Prove $|f'(z)|\leq{\frac{1}{2 \operatorname{Im}z}}$ Could you help me with proving:
Let $f$ be an analytic function defined in on upper half plane(UHP). Suppose that $|f(z)|<1$ for all $z$ in UHP. Prove that for every $z$ in UHP
$$
|f'(z)|\leq{\frac{1}{2\operatorname{Im} z}} .
$$
I guess that I need to use Cauchy's estimate but I am not sure how to get $2 \operatorname{Im} z$ bound there.
Thank you in advance.
 A: Let $T(z) = \frac{z-i}{z+i}$. $T$ is a biholomorphic map from the upper half-plane to the unit disk $\mathbb{D}$. Define $g \colon \mathbb{D} \to \mathbb{D}$ as $g = f\circ T^{-1}$. By the Schwarz-Pick lemma, we have
$$\frac{\lvert g'(w)\rvert}{1 - \lvert g(w)\rvert^2} \leqslant \frac{1}{1-\lvert w\rvert^2}$$
for all $w\in\mathbb{D}$, which we can weaken to $$\lvert g'(w)\rvert \leqslant \frac{1}{1-\lvert w\rvert^2}.\tag{$\ast$}$$
Now by the chain rule we have
$$f'(z) = (g\circ T)'(z) = g'(T(z))\cdot T'(z).$$
Further, $T'(z) = \frac{2i}{(z+i)^2}$, and hence by the weakening $(\ast)$ of the Schwarz-Pick bound, we have
\begin{align}
\lvert f'(z)\rvert &\leqslant \frac{1}{1 - \lvert T(z)\rvert^2}\cdot \frac{2}{\lvert z+i\rvert^2}\\
&= \frac{2}{\lvert z+i\rvert^2 - \lvert z-i\rvert^2}\\
&= \frac{2}{(\lvert z\rvert^2 - iz + i \overline{z} + 1) - (\lvert z\rvert^2 +iz - i\overline{z} + 1)}\\
&= \frac{2}{-2iz + 2i\overline{z}}\\
&= \frac{1}{-i(z-\overline{z})}\\
&= \frac{1}{-i(2i\operatorname{Im} z)}\\
&= \frac{1}{2\operatorname{Im} z}.
\end{align}
A: Partial answer: By Cauchy estimate
$$
|f^\prime(z)| \le \frac{1}{R}\max_{t \in \partial D_R(z)}|f(t)|<\frac{1}{R}.
$$
holds for all $R>0$ for which $f$ is holomorphic in a neighborhood of the closed disk $\overline{D_R(z)}$. It implies that
$$
|f^\prime(z)| \le \frac{1}{|z|}.
$$
Now, if $|z|\ge 2\mathrm{Im}z$, i.e. $|\mathrm{Re}z| \ge \sqrt{3}\mathrm{Im}z$, we are done. 
