Csdrhrt

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.

$f(z)=frac {2z+1} {z+2}$ 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 $g(z)=f(z)$.

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
$$f(z)=e^{itheta}frac{z-a}{1-bar{a}z}$$
with $|a|<1$ and $thetainmathbb{R}$.
Now
$$
begin{cases}f(0)=-e^{itheta}a=1/2\
f'(0)=e^{itheta}(1-|a|^2)=3/4
end{cases}
Leftrightarrow
begin{cases}a=-1/2\
e^{itheta}=1
end{cases}
$$
and we may conclude that $f$ exists and it is unique:
$$f(z)=frac{2z+1}{z+2}.$$

The Mobius tfm $zmapsto frac{1+2z}{2+z}$ 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 $zmapsto frac{z-a}{1-overline{a}z}e^{itheta}$, 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 $zmapsto frac{z-a}{1-overline{a}z}rho$ with $|rho|<1$.