holomorphic on right half plane Could any one tell me how to solve this one?
Let $f$ be holomorphic function in the right half plane, with $|f(z)|<1$ for all $z\in \{z:Re(z)>0\}$, $f(1)=0.$ Find out largest possible value of $|f(2)|.$
 A: Let $\phi(z) = \frac{1-z}{1+z}$. $\phi$ is a Möbius transformation that maps $U$, the open unit disk, into the right half plane. Let $\tilde{f} = f \circ \phi$. Then $\tilde{f}$ maps $U$ into itself, and $\tilde{f}(0) = f(1) = 0$. Hence the Schwartz lemma tells us that $|\tilde{f}(z)| \le |z|$ for $z \in U$. We have $\phi(-\frac{1}{3}) = 2$, hence this tells us that $|f(2)| = |\tilde{f}(-\frac{1}{3})| \le \frac{1}{3}$.
If we let $f(z) = \phi(z)$, we get $f(2) = -\frac{1}{3}$, hence the maximum is attained.
A: Conformally map it into the unit disk. You get a map $\mathbb D\rightarrow \mathbb D$, so the Schwarz lemma applies.
You can also send the image of $1$ in the unit disk to $0$ using a Blaschke factor (if necessary). If, after all of these maps, $2$ is sent to the point $w$, the Schwarz lemma tells you that $|f(w)|\le |w|$. Now just unwrap all of the maps to get the answer. 
A: Consider the map $F:\mathbb{D}\rightarrow\mathbb{H},$ defined by $F(z)=\frac{1+z}{1-z}$, where $\mathbb{H}$ denotes the right half plane. Note that $F(0)=1$. Now compose $F$ with that of $f$. The map $f\circ F$ is a map from disc to itself which fixes the origin. Use Schwarz Lemma to show that the maximum value is $\frac{1}{3}.$
