Let $f : \mathbb C \to \mathbb C$ be an entire function with a separated zero set, i.e. there is a $\delta>0$ s.t. $|z-z'| > \delta$ for every distinct zeros of $f$. Further, suppose that all zeros of $f$ are simple and
$$
|f(z)| \geq C\_\varepsilon e^{a|z|^2}, \quad z \in Z\_\varepsilon
$$
where $Z\_\varepsilon$ denotes all complex numbers with distant $\geq \varepsilon$ to the zero set of $f$. Hence, on such a set, $f$ grows (up to constants) precisely like $e^{a|z|^2}$.


I was wondering if this implies a lower bound on $|f'(z)|$ for zeros $z$ of $f$. Intuitively, $|f'(z)|$ should grow rather fast if the zero $z$ is far away from the origin. Are there results in this direction or is there a way to show a growth estimate?



Lower bounding the derivative of a simple zero of an analytic function
Sure enough, the idea is that if $f(z\_0) = 0$ then $g(z) = \frac{f(z)-f(z\_0)}{z-z\_0}$ is an entire function which is non-vanishing and for which we know the lower bound on the circle $|z-z\_0| =\frac{\delta}{2} = r$. Moreover, $g(z\_0) = f'(z\_0)$. But then on this disk the function $u(z) = \log g(z)$ is harmonic, thus we have the mean-value property:


$$\log |f'(z\_0)| = \frac{1}{2\pi r}\int\_{|z-z\_0| = r} \log |g(z)||dz| \ge \frac{1}{2\pi r} \int\_{|z-z\_0|=r} a|z|^2 + \log (C\_r) - \log |z-z\_0||dz|.$$


Last two terms will give us a constant independent of $z\_0$, so we are left with the integral


$$\frac{1}{2\pi r}\int\_{|z-z\_0|=r} |z|^2|dz| = \frac{1}{2\pi r}\int\_{|w| = r} |w|^2 + |z\_0|^2 + 2Re(w\bar{z\_0})|dz| = \frac{r}{2\pi} + |z\_0|^2,$$
where the integral $\int\_{|w| = r}2Re(w\bar{z\_0})|dz|$ is zero by, say, symmetry.


Collecting everything we have $\log |f'(z\_0)| \ge c + a|z\_0|^2$ for some absolute constant $c$, thus $|f'(z\_0)| \ge e^c e^{a|z\_0|^2}$. This should be optimal up to possibly a power of $z\_0$.


Although I used some shortcuts based on the chosen function $|z|^2$, this method works in a very big generality, say for all functions of finite order, you just have to play with the radius of the circle you consider.


