Existance of an analytic function on unit disc Is there an analytic function $f:B_1(0)\to B_1(0)$ such that $f(0)=1/2$ and $f^{\prime}(0)=3/4$? If it exists, is it unique?
The answer to the first part of the question is affirmative. We can use Scharz Pick lemma to find a Mobius transformation $f$ satisfying the condition. But is the function unique? I got stuck here. Please help. 
 A: $f(z)=\frac {2z+1} {z+2}$ is one function with these properties. If $g$ is another such function define $h(z)=\phi (g(z))$ where $\phi (z)=\frac {z-\frac 1 2} {1-\frac 1 2 z}$. You can verify that $h$ maps $B(0,1)$ into itself and vanishes at $0$. A simple calculation shows $h'(0)=1$. By Schwarz Lemma we get $h(z)=z$ for all $z$ from which we get $g(z)=f(z)$. 
A: By Schwarz-Pick Lemma, for all $|z|<1$,
$$|f'(z)|\leq \frac{1-|f(z)|^2}{1-|z|^2}.$$
Note that if equality holds at some point then
$f$ must be an analytic automorphism of the unit disc.
Since for $z=0$ the above equality holds then
 $$f(z)=e^{i\theta}\frac{z-a}{1-\bar{a}z}$$
with $|a|<1$ and $\theta\in\mathbb{R}$.
Now 
$$
\begin{cases}f(0)=-e^{i\theta}a=1/2\\
f'(0)=e^{i\theta}(1-|a|^2)=3/4
\end{cases}
\Leftrightarrow
\begin{cases}a=-1/2\\
e^{i\theta}=1
\end{cases}
$$
and we may conclude that $f$  exists and it is unique:
$$f(z)=\frac{2z+1}{z+2}.$$
A: The Mobius tfm $z\mapsto \frac{1+2z}{2+z}$ does the job. If you want $f$ to be a bijection, then the answer is yes because the only transforms that map the disk into itself bijectively are $z\mapsto \frac{z-a}{1-\overline{a}z}e^{i\theta}$, and $a,\theta$ are determined by the conditions. There is something special about the choice $f'(0)=\frac{3}{4}$, because otherwise you could choose $z\mapsto \frac{z-a}{1-\overline{a}z}\rho$ with $|\rho|<1$.
