Let $f$ be analytic on the unit disk $D = \{z : |z| ≤ 1\}$ and suppose $Im(f(z)) > 0$ for $z ∈ D$ and $f(0) = i$. Prove that $|f ' (0)| ≤ 1$. Let $f$ be analytic on the unit disk $D = \{z : |z| ≤ 1\}$ and suppose $Im(f(z)) > 0$ for
$z ∈ D$ and $f(0) = i$. Prove that $|f
'
(0)| ≤ 1$. For what functions do we have equality?
I'm not sure how to go about this. Any solutions or hints are greatly appreciated.
 A: Hint: Let $g(w) = (w-i)/(w+i)$ Then $g$ maps the open upper half plane into $\mathbb D$ and $g(i)=0.$ Consider $g\circ f.$
A: The statement is false. Set 
$$
f(z)=i\frac{1-rz}{1+rz}
$$
defined on the disk $D_{1/r}:= \{ z: |z|<1/r\}$ where $r$ is chosen to be $0<r<1$ and $1/r$ close enough to $1$. Then $f$ is holomorphic on $D_{1/r}$, hence on $D$. It maps $D_{1/r}$ onto the upper half plane, $f(0)=i$ but $f'(0)=-2ri$ so that 
$$
|f'(0)|=2r>1
$$ 
if $r$ was taken to be greater than $1/2$. 
However, if $f$ is holomorphic on $\mathbb D:= \{ z: |z|<1\}$, $\text{Im} (f(z))>0$ for all $z\in \mathbb D$ and $f(0)=i$, then one can deduce that $|f'(0)|\le 2$. The proof follows from the 
Schwarz lemma: If a holomorphic function $f:\mathbb D\to \mathbb D$ satisfies $f(0)=0$, then $|f(z)|\le |z|$ on $\mathbb D$ and $|f'(0)|\le 1$. 
The proof of this theorem can be found in any textbook so I omit it. Let $h:=g\circ f$ where 
$$
g(z)=\frac{z-i}{z+i}.
$$
Then $g$ is holomorphic and it maps upper half plane onto $\mathbb D$. Hence $h$ is holomorphic, $h(0)=g(i)=0$ and $|h(z)|<1$ on $\mathbb D$. Applying Schwarz's lemma to $h$, we conclude that 
$$
|h'(z)|=\left| \frac{2i}{(f(z)+i)^2}\cdot f'(z)\right|\le 1.
$$
Consequently, $|f'(0)|\le |2i|=2$.
