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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 $|z|<1$,
$$|f'(z)|leq frac1-1-.$$
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
$$f(z)=e^ithetafracz-a1-baraz$$
with $|a|<1$ and $thetainmathbbR$.
Now
$$
begincasesf(0)=-e^ithetaa=1/2\
f'(0)=e^itheta(1-|a|^2)=3/4
endcases
Leftrightarrow
begincasesa=-1/2\
e^itheta=1
endcases
$$
and we may conclude that $f$ exists and it is unique:
$$f(z)=frac2z+1z+2.$$

The Mobius tfm $zmapsto frac1+2z2+z$ 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 $zmapsto fracz-a1-overlineaze^itheta$, and $a,theta$ are determined by the conditions. There is something special about the choice $f'(0)=frac34$, because otherwise you could choose $zmapsto fracz-a1-overlineazrho$ with $|rho|<1$.