Existance of an analytic function on unit disc

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $|\rho|<1$.