Bounds of $|f(z)|$ for specific z via Schwarz Lemma and the such I'm trying to brush up on some complex analysis for a qualifying exam. I'm quickly realizing that I have never done many applications of Schwarz. In particular I don't have much experience with those types of questions that ask to find a bound for $|f(z_0)|$ for some specified $z_0$. So my questions are
(1) Can you suggest any texts, preferably accessible online, that gives examples of such problems with good explanations?
(2) So far, in reviewing past quals, I have only come upon one of these questions, so I have only one example to illustrate the type of problem. I think I was able to solve it, but I was mostly fumbling through it, are there general ways to approach it that would be better?
The question was, suppose that $f(z)$ is a holomorphic on $|z| <1$ and satisfies $|f(z)| < 1$ and $f(1/2)$ = 0. Use Schwarz Lemma to show that $|f(4/5)| \le 1/2$.  
I know that in general $\frac{a-z}{1-\bar{a}z}$ maps $a$ to $0$ and maps the unit disk to the unit disk. So $g(z) = f(\frac{1/2-z}{1-\bar{1/2}z})$ has $g(0)=0$. And for all $z$ s.t. $|z|<1$, since that above function maps the unit disk to the unit disk $\frac{1/2-z}{1-\bar{1/2}z} \in$ the unit disk. Thus $|g(z)| = |f(\frac{1/2-z}{1-\bar{1/2}z})| \le 1$. So Schwarz lemma applies to $g$ and we can conclude $|g(z)| \le |z|$. But $z = 1/2 \rightarrow \frac{1/2-z}{1-\bar{1/2}z}=4/5$ so the desired bound follows. 
Can you suggest better approaches? Thanks!
 A: Instead of introducing auxiliary Möbius transformations every time, one could prove the general Schwarz-Pick lemma 
$$\left|\frac{f(a)-f(b)}{1-f(a)\overline{f(b)}}\right|\le \left|\frac{a-b}{1-a\overline{b}}\right|,\quad a,b\in\mathbb D \tag1$$
valid for any holomorphic map $f:\mathbb D\to\mathbb D$.  The proof is, of course, an application of Schwarz lemma to the composition of $f$ with Möbius transformations.
You can think of the quantity $d(a,b)=\left|\frac{a-b}{1-a\overline{b}}\right|$ as a kind of distance adapted to the geometry of the unit disk.  Then (1) simply says that this distance does not increase under $f$. In your case, 
$$d(f(4/5),f(1/2))\le d(4/5,1/2)=1/2$$
and since $f(1/2)=0$, the left side simplifies to $d(f(4/5),0)=|f(4/5)|$. 

If, instead of  $f(1/2)=0$,  you were told that $f(1/2)=1/3$, extra work would be needed to decipher the inequality 
$$\left|\frac{f(4/5)-1/3}{1-f(4/5)/3}\right|\le \frac12$$
Geometric interpretation helps here: the inequality describes a non-Euclidean disk with non-Euclidean center $1/3$. Its farthest point from the origin  $0$ will be found on the radius through $1/3$. This reduces the task to solving the equation
$$\frac{x-1/3}{1-x/3}=\frac12$$
for real $x$. The solution is $5/7$. Thus, $|f(4/5)|\le 5/7$ in this version of the problem.

The quantity $\rho(a,b)=\tanh^{-1}d(a,b)$ is known as the Poincaré metric on the unit disk, and has a natural generalization to other domains in $\mathbb C$. The   Schwarz-Pick lemma generalizes accordingly: a holomorphic map $f:\Omega_1\to\Omega_2$ is a contraction with respect to the hyperbolic metrics of $\Omega_1$ and $\Omega_2$.
The hyperbolic metric and geometric function theory by Beardon and  Minda is an introductory article which explains all this and more; probably more than you need to know for a qualifying exam. 
