Find $\sup\limits_{f \in F}|f(2i)|$ May you help me with this question:
Let $F$ a familiy of analytic functions at $H=\{z:\mathrm{Im}z>0\}$ that satisfies $f(i)=0$ and $|f(z)|\leq 1$. 
I want to find $\sup\limits_{f \in F}|f(2i)|$.
I think that the solution is based on Möbius transformation. 
Thank you.
 A: Hint: Map the upper half plane to the unit disk.  Where does the point $i$ go?  What theorems do we have to deal with this?  (Schwarz's lemma)  Try to remember the natural isomorphism between these two, and that some theorems are more easily stated for the unit disk then the upper half plane.
Hint 2:  Let $\mathbb{H}$  refer to the upper half plane, and $\mathbb{D}$  the open unit disk. Consider the function $$g:\mathbb{H}\rightarrow\mathbb{D} \text{ defined by }g(z)=\frac{z-i}{z+i}.$$ Then this is a conformal isomorphism between $\mathbb{H}$ and $\mathbb{D}$ and $$g^{-1}(z)=i\frac{1+z}{1-z}.$$ If $f\in F$ , then $$h:=g\circ f\circ g^{-1}:\mathbb{D}\rightarrow\mathbb{D}$$  will be analytic, and $h(0)=0.$ Applying Schwarz's lemma to $h$, we find that $|h(z)|\leq z$.  Now, calculate where the point $2i$ is mapped to on the unit disk, and see what you can conclude.
Hint 3:  We can calculate that $z=\frac{1}{3}$ gives $g(z)=2i$.  Hence $$|g\left(f\left(2i\right)\right)|\leq \frac{1}{3}$$ or in other words $$\biggr|\frac{f(2i)-i}{f(2i)+i}\biggr| \leq \frac{1}{3}.$$  Playing around with this, we will find a disk in the upper half plane in which $f(2i)$ must lie in.  Then show that the point with the highest norm inside this disk does correspond to some function, and conclude that the point is the supremum of the norms of $f(2i)$ over all $f$.
