Definition: Given a lattice $\Gamma \subset \mathbb C$, a Theta function $\vartheta: \mathbb C \to \mathbb C$ is a holomorphic function with the following property:
$$ \vartheta(z+ \gamma) = e^{2i\pi a_{\gamma}z+b_{\gamma}}\vartheta(z)$$
for every $\gamma \in \Gamma$, and $ a_{\gamma},b_{\gamma} \in \mathbb C$.
Exercise: A Theta function never vanishes iff $\vartheta(z) = e^{p(z)}$ with $p(z)$ a polynomial of degree at most 2.
Thoughts/Hint: The "only if" part is trivial. The hint is: show that $\log(\vartheta(z))=O(1 + |z|^2)$. I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don't get where the square comes from.
Thanks!
Look at the derivative of $\log \vartheta$, i.e. at the function
$$q \colon z \mapsto \frac{\vartheta'(z)}{\vartheta(z)}.$$
By assumption, $q$ is an entire function, and from the property of $\vartheta$ we find
$$\vartheta'(z+\gamma) = 2\pi i a_\gamma e^{2\pi i a_\gamma z + b_\gamma}\vartheta(z) + e^{2\pi i a_\gamma z + b_\gamma}\vartheta'(z),$$
whence
$$q(z + \gamma) = 2\pi i a_\gamma + q(z)$$
for each $\gamma \in \Gamma$.
From this relation deduce that $q$ is a polynomial of degree $\leqslant 1$. (Consider a fundamental parallelogram of $\Gamma$, and a corresponding basis $\omega_1,\, \omega_2$ of $\Gamma$.) Then, since we have $p' = q$ for $p$ such that $\vartheta(z) \equiv e^{p(z)}$, it follows that such a $p$ is a polynomial of degree $\leqslant 2$.