Finding an analytic function satisfying given two conditions. 
Does there exists an analytic function $f:D\to D$ such that $f(1/2)=1/2$ and $f'(1/2)=-1$ ? If exists then find such a function. where , $D=\{z\in \mathbb C:|z|<1\}.$ 

I found that such a function exists, as $$|f'(z)|\le \frac{1-|f(z)|^2}{1-|z|^2}$$ holds good. But I am unable to find such a function..Please help to construct such a function.
Are there any particular rule to find such a function or it can be done only by trial ?
 A: The Möbius Transformation: $\displaystyle \mu_a(z) = \frac{z - a}{1-\bar{a}z}$, for $|a| < 1$, it gives an injective mapping of $D = \{z: |z| < 1\}$ onto itself with inverse $\mu_{-a}(z)$,
satisfies $\mu_a'(0) = 1-|a|^2$ and $\displaystyle \mu_a'(a) = \frac{1}{1-|a|^2}$
As @Blazej commented, check that $\mu_a$ maps $\partial D$ onto itself.
Given an analytic function $f:D \to D$, with $\displaystyle f\left(\frac{1}{2}\right) =\frac{1}{2}$ and $\displaystyle f'\left(\frac{1}{2}\right) = -1$,
Define, $g(z) := \mu_{1/2}\circ f \circ \mu_{-1/2} (z)= \mu_{1/2}(f(\mu_{-1/2}(z)))$
Then $g$ maps $D$ into $D$ and satisfies $g(0) = 0$, and as a consequence of the Schwarz Lemma we infer that: $$|g'(0)| \le 1$$
Now, applying the chain rule see that:  
$\displaystyle \begin{align} g'(0) &= (\mu_{1/2}\circ f)'(\mu_{-1/2}(0)).\mu'_{-1/2}(0) \\&= (\mu_{1/2}\circ f)'\left(\frac{1}{2}\right).\mu'_{-1/2}(0)\\&= \mu'_{1/2}\left(\frac{1}{2}\right)f'\left(\frac{1}{2}\right)\left(1 - \frac{1}{4}\right) = f'\left(\frac{1}{2}\right) = -1\end{align}$
This corresponds to the equality case in the Schwarz Lemma, hence $g$ must be of the form $$g(z) = -z$$ in $D$.
Hence, $$f(z) = \mu_{-1/2}\left(-\mu_{1/2}(z)\right).$$
