\ge R$.