The Jensen’s inequality bounds the number of zeros of an analytic function in a small disc in terms of size of the function in a slightly larger disc. Although Jensen’s inequality is a simple consequence of Jensen’s formula but we will provide a different proof based on Blaschke factors.
Theorem. Let $f$ be a function that is analytic on an open set containing the closed disc $\overline{D(0, R)}$. If $|f(z)| \leq M$ for $z \in \overline{D(0, R)}$ and $f(0) \neq 0$, then for $r < R$ the number of zeros of $f$ counted with multiplicity in the disc $\overline{D(0, r)}$ does not exceed
\[
\frac{\log (M/|f(0)|)}{\log (R/r)}.
\]
Proof. Let $z_1, \dots, z_K$ be the zeros of $f$ in the open disc $|z| < R$ counted with multiplicity. Note that there can only be finitely many zeros of $f$ in the disc $D(0, R)$ since a complex analytic function that is not identically zero cannot have infinitely many zeros in a compact set. We now define
\[
g(z) = f(z) \prod_{k = 1}^{K} \frac{R^2- z\overline{z}_k}{R(z- z_k)}.
\] Note that $g$ is analytic on the domain of $f$. Each factor in the product above in the definition of $g$ has the property that it has a pole at $z_k$ and that it is $1$ whenever $|z| = R$. The later can be seen by observing that
\[ \left| \frac{R^2- z \overline{z}_k}{R(z- z_k)} \right|^2 = \frac{R^4- 2 R^2 \text{Re}(z \overline{z}_k) + |z|^2|z_k|^2}{R^2 |z|^2- 2 R^2 \text{Re}(z \overline{z}_k) + R^2 |z_k|^2} = 1 \] as $|z| = R$. Thus for $|z| = R$ we have
\[
|g(z)| = |f(z)| \prod_{k = 1}^{K} \left| \frac{R^2- z \overline{z}k}{R(z- z_k)} \right| = |f(z)| \leq M.
\] Hence by the maximum modulus principle we have $|g(0)| \leq M$ and consequently
\[
|g(0)| = |f(0)| \prod_{k = 1}^{K} \frac{R}{|z_k|} \leq M.
\] Each of the factor in the product above is $\geq 1$ as $|z_k| < R$. Moreover, if $|z_k| \leq r$, then $R/|z_k| \geq R/r$. Let $L$ be the number of zeros of $f$ in the disc $\overline{D(0, r)}$ counted with multiplicity. Then we obtain that $|g(0)| \geq |f(0)| (R/r)^L$. This leads to the desired bound.
Using Jensen’s inequality it can be easily proven that a polynomial of degree $n$ with complex coefficients has at most $n$ roots counted with multiplicity.
We now proceed to prove a result which provides bounds for a complex analytic function and its derivatives based on one-sided bounds for its real or imaginary part in a slightly larger region. The key idea is to bound Taylor coefficients, the modulus of which can be expressed as a linear combination of integrals in Cauchy integral formulæ.
Theorem. (Borel-Carathéodory lemma) Suppose that $h$ is analytic in an open set containing the disc $\overline{D(0, R)}$, that $h(0) = 0$, and that $\text{Re}( h(z)) \leq M$ for $z \in \overline{D(0, R)}$. If $r < R$, then for $z \in \overline{D(0, r)}$ we have
\[
|h(z)| \leq \frac{2Mr}{R- r}
\] and
\[
|h'(z)| \leq \frac{2MR}{(R- r)^2}.
\]
Proof. Note that it suffices to show that
\[
\left| \frac{h^{(k)}(0)}{k!} \right| \leq \frac{2M}{R^k}
\] for $k \geq 1$, for then we have
\[
|h(z)| \leq \sum_{k = 1}^{\infty} \left| \frac{h^{(k)}(0)}{k!} \right| r^k \leq 2M \sum_{k = 1}^{\infty} \left( \frac{r}{R} \right)^k = 2M \left( \frac{r/R}{1- r/R} \right) = \frac{2Mr}{R- r}
\] and
\[
|h'(z)| \leq \sum_{k = 1}^{\infty} \frac{|h^{(k)}(0)|kr^{k-1}}{k!} \leq \frac{2M}{R} \sum_{k = 1}^{\infty} k \left( \frac{r}{R} \right)^{k-1} = \frac{2M}{R} \frac{1}{(1- r/R)^2} = \frac{2MR}{(R -r)^2}
\] By Cauchy’s integral formula we have
\[
\int_{0}^{1} h(Re(\theta)) \, d \theta = \frac{1}{2 \pi i} \oint_{C(0, R)} \frac{h(z)}{z} \, dz = h(0) = 0. \tag{1}
\] If $k > 0$, then we have
\[
\int_{0}^{1} h(Re(\theta))e(k \theta) \, d\theta = \frac{R^{-k}}{2 \pi i} \oint_{C(0, R)} h(z) z^{k- 1} \, dz = 0
\] by Cauchy’s theorem as $h(z) z^{k-1}$ is analytic on the disc $\overline{D(0, R)}$ and
\[
\int_{0}^{1} h(R e (\theta)) e(-k \theta) \, d \theta = \frac{R^{k}}{2 \pi i} \oint_{C(0, R)} h(z) z^{-k-1} \, dz = \frac{R^k h^{k}(0)}{k!}.
\] For $k > 0$ and for any real $\phi$ we have
\[
\int_{0}^{1} h(R e(\theta)) e(k\theta + \phi) \, d \theta = e(\phi) \int_{0}^{1} h(Re(\theta)) e(k \theta) \, d \theta = 0 \tag{2}
\] and
\[
\int_{0}^{1} h(Re(\theta)) \overline{e(k \theta + \phi)} \, d\theta = e(- \phi)\int_{0}^{1} h(Re(\theta)) e(-k \theta) \, d\theta = \frac{R^k e(-\phi) h^{(k)}(0)}{k!}. \tag{3}
\] Adding (1), (2) and (3) we end up with
\[
2\int_{0}^{1} h(Re(\theta))(1 + \cos 2 \pi( k \theta + \phi)) \, d\theta = \frac{R^k e(-\phi) h^{(k)}(0)}{k!}.
\] Now taking the real parts we obtain
\[
\text{Re}\left(\frac{R^k e(- \phi) h^{(k)}(0)}{k!} \right) \leq 2M \int_{0}^{1} (1 + \cos 2\pi (k \theta + \phi)) \, d\theta = 2M.
\] Choosing $\phi$ to be such that $e(-\phi) h^{(k)}(0) = |h^{(k)}(0)|$ we see that $|h^{(k)}(0)|/k! \leq 2M R^{-k}$.