The existence of a holomorphic map from disk to disk with given value and derivative at a point I was trying to solve the follwing question from past papers of a competitive exam.
Let $\mathbb D=\{z\in \mathbb C: |z|<1\}$ Which of the following are correct?


*

*$\exists$ a holomorphic function $f:\mathbb D \to \mathbb D$ with $f(0)=0,f'(0)=2$

*$\exists$ a holomorphic function $f:\mathbb D \to \mathbb D$ with $f(\frac34)=\frac34,f'(\frac23)=\frac34$

*$\exists$ a holomorphic function $f:\mathbb D \to \mathbb D$ with $f(\frac34)=-\frac34,f'(\frac34)=-\frac34$

*$\exists$ a holomorphic function $f:\mathbb D \to \mathbb D$ with $f(\frac12)=-\frac12,f'(\frac14)=1$


It may have more than one option correct.
What I found is that 1. can not be true since Schwarz lemma gives $|f'(0)|\le 1$ 
But I am unable to conclude anything about the other options because I could not use Schwarz lemma in other options since $f(0)$ is not $0$ there.
I am very new to complex analysis. So may be the question is not too hard to answer but I am completely stuck.  
 A: In the cases 2 and 3 you can construct the function explicitly, for instance, by taking $f(z) = 3/4\, z + 3/16$ and $f(z) = -3/4\, z - 3/16$, respectively.
To see that these functions map the disk to itself, we note that 
for $|z|<1$, $|f(z)|\leq \frac{3}{4}|z| + \frac{3}{16} \leq \frac{15}{16}<1$.
The fourth case seems to be more difficult. Below is a brute force way to show that the statement is false, though there may be some elementary argument showing the same.
We will use Schwarz-Pick theorem 

If $f:\mathbb{D}\to \mathbb{D}$ is holomorphic, then for all $z_1, z_2 \in \mathbb{D}$, 
  $$ \left|\frac{f(z_1) - f(z_2)}{1-\overline{f(z_1)} \, f(z_2)} \right|\leq \left|\frac{z_1-z_2}{1-\bar{z}_1  z_2 } \right|
$$
  and for all $z\in \mathbb{D}$,
  $$ \frac{|f'(z)|}{1-|f(z)|^2}\leq \frac{1}{1-|z|^2}. $$
  If equality holds throughout one of these inequalities, then $f$ is a Möbius automorphism of $\mathbb{D}.$

Let us first show that $f(1/4)=-1/4.$
By applying the 2nd Schwarz-Pick inequality to $f'(1/4)=1$ we first find that $|f(1/4)|\leq 1/4$.
Then we apply the 1st Schwarz-Pick inequality to $f(1/2)=-1/2$ and find that
$$
\left|\frac{1+2 \,f(1/4)}{2 + f(1/4)}\right| \leq \frac{2}{7}.
$$
Now, if we look at the function
$$
g(w) = \frac{1+2 w}{2 + w},
$$
we note that $|g(w)|\geq 2/7$ in the region $|w|\leq 1/4$, and equality holds only for one point, $w=-1/4$ (you can for instance use the fact that an analytic function attains its minimum on the boundary of a domain, then substitute $w = e^{i\theta}/4$ and apply calculus tools). Therefore, $f(1/4)=-1/4$.
The problem of finding a holomorphic function $f:\mathbb{D} \to \mathbb{D}$ with $f(1/2) = -1/2$  and $f(1/4) = -1/4$ is a simple case of Nevanlinna-Pick interpolation problem. 
 The determinant of the Pick matrix is zero, hence the solution must be unique, i.e. $f(z) = -z$. However, in that case $f'(1/4) \neq 1$, therefore the fourth statement is false.
