Why is there no holomorphic function $f:D \rightarrow D$ such that $f(1)=1, f(2)=4$? Why is there no holomorphic function $f:D \rightarrow D$ such that $f(1)=1, f(2)=4$? 
Here, $D = \{ z\in \mathbb{C}:Re(z) > 0\}$.
Would you give me any hint? Thanks in advance.
 A: Consider the diagram, which transforms the given function $f:D\to D$ into a function $g:U\to U$, where $U$ is the unit disk centered in the origin:
$\require{AMScd}$
\begin{CD}
    D @>f>> D\\
    @V \mu V V @VV \mu V\\
    U @>>g> U
\end{CD} 
with 
$$\mu(z)=\frac{z-1}{z+1}\ .$$
The Möbius transformation $\mu$ above is associated to the matrix $\begin{bmatrix}1 & -1\\1&1\end{bmatrix}$, it has as inverse the Möbius transformation associated to the matrix $\begin{bmatrix}1 & 1\\-1 & 1\end{bmatrix}$. It is relatively simple to show that $\mu$ is biholomorphic.
We have $\mu(1)=0$, $\mu(2)=1/3$, $\mu(4)=3/5$. This means that the holomorphic function $g$ satisfies $g(0)=0$, and $g(1/3)=3/5$.
This contradicts $|g(z)|\le |z|$, Schwarz Lemma on $U$.
A: If $D$ were the unit disc this would amount to a function $f:\>D\to D$ with $f(0)=0$ and $f\bigl({1\over4}\bigr)={1\over2}$ (or similar data).
A: Via the map $z \mapsto \frac{z-1}{z+1}$, such an $f$ would give you a map $g$ from the unit disc to itself with $g(0) = 0$ and $g(1/3) = 3/5$. Now use the Schwarz lemma.
A: Filling in some details: the map $z\mapsto \frac{1}{z+1}$ sends the right halfplane $\text{Re}(z)>0$ into the open disk having the segment $(0,1)$ as a diameter, hence the map $z\mapsto 2\left(\frac{1}{z+1}-\frac{1}{2}\right)=\frac{z-1}{z+1}$ sends the right halfplane into $\|z\|<1$. If $f$ is a holomorphic function on the right halfplane whose range is contained in the right halfplane, then 
$$ g(w) = \frac{f\left(\frac{1+w}{1-w}\right)-1}{f\left(\frac{1+w}{1-w}\right)+1}$$
is a holomorphic function on $\|z\|<1$ whose range is contained in $\|z\|<1$. If $f(1)=1$ and $f(2)=4$ we have $g(0)=0$ and $g\left(\frac{1}{3}\right)=\frac{3}{5}$. In particular $h(z)=\frac{g(z)}{z}$ is a holomorphic function on the open unit disk such that $h\left(\frac{1}{3}\right)=\frac{9}{5}$. By the maximum modulus principle it follows that for any $\varepsilon\in\left(0,\frac{2}{3}\right)$ we have
$$ \max_{\|z\|=1-\varepsilon}\|h(z)\|\geq \tfrac{9}{5} $$
hence for any $\rho\in\left(\frac{1}{3},1\right)$ we have some $z$ with $\|z\|=\rho$ such that $\|g(z)\|\geq \frac{9}{5}\|z\|$.
This contradicts the fact that $g$ maps the open unit disk into itself.
In particular there is no holomorphic function from the right halfplane into itself such that $f(1)=1$ and $f(2)=4$.
