Typo in complex Schwarz Lemma analysis problem Texas A&M U has on their August 2012 qualifying exam this problem: 
let $f$ be analytic on $\mathbb{D}$ with $|f(z)|\leq 1$. Then we have 
$$\frac{|f(0)|-|z|}{1-|f(0)||z|}\leq |f(z)| \leq \frac{|f(0)|+|z|}{1+|f(0)||z|}$$
However, I am only getting this result:
$$\frac{|f(0)|-|z|}{1+|f(0)||z|}\leq |f(z)| \leq \frac{|f(0)|+|z|}{1-|f(0)||z|}$$
How are these related? Is there a typo in Texas' problem? I find that unlikely.
 A: Suppose that $|f(z)|<1$ for $z$ with $|z|<1$ and consider $$
g(z)=\frac{f(z)-c}{1-\bar{c}f(z)},$$ where $c=f(0).$
Then $g(z)$ is analytic on $\mathbb{D}$ and  satisfies $|g(z)|<1, g(0)=0.$ By Schwarz lemma
 we get $|g(z)|\le |z|$, that is, $$
\left|\frac{f(z)-c}{1-\bar{c}f(z)}\right|\le |z|.
$$
Write $w=f(z), r=|z|$ for simplicity. Then we have
\begin{align}
\left|\frac{w-c}{1-\bar{c}w}\right|&\le r,\\
1-\left|\frac{w-c}{1-\bar{c}w}\right|^2&\ge 1-r^2,\\
|1-\bar{c}w|^2-|w-c|^2&\ge (1-r^2)|1-\bar{c}w|^2.
\end{align}
Since
\begin{align}
LHS&=1-|c|^2-|w|^2-|c|^2|w|^2,\\
RHS&\ge (1-r^2)(1-|c||w|)^2\\
&=1-2|c||w|+|c|^2|w|^2-r^2(1-|c||w|)^2,
\end{align}
we have
\begin{align}
0&\ge |w|^2-2|w||c|+|c|^2- r^2(1-|c||w|)^2\\
&=(|w|-|c|)^2-r^2(1-|c||w|)^2\\
&=\left(|w|-|c|+r(1-|c||w|)\right)\left(|w|-|c|-r(1-|c||w|)\right)\\
&=\left((1-|c|r)|w|-(|c|-r)\right)\left((1+|c|r)|w|-(|c|+r)\right)
\end{align}
Therefore we have
$$
\frac{|c|-r}{1-|c|r}\le |w|\le \frac{|c|+r}{1+|c|r}.$$
The proof is complete.
