Prove $|f'(z)|\leq\frac{Re(f(z))}{Re(z)}$ for all $z$ in the right half plane I am reviewing complex analysis for the final exam, but I am stuck in this question without any ideas...
Let $D=\{z\in\mathbb{C}|Re(z)>0\}$ and $f:D\rightarrow D$ a holomorphic function. 
Prove that $|f'(z)|\leq\frac{Re(f(z))}{Re(z)}$ for all $z\in D$
I have tried many things. 
Firstly, I tried to use Cauchy's inequality. We could find a disc centered at $z_{r}$ with radius $r$ such that $|Re(z_{r})-0|<r$. However, it is hard for me to even find $sup_{z\in D_{r}(z_{r})}|f(z)|$. Then, it is even hard to find the real part of $f(z)$ under this formula.
Secondly, I tried to use conformal map $F(T_{1}(f(z)))$ where $T_{1}(z)=iz$ and $F(z)=\frac{i-z}{i+z}$. Clearly $T_{1}(z)$ is the rotation conformally mapping the right half plane to the upper half plane, and $F(z)$ is the standard conformal map mapping the upper half plane to the unit disc. Then, I tried to apply Cauchy's inequality to the composition map, but I got stuck.
I have no idea about how to solve this question.
I am really really grateful for any hints and explanations!!
 A: You are on the right track. Just need to turn to Schwarz lemma for help.
For the sake of convention, I will use $\mathbb{H}$ to denote your domain $D$, i.e., the right half plane, and $\mathbb{D}$ to denote the unit disk.
Define
$$
T_{z_0}:\mathbb{H}\to\mathbb{D},\quad z\mapsto\frac{z-z_0}{z+\overline{z_0}},
$$
which maps $z_0\in\mathbb{H}$ to $0\in\mathbb{D}$. Obviously, its inverse reads
$$
T_{z_0}^{-1}:\mathbb{D}\to\mathbb{H},\quad z\mapsto\frac{z_0+\overline{z_0}z}{1-z},
$$
which maps $0\in\mathbb{D}$ to $z_0\in\mathbb{H}$.
Consider the composition (where $z_0\in\mathbb{H}$ is a fixed parameter)
$$
g=T_{f(z_0)}\circ f\circ T_{z_0}^{-1}:\mathbb{D}\to\mathbb{D},
$$
which maps $0\in\mathbb{D}$ to $0\in\mathbb{D}$. Thanks to this fact, together with that $g$ is obviously holomorphic, Schwarz lemma applies. Therefore,
$$
\left|g'(z)\right|\le 1
$$
holds for all $z\in\mathbb{D}$. Specifically, since $0\in\mathbb{D}$, we have
$$
\left|g'(0)\right|\le 1.
$$
This inequality suffices to complete this proof.
The rest of this proof involves tedious calculation, caused by the complexity of
$$
g(z)=T_{f(z_0)}\circ f\circ T_{z_0}^{-1}(z)=\frac{f\Bigl(\frac{z_0+\overline{z_0}z}{1-z}\Bigr)-f(z_0)}{f\Bigl(\frac{z_0+\overline{z_0}z}{1-z}\Bigr)+\overline{f(z_0)}}.
$$
Nevertheless, as you carry out each step carefully, you will eventually obtain that
$$
\left|g'(0)\right|\le 1\iff\left|f'(z_0)\right|\le\frac{\Re\left(f(z_0)\right)}{\Re\left(z_0\right)}.
$$
Finally, thanks to the arbitrariness of $z_0\in\mathbb{H}$, it is an immediate result that
$$
\left|f'(z)\right|\le\frac{\Re\left(f(z)\right)}{\Re\left(z\right)}
$$
holds for all $z\in\mathbb{H}$.
Hope this could be helpful for you.
