Bound on imaginary parts of family of analytic functions Question: Let $ F$ be the set of holomorphic maps $f$ from the unit disc into the upper half plane, such that $f(0)=i$. Show that the supremum of the imaginary parts, $\sup_{f\in F}$ Im[$f(\frac i 2)$] is bounded, and find the supremum.
 A: You're almost there I think. 
Like you mention, you can reduce the problem to holomorphic maps $h : \Delta \to \Delta$, using a Möbius transformation $m$ mapping the upper half-plane to the unit disk. Chose your favorite such transformation and call it $m$, and call $z_0 = m(i/2)$.
By Schwartz's lemma, $|h(z_0)| < |z_0|$; what can you deduce about 
$|f(i/2)|$ (using $m$) ?
A: Let $g(z)=\frac{z-i}{z+i}$ from $H$ to $D$ is isomorphism. Hence, $h=g\circ f$ is analytic from $D$ to $D$ and $h(0)=0$. By Schwartz lemma $|h(z)|\le|z|$ for all $z\in D$. $|h(i/2)|\le1/2$ and $|\frac{f(i/2)-i}{f(i/2)+i}|\le1/2$. Let $f(i/2)=a+bi$. Then, $a^2+(b-1)^2\le\frac{1}{4}(a^2+(b+1)^2)$. $3b^2-10b+3+3a^2\le0$. We can view this as a quadratic equation of $b$. Hence, $b\le\frac{5+\sqrt{25-9(1+a^2)}}{3}$. Thus, $b$ is bounded by $3$. Maximum is achieved for $a=0$. So we want $f$ from $D$ to $H$ satisfying $f(0)=i$ and $f(i/2)=3i$. We at least need $|h(i/2)|=|i/2|$. By Schwartz Lemma this means that $h(z)=\alpha z$ for some $|\alpha|=1$. Indeed, $\frac{f(z)-i}{f(z)+i}=\alpha z$, and then $f(z)=\frac{1+\alpha z}{1-\alpha z}i$. $f(i/2)=3i$ implies that $\alpha=-i$. So, $f(z)=\frac{1-iz}{1+iz}i$. We can check that this $f$ satisfying all requirements, and thus maximum $3$ can be achieved. So, $3$ is the supremum.
